---
author:
- 'U. Fidalgo and J. Olson'
title: Convergent non complete interpolatory quadrature rules
---

Introduction {#sec:1}
============

Let $\mathcal{C}([-1,\,1])$ be the set of all continuous functions defined on $[-1,\,1]$. Given an $n$-tuple of nodes ${\bf x}_n=(x_{1,n},\ldots,x_{j,n})$ satisfying $-1<x_{1,n}<x_{2,n}<\cdots<x_{n,n}<1$, we consider [*integration rules*]{} $$\label{rule}
I_n[f]= \sum_{j=1}^n w_{j,n} f(x_{j,n}), \quad f \in \mathcal{C}[-1,\,1]$$ associated to the integrals $$\label{integrals}
\displaystyle I(f)=\int_{-1}^1 f(x)\, \mbox{d} \lambda_0(x), \quad \mbox{where} \quad  \frac{\mbox{d} \lambda_0(x)}{\mbox{d} x}=\frac{1}{\pi \sqrt{1-x^2}}.$$ The numbers $w_{j,n},$ $j=1,\ldots,n$ are called weights.

An integration rule $I_n[\cdot]$ is said to be interpolatory if there exists a number $m \in \left\{0,1,\ldots, 2n-1\right\}$, such that the following equality holds for every polynomial $p$ with degree $\leq m$ (we denote $p \in \Pi_m$): $$\label{interpolatoryq}
I_n[p]=\int_{-1}^1 p(x)\, \mbox{d} \lambda_0(x).$$ When the equality (\[interpolatoryq\]) holds for certain $m$, and is not extendable for all polynomials with degree $m+1$, we say that $I_n[\cdot]$ is an interpolatory quadrature rule with $m$-degree of exactness. When $m=2n-1$ $I_n$ is the Gaussian quadrature rule.

Consider a sequence of interpolatory quadratures $\displaystyle \left\{I_n\right\}_{n \in \Lambda}$ constructed with the following schemes of nodes and weights $$\label{nodessequence}
  {\bf x}=\left\{{\bf x}_n=(x_{1,n},\ldots,x_{n,n})\right\}_{n\in \Lambda} \quad \mbox{and} \quad {\bf w}=\left\{{\bf w}_n=(w_{1,n},\ldots,w_{n,n})\right\}_{n\in \Lambda},$$ respectively.

We say that $\displaystyle \left\{I_n\right\}_{n \in \Lambda}$ is convergent if $$\label{conv}
\lim_{n\in\Lambda} I_n[f]=\int_{-1}^1 f(x)\, \mbox{d} \lambda_0(x),\quad \mbox{for all} \quad f\in C[-1,1].$$ According to a classical result of Pólya [@DR page 130], when $\displaystyle  m(n) \to \infty$ as $n\to \infty $, the equality holds true if and only if $ \displaystyle \sup_{n\in \Lambda} \sum_{j=1}^n \left|w_{j,n}\right| < \infty.$ This condition is satisfied if the weights $w_{j,n}$ are all positive. From (\[interpolatoryq\]) we observe that $$\label{polyacond}
\sum_{j=1}^n w_{j,n}=\int_{-1}^1\, \frac{\mbox{d} x}{\pi \,\sqrt{1-x^2}}=1 < \infty.$$ In the Gaussian quadrature rule (maximum degree of exactness $m(n)=2n-1$) the weights $w_{j,n}$, $j=1,\ldots, n$ are all positive and the convergence of the rule is guarantied. However the nodes are all fixed. For each $n \in \mathbb{N}$ the points of evaluation $x_{j,n}$, $j=1,\ldots,n$ must be the roots of the $n$th orthogonal polynomial with respect to $\lambda_0$ (see for instance [@Sze]). This is the Chebyshev polynomial with degree $n$. This means that if we do not have the value of $f$ at each point $x_{j,n}$ the calculus gets stuck. It is convenient to have more flexibility in the distribution of the evaluation nodes. We study convergent interpolatory integration rules with orders of exactness $m <2n-1$.

The authors of [@BLS2] analyze a wide class of interpolatory quadrature rules with $m(n)$ degrees of exactness behaving as follows $$\label{limita}
\displaystyle \lim_{n \to \infty} \frac{m(n)}{2 n}=a \in [0,\, 1].$$ They characterized all possible weak\*-limit points of the sequence of counting measures associated with distribution of nodes corresponding to a convergent scheme.

A sequence $\left\{\nu_n\right\}_{n\in\mathbb{N}}$ of measures is said to converge weakly to the measure $\nu$ provided that there exists a compact set $K$ containing the support of $\nu$ and of each $\nu_n$, and that $$\lim_{n\in\mathbb{N}}\int f d\nu_n=\int f d\nu$$ for each continuous function $f$ on $K$. In such a case, we write $\nu_n\overset{*}{\to}\nu$. We say that $\nu$ is a weak\*-limit of the sequence $\left\{\eta_n\right\}_{n\in\mathbb{N}}$ if some subsequence of $\left\{\nu_n\right\}_{n\in \Lambda\subset \mathbb{N}}$ is weakly convergent to $\nu$.

Set two schemes of numbers as in (\[nodessequence\]) associated to an interpolatory quadrature rule $\left\{I_{n}\right\}_{n\in \Gamma}$ where the degree of exactness satisfies (\[limita\]) for certain $a\in [0,\, 1]$. We also consider its corresponding sequence $\left\{\eta_n\right\}_{n\in \mathbb{N}}$ of probability counting measures $$\label{zerocountingmeasures}
\eta_n:=\frac{1}{n} \sum_{j=1}^n \delta_{x_{j,n}}, \quad n \in \mathbb{N}.$$ According to [@BLS1], if the rule $\{I_n\}_{n \in \Lambda}$ is convergent then every weak\*-limit $\nu$ of the the sequence $\left\{\eta_n\right\}_{n\in\mathbb{N}}$ satisfies that $$\label{sigmainequality}
 \nu \geq  a\, \lambda_0.$$ Also from [@BLS1] we have that this necessary condition is not sufficient. Theorem \[main\] states conditions of convergence on the distribution of nodes.

Let us introduce some previous notation. Set $\mathcal{K}_1$ and $\mathcal{K}_2$ two compact subsets of the complex plane $\mathbb{C}$. Let $\displaystyle \mbox{dist}(\mathcal{K}_1, \mathcal{K}_2)=\min \left\{\left|\left|x-y\right|\right\|: x \in \mathcal{K}_1\,\, \mbox{and}\,\, y\in \mathcal{K}_2\right\}$ denote the distance between $\mathcal{K}_1$ and $\mathcal{K}_2$. Consider a compact set $K \subset \mathbb{C} \setminus [-1,\,1]$, and a measure $\mu$ supported on $K$. A measure $\widetilde{\mu}$ supported on $[-1,\,1]$ is said to be the balayage of $\mu$ if they have the same total variation $||\mu||=||\widetilde{\mu}||$ and their logarithmic potentials coincide on $[-1,\,1]$. This is $$V^{\widetilde{\mu}}(x)=\int \log \frac{1}{|x-t|} \mathrm{d} \widetilde{\mu}(t)=\int \log \frac{1}{|x-\zeta|} \mbox{d} \mu (\zeta)=V^{\mu}(x), \quad x\in [-1,\,1].$$ In [@ST Section II.4] we can find a deep study about balayage of measures. We are now ready to state the main result of this paper:

\[main\] Fix a number $\kappa \in \mathbb{N}$ and a probability discrete measure $$\sigma=\frac{1}{\kappa}\sum_{k=1}^{\kappa} \delta_{\zeta_k}, \quad \zeta_k \subset \mathbb{C}\setminus [-1,\,1], \quad k=1,\ldots, \kappa.$$ Assume that $\sigma$ is symmetric with respect to $\mathbb{R}$ with $\displaystyle \mathrm{dist}\left(\left\{\zeta_1,\ldots,\zeta_{\kappa}\right\},[-1,\,1]\right)>1$. Denote $\widetilde{\sigma}$ the balayage measure associated to $\sigma$ supported on the interval $[-1,\,1]$. Given a rational number $a \in [0,\, 1]$, consider a subsequence $\Lambda \subset \mathbb{N}$ such that for each $n\in \Lambda$, $\displaystyle 2 \frac{1-a}{\kappa} n \in \mathbb{N}$. Let $\displaystyle {\bf x}=\left\{{\bf x}_n=\left(x_{1,n}, \ldots, x_{n,n}\right)\right\}_{n\in \Lambda}$ be a scheme of nodes. If for each $j=1,\ldots n,$ $n\in \Lambda$ there are two constants $A\geq 0$ and $\ell>0$ satisfying $$\label{maincond}
\left|(1-a)\pi \int^1_{x_{j,n}} \mathrm{d} \widetilde{\sigma}(t) -a \arccos x_{j,n}-\frac{2j-1}{2n}\pi\right|\leq A  e^{-\ell n},$$ then there always exist weights $\displaystyle {\bf w}=\left\{{\bf w}_n=(w_{1,n}, \ldots, w_{n,n})\right\}_{n\in \mathbb{N}}$, where $\displaystyle \left\{I_n\right\}_{n\in \Lambda}$ corresponding to ${\bf x}$ and ${\bf w}$ is convergent.

In Section \[examples\] we give some explicit schemes that satisfy the relation (\[maincond\]). The statement of Theorem \[main\] is proved in Section \[proofmaintheorem\]. In such proof we use results coming from the orthogonal polynomials theory that are analyzed in Section \[sec:2\] and Section \[asymptotic analysis\]. In Section \[sec:2\] we study algebraic properties of families of orthogonal polynomials and their connections with convergent conditions of non-complete interpolatory quadrature rules. In Section \[asymptotic analysis\] we describe the strong asymptotic behavior of an appropriated family of orthogonal polynomials with respect to a varying measure.

Some explicit convergent schemes of nodes {#examples}
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We consider three particular cases where the inequality (\[maincond\]) holds. In the three situations the measure $\sigma=\delta_{\zeta}$ corresponds to a Dirac delta supported on a point belonging to the real line $\zeta>2$. Hence the situations are when $a$ takes the values $0,$ $1/2,$ and $1$.

According to [@ST Section II.4 equation (4.46)], the balayage measure of $\sigma=\delta_{\zeta}$ on $[-1,\,1]$ has the following differential form $$\label{zetadeltabalayage}
\mathrm{d} \widetilde{\sigma}(t)=\frac{\sqrt{\zeta^2-1}}{\pi (\zeta-t)\sqrt{1-t^2}} \, \mathrm{d}\, t.$$ We study the function $$I_a(x)=(1-a)\pi \int^1_{x} \mathrm{d} \widetilde{\sigma}(t)=(1-a) \sqrt{\zeta^2-1} \int^1_{x} \frac{\mathrm{d}\, t}{(\zeta-t)\sqrt{1-t^2}}.$$ Taking the change of variables $t=\cos \theta$ and taking into account $\zeta>2$ ($\varphi(\zeta)>2$ implies that $\arg (1-\varphi(\zeta))=\pi$), we have that $$I_a(x)=(1-a)\, \left( -\arccos x +2 \arg \left(e^{i \arccos x}-\varphi(\zeta)\right)-2\pi \right).$$ In this situation the condition of convergence (\[maincond\]) in Theorem \[main\] acquires the following form $$\label{mainconddelta}
\left|\arccos x_{j,n}+2(1-a)\, \left[\pi- \arg \left(e^{i \arccos x}-\varphi(\zeta)\right) \right] +\frac{2j-1}{2n}\pi\right|\leq A  e^{-\ell n}.$$ Then a scheme $\displaystyle {\bf x}=\left\{{\bf x}_n=\left(x_{1,n}, \ldots, x_{n,n}\right)\right\}_{n\in \Lambda}$ that satisfies the following relation is convergent $$\arccos x_{j,n}+2(1-a)\, \left[\pi- \arg \left(e^{i \arccos x}-\varphi(\zeta)\right) \right]=-\frac{2j-1}{2n}\pi-A e^{-\ell n}=\kappa_{j,n},$$ with $A>0$ and $\ell>0$. This means that $$\cos \left(\arccos x_{j,n}+2(1-a)\, \left[\pi- \arg \left(e^{i \arccos x}-\varphi(\zeta)\right) \right]\right)=\cos \kappa_{j,n}.$$ Using the cosine addition formula we have that $$x_{j,n}\cos \left\{2(1-a)\, \left[\pi- \arg \left(e^{i \arccos x}-\varphi(\zeta)\right) \right]\right\}$$ $$\label{generalexample}
-\sqrt{1-x_{j,n}^2}\sin \left\{2(1-a)\, \left[\pi- \arg \left(e^{i \arccos x}-\varphi(\zeta)\right) \right]\right\}=\cos \kappa_{j,n}.$$

First we consider the situation $a=1$. In this case the expressions in (\[generalexample\]) become $$\label{aonesol}
x_{j,n}=\cos \kappa_{j,n}, \quad j=1,\ldots, n, \quad n\in \Lambda.$$ The nodes are close to the zeros of the Chebyshev polynomials. That’s why the term corresponding to the $\sigma$’s influence in (\[generalexample\]) vanishes when $a=1$.

Let us analyze now the case $a=1/2$. We consider the following identities $$\label{condcos}
\cos \left[\pi-\arg \left(e^{i \arccos x_{j,n}}-\varphi(\zeta)\right)\right]=\frac{\varphi(\zeta)-x_{j,n}}{\sqrt{\varphi^2(\zeta)-2 \varphi(\zeta) x_{j,n}+1}}$$ and $$\label{condsin}
\sin \left[\pi-\arg \left(e^{i \arccos x_{j,n}}-\varphi(\zeta)\right)\right]=\frac{\sqrt{1-x_{j,n}^2}}{\sqrt{\varphi^2(\zeta)-2 \varphi(\zeta) x_{j,n}+1}}.$$ Substituting (\[condcos\]) and (\[condsin\]) in (\[generalexample\]) we arrive at the quadratic equations: $$x_{j,n}^2-\frac{2\sin^2\kappa_{j,n}}{\varphi(\zeta)}\, x_{j,n}+ \frac{\sin^2\kappa_{j,n}}{\varphi^{2}(\zeta)} -\cos^2 \kappa_{j,n}=0, \quad j=1,\ldots, n, \quad n\in \Lambda.$$ For each $j=1,\ldots, n,$ $ n\in \Lambda$ we obtained the following solutions $$\label{ahalfsol}
x_{j,n}=\frac{1}{\varphi(\zeta)}\left[\sin^2 \kappa_{j,n}+ \cos \kappa_{j,n} \sqrt{\varphi^2(\zeta)-\sin^2 \kappa_{j,n}}\right].$$ During the process of finding these above solutions we introduce some extra solutions that we removed. Observe that when $\zeta$ tends to $\infty$ the expressions in (\[ahalfsol\]) reduce to (\[aonesol\]). This is in accordance with the fact that $\widetilde{\sigma}$ approaches $\lambda_0$ as $\zeta \to \infty$, see (\[zetadeltabalayage\]), hence we only considered the positive branch of the square root in (\[ahalfsol\]).

Finally take $a=0$. From (\[generalexample\]) we have that $$x_{j,n}\cos \left\{2\, \left[\pi- \arg \left(e^{i \arccos x}-\varphi(\zeta)\right) \right]\right\}$$ $$-\sqrt{1-x_{j,n}^2}\sin \left\{2\, \left[\pi- \arg \left(e^{i \arccos x}-\varphi(\zeta)\right) \right]\right\}=\cos \kappa_{j,n}.$$ We use the conditions (\[condcos\]) and (\[condsin\]), and obtain the following expression $$x_{j,n}=\frac{2 \zeta \cos \kappa_{j,n}}{\varphi(\zeta) + 2 \cos \kappa_{j,n}}, \quad j=1,\ldots, n, \quad n\in \Lambda.$$ Taking into account that $\displaystyle \varphi(\zeta)=\zeta+\sqrt{\zeta^2+1}$ we see that the above expression is reduced to (\[aonesol\]) when $\zeta$ goes to infinity.

Connection with orthogonal polynomials {#sec:2}
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Let $\mu$ be a positive finite Borel measure with infinitely many points in its support $\mbox{supp}(\mu)$. Set $\Delta$ denoting the least interval which contains $\mbox{supp}(\mu)$. A collection of monic polynomials $\displaystyle \left\{q_{\mu,n}\right\}_{n \in \mathbb{Z}_+},$ $\mathbb{Z}_+=\left\{0,1,\ldots\right\}$ is the family orthogonal polynomials with respect to $\mu$ if its elements satisfy the following orthogonality relations $$\label{ortoargentino}
0=\int  x^{\nu} q_{\mu,n}(x) \, \mbox{d} \mu(x), \quad \nu=0,1,\ldots, n-1, \quad n\in \mathbb{Z}_+.$$ Each $q_{\mu,n}$ has $n$ single roots lying in the interior of $\Delta$ (we denote $\displaystyle \overset{\circ}{\Delta}$) such that it vanishes at most once in each interval of $\displaystyle \Delta \setminus \mbox{supp}(\mu)$ (see [@chi Theorem 5.2] or [@freud Chapter 1]). We also know that $q_{\mu,n+1}$ and $q_{\mu,n}$ interlace their zeros. In [@We] B. Wendroff proved that given two polynomials $P_n$ and $P_{n+1}$, with $\deg P_{n+1}=\deg P_n+1=n+1$, that interlace zeros, there always exist measures $\mu$ such that $P_n=q_{\mu,n}$ and $P_{n+1}=q_{\mu,n+1}$. Now we find some of these measures.

We say then a polynomial $\displaystyle P_n(x)=\prod_{j=1}^n\left(x-x_{j}\right)$ of degree $n$ is admissible with respect to the measure $\mu$, if its roots are all simple, lying in $\displaystyle \overset{\circ}{\Delta}$, with at most one zero into each interval of $\displaystyle \Delta \setminus \mbox{supp}(\mu)$. The system of nodes $(x_{1},\ldots,x_{n})$ is also said to be admissible with respect to $\mu$.

\[interwendroff\] Let $\displaystyle P_n(x)=\prod_{j=1}^n \left(x-x_{j}\right)$ and $\displaystyle \widetilde{P}_{n}(x)=\prod_{j=1}^{n-1} \left(x-\widetilde{x}_{j}\right)$ be two admissible polynomials with respect to $\mu$ that satisfy $\displaystyle x_1<\widetilde{x}_1<x_2< \cdots < \widetilde{x}_{n-1}<x_n.$ Then there exists a positive integrable function $\rho_n$ with respect to $\mu$ ($\rho_n$ is a weight function for $\mu$) such that for the measure $\mu_n$ which differential form $\displaystyle \mbox{d} \mu_n(x)=\rho_n(x) \, \mbox{d} \mu(x),$ $x \in \mbox{supp} (\mu)$, $P_n\equiv q_{\mu_n,n}$ and $\widetilde{P}_{n}\equiv q_{\mu_n, n-1}$ are the $n$-th and $n-1$-th monic orthogonal polynomials with respect to $\mu_n$, respectively.

In the proof we follow techniques used in [@kroo].

Consider $\Phi$ a set of weight functions such that for every constant $\alpha >0$ it satisfies:

-   $\rho \in \Phi \implies \alpha \rho \in \Phi.$

-   $(\rho, \widetilde{\rho})\in \Phi^2= \Phi \times \Phi \implies \alpha \rho +(1-\alpha)\widetilde{\rho} \in \Phi,$ $\alpha \leq 1.$

-   If a polynomial $Q$ satisfies $\displaystyle \int Q(x) \rho(x) \, \mbox{d} \mu(x) >0$ for all $\rho \in \Phi,$ then $Q \geq 0$ in $\mbox{supp} (\mu)$.

Two examples of sets of weight functions satisfying the above conditions are the positive polynomials and positive simple functions in [@rud Definition 1.16]. In general, the positive linear combinations of a Chevyshev system (see [@KN Chapter II]) conform a set as $\Phi$. Examples of Chevyshev systems can be found in [@nik] (also in [@FMM]).

Given $\rho \in \Phi$ we set $${\bf v}_{\rho}=\left(\int \widetilde{P}_n(x)\rho(x) \mbox{d} \mu(x), \ldots, \int x^{n-2} \widetilde{P}_n(x)\rho(x) \mbox{d} \mu(x),\right.$$ $$\left.\int P_{n}(x)\rho(x) \mbox{d} \mu(x), \ldots,\int x^{n-1} P_{n}(x)\rho(x) \mbox{d} \mu(x)\right)\in \mathbb{R}^{2n-1}.$$ Let us focus on $\mathcal{K}=\left\{{\bf v}_{\rho}:\rho \in \Phi\right\}$. Proving Lemma \[interwendroff\] reduces to showing that $\mathcal{K}$ contains the origin. From condition $(i)$ we have that the origin belongs to $\mathcal{K}$’s closure, $\overline{\mathcal{K}}$. Since $\mathcal{K}$ is open we need to prove the origin is an interior point. We proceed by contradiction. Suppose that the origin belongs to the boundary of $\mathcal{K}$. This is ${\bf O}\in \partial \mathcal{K}= \overline{K}\setminus \mathcal{K}$. There exists a hyper-plane $\mathcal{A}$ that touches tangentially $\partial \mathcal{K}$ at ${\bf O}$. On the other hand we have that condition (ii) implies that $\mathcal{K}$ is convex, then there exists a vector $\displaystyle {\bf a}=\left(a_{0,n-1}, \ldots, a_{n-2,n-1},a_{0,n},\ldots,a_{n-1,n}\right)$ which is orthogonal with respect to $\mathcal{A}$ in the sense of the standard inner vector product (${\bf a}\cdot {\bf u}=0$, for all ${\bf u}\in \mathcal{A}$), and for each ${\bf v}_{\rho}\in \mathcal{K}$, ${\bf v}_{\rho} \cdot {\bf a}>0$. So the polynomials $$p_{n-1}(x)=a_{0,n-1}+a_{1,n-1}x+ \ldots+a_{n-2,n-1}x^{n-2}$$ and $$p_{n}(x)=a_{0,n}+a_{1,n}x+ \ldots+a_{n-1,n}x^{n-1}$$ satisfy that $$0< \int \left[p_{n-1}(x)\widetilde{P}_{n}(x)+p_{n}P_{n}(x)\right]\, \rho(x) \, \mbox{d} \mu(x), \quad \mbox{for all} \quad \rho \in \Phi.$$ According to condition (iii) the polynomial $\mathcal{P}(x)=p_{n-1}(x)\widetilde{P}_n(x)+p_{n}P_{n}(x)$, with real coefficients, must be non-negative in $\mbox{supp}(\mu)$. However we shall prove that this is impossible, arriving then to a contradiction.

Assume that $\mathcal{P}(x)=p_{n-1}(x)\widetilde{P}_n(x)+p_{n}P_{n}(x)$ does not change sign in $\mbox{supp}(\mu)$. Suppose that there is a point $t\in \mbox{supp}(\mu)$, such that $t=x_k$ $\displaystyle k\in \left\{1, \ldots, n\right\}$ satisfying that $\mathcal{P}(x_k)=0$, then taking into account that $\widetilde{P}_{n}$ and $P_{n}$ interlace zeros, we have that $p_{n-1}(x_k)=0$. Also, since $\mathcal{P}$ is non-negative on $\mbox{supp}(\mu)$, we have that $t=x_k$ is a zero of multiplicity even for $\mathcal{P}$. Consider $\displaystyle S=\left\{t_1, \ldots, t_{\ell}\right\}\subset \left\{x_1,\ldots,x_{n}\right\}$ the set of all points where $P_{n}$ and $\mathcal{P}$ vanishe at same time. Then we can write $$\label{positivepol}
\mathcal{P}=q(x)\prod_{i=1}^{\ell} \left(x-t_{i}\right)^{2 d_i}, \quad d_i\in \mathbb{N}, \quad i=1,\ldots,\ell,$$ where $q$ is a polynomial with positive values at every root of $P_{n}$. We also write $$\label{pnexpand}
p_{n-1}(x)=\widetilde{p}(x)\prod_{i=1}^{\ell} \left(x-t_{i}\right).$$ The polynomial $\widetilde{p}$ has degree $\deg p_n-\ell$. Since $\widetilde{P}_{n}$ and $P_n$ interlace zeros, we have that $$\label{quotient}
\frac{\widetilde{P}_n(x)}{P_{n}(x)}=\sum_{j=1}^{n} \frac{\lambda_{j}}{x-x_{j}}, \quad \lambda_{j}>0, \quad j=1,\ldots,n.$$ Observe that $$\frac{\displaystyle \prod_{i=1}^{\ell}\left(x-t_i\right)}{P_{n}(x)}=\frac{1}{\displaystyle  \prod_{i=1}^{\ell}\left(x-t_i\right)^{2d_i-1}q(x)}\left[p_{n-1}(x)\sum_{j=1}^{n} \frac{\lambda_{j}}{x-x_{j}}+p_{n}(x)\right].$$ This means that the above function satisfies that $$\frac{1}{\displaystyle  \prod_{i=1}^{\ell}\left(z-t_i\right)^{2d_i-1}q(z)}\left[p_{n-1}(z)\sum_{j=1}^{n} \frac{\lambda_{j}}{z-x_{j}}+p_{n}(z)\right]=\mathcal{O}\left(\frac{1}{z^{n-\ell}}\right) \quad \mbox{as} \quad z \to \infty,$$ which is a holomorphic functions on $\displaystyle \overline{\mathbb{C}}\setminus \left(\left\{x_1,\ldots,x_{n}\right\}\setminus S\right).$ For each $\nu=0,\ldots, n-\ell-2$ we have then $$\frac{z^{\nu}}{\displaystyle  \prod_{i=1}^{\ell}\left(z-t_i\right)^{2d_i-1}q(z)}\left[p_{n-1}(z)\sum_{j=1}^{n} \frac{\lambda_{j}}{z-x_{j}}+p_{n}(z)\right]=\mathcal{O}\left(\frac{1}{z^{2}}\right) \quad \mbox{as} \quad z\to \infty,$$ also holomorphic functions on $\displaystyle \overline{\mathbb{C}}\setminus \left(\left\{x_1,\ldots,x_{n}\right\}\setminus S\right).$ Set the elements $\displaystyle y_j \in \left\{x_1,\ldots,x_{n}\right\}\setminus S$, $j=1, \ldots, n-\ell$ with $y_1 < y_2 < \cdots < y_{n-\ell}$, and $\widetilde{\lambda}_j$, $j=1, \ldots, n-\ell$ the coefficients $\lambda$’s defined in (\[quotient\]) corresponding to points $y_j$. Also let $\lambda_j^{\prime}$ denote the $\lambda$’s of $t_j$, $j=1, \ldots, \ell$. Call $F$ the set of the roots of the polynomial $q$ defined in (\[positivepol\]). Consider a closed integration path $\Gamma$ with winding number $1$ for all its interior points. Denote $\mbox{Ext} (\Gamma)$ and $\mbox{Int}(\Gamma)$ the unbounded and bounded connected components respectively of the complement of $\Gamma$. Take $\Gamma$ so that $I\subset \mbox{Int}(\Gamma)$ and $F\subset \mbox{Ext}(\Gamma)$. From Cauchy’s Theorem and the above two conditions, it follows that $$0=\frac{1}{2 \pi i} \int_{\Gamma} \frac{z^{\nu}}{\displaystyle  \prod_{i=1}^{\ell}\left(z-t_i\right)^{2d_i-1}q(z)}\left[p_{n-1}(z)\sum_{j=1}^{n} \frac{\lambda_{j}}{z-x_{j}}+p_{n}(z)\right] \, \mbox{d} z$$ $$=\frac{1}{2 \pi i} \int_{\Gamma} \frac{\displaystyle z^{\nu} p_{n-1}(z)\sum_{j=1}^{n-\ell} \frac{\widetilde{\lambda}_{j}}{z-y_{j}}\,  \mbox{d} z}{\displaystyle  \prod_{i=1}^{\ell}\left(z-t_i\right)^{2d_i-1}q(z)} + \frac{1}{2 \pi i} \int_{\Gamma} \frac{z^{\nu} p_{n}(z) \, \mbox{d} z}{\displaystyle  \prod_{i=1}^{\ell}\left(z-t_i\right)^{2d_i-1}q(z)} .$$ Since $\displaystyle \frac{z^{\nu} p_{n}(z)}{\displaystyle  \prod_{i=1}^{\ell}\left(z-t_i\right)^{2d_i-1}q(z)}  \in \mathcal{H} \left(\mbox{Int}(\Gamma)\right)$ the second term vanishes. From (\[pnexpand\]), using the Cauchy integral formula, we obtain: $$0=\sum_{j=1}^{n-\ell} y_j^{\nu}\widetilde{p}(y_j)\frac{\widetilde{\lambda}_{j}}{\displaystyle  \prod_{i=1}^{\ell}\left(y^j-t_i\right)^{2(d_i-1)}q(y_j)}=0, \quad \nu=0,\ldots, n-\ell-1.$$ Taking into account that for each $j=1, \ldots, n-\ell$, $\displaystyle \frac{\widetilde{\lambda}_{j}}{\displaystyle  \prod_{i=1}^{\ell}\left(y^j-t_i\right)^{2(d_i-1)}q(y_j)}>0$, we conclude that the above orthogonality relations imply that $\widetilde{p}$ must change sign at least $n-\ell$ times, hence $\deg \widetilde{p} \geq n-\ell$. Since $\deg \widetilde{p}= \deg p_{n-1}-\ell\leq n-\ell-1$ we arrive at a contradiction which completes the proof.

Consider a monic polynomial $\displaystyle P_n(x)=\prod_{j=1}^n(x-x_{j})$ with degree $n\in \mathbb{N}$ which is $\mu$ admissible. We say that a weight function $\rho_n$ on $\mbox{supp}(\mu)$ is orthogonal with respect to $P_n(x)$ and $\mu$ if $P_n \equiv q_{\mu_n,n}$, where $\mbox{d} \mu_n(x)=\rho_n(x) \, \mbox{d} \mu(x)$, $x\in \mbox{supp}(\mu)$. We also say that $\rho_n$ is orthogonal with respect to ${\bf x}_n=(x_{1}, \ldots, x_{n})$ and $\mu$. A sequence of weight functions $\left\{\rho_n\right\}_{n\in \mathbb{N}}$ is a family of orthogonal weight functions with respect to the sequence of polynomials $\left\{P_n\right\}_{n\in \mathbb{N}}$, if for each $n\in \mathbb{N}$, $P_n\equiv q_{\mu_n,n}$.

Let $q_{m(n)}$ be an arbitrary polynomial with degree $\deg q_{m(n)}(x)=2n-m(n)-1$ being positive on $[-1,\,1]$. Let $\mu_n$ denote the measure with differential form $\mbox{d} \mu_n(x) = q_{m(n)}^{-1}(x) \mbox{d} \mu(x)$, $x\in \mbox{supp}(\mu)$. Set a system of nodes ${\bf x}_n=(x_{1}, \ldots, x_{n})$ such that $\displaystyle P_n(x)=\prod_{j=1}^n(x-x_{j})=q_{\mu_n,n}$. This means that ${\bf x}_n$ is the system of $n$ nodes corresponding to the Gaussian quadrature rule for the measure $\mu_n$. Given an arbitrary polynomial $p\in \Pi_{m(n)}$, we have that $$\label{supergauss}
q_{m(n)}(x)p(x)-\sum_{j=1}^n q_{m(n)}(x_{j})p(x_{j})L_{j,n}(x)=q_{\mu_n,n}(x)\mathcal{P}_{n-1}(x),$$ where $\displaystyle L_{j,n}(x):=\prod_{\underset{k\not=j}{k=1}}^n \frac{x-x_{k}}{x_{j}-x_{k}},\quad j=1,\ldots,n,$ and $ \mathcal{P}_{n-1}$ is a certain polynomial with $ \displaystyle \deg \mathcal{P}_{n-1}=n- (m(n)-\deg p)-1\leq n-1.$

Observe that $\displaystyle \int p(x) \, \mbox{d}\,  \mu (x)= \int q_{m(n)}(x)p(x) \mbox{d} \mu_n(x).$ Hence from (\[supergauss\]) we obtain $$\int p(x) \, \mbox{d}\,  \mu (x)- \int \sum_{j=1}^n q_{m(n)}(x_{j})p(x_{j})L_{j,n}(x)\,  \mbox{d}\,  \mu_n (x)=\int q_{\mu_n,n}(x)\mathcal{P}_{n-1}(x)\, \mbox{d}\, \mu_n(x),$$ which vanishes because $q_{\mu_n,n}$ satisfies the orthogonality relations for $\mu_n$ as in (\[ortoargentino\]). We conclude then $$\int p(x)\, \mbox{d}\, \mu(x) =\sum_{j=1}^n p(x_{j,n})q_{m(n)}(x_{j,n}) \int L_{j,n}(x)\frac{\mbox{d}\,  \lambda_0 (x)}{q_{m(n)}(x)}=\sum_{j=1}^n w_{j,n} p(x_{j,n}).$$ This is an interpolatory integration rule with degree of exactness $m(n)$, where the weights can be defined via $$\label{lambda}
w_{j,n}=q_{m(n)}(x_{j})\int L_{j,n}(x)\, \mbox{d} \mu_n(x)=q_{m(n)}(x_{j}) \widetilde{w}_{j,n}, \quad j=1,\ldots,n.$$ The numbers $\widetilde{w}_{j,n}$, $j=1, \ldots,n$ are the weights corresponding to a Gaussian quadrature rule, which are all positive. Since $q_{m(n)}$ is also positive the weights $w_{j,n}>0$. According to Pólya’s condition a sequence of these rules of integration is convergent.

Let us consider $\displaystyle {\bf x}=\left\{{\bf x}_n=\left(x_{1,n}, \ldots, x_{n,n}\right)\right\}_{n \in \mathbb{N}}$ an admissible scheme of nodes for a measure $\mu$, and take a corresponding family of orthogonal weights $\left\{\rho_n\right\}_{n \in \mathbb{N}}$. For each $n$, $\mu_n$ denotes the measure with differential form $\mbox{d} \mu(x)= \rho_n(x) \mbox{d} \mu(x)$, and introduce its family of orthonormal polynomials $\displaystyle \left\{p_{\mu_n,j}\right\}_{j\in \mathbb{Z}_+}$. This means that $\displaystyle p_{\mu_n,j}\equiv q_{\mu_n,j}/ \left|\left|q_{\mu_n,j}\right|\right|_{2, \mu_n},$ $j\in \mathbb{Z}_+$ where $||\cdot||_{2,\mu_n}$ denotes the $\mbox{L}_2$ norm corresponding to the measure $\mu_n$.

Given a function $f \in \mbox{L}_{2,\mu_n}$ and $j\in \mathbb{Z}_+$ we consider the $j$-th partial sum of the Fourier series corresponding to $f/\rho_n$ on the bases $\displaystyle \left\{p_{\mu_n,j}\right\}_{j\in \mathbb{Z}_+}$: $$S_{f, \mu_n, j}=\sum_{k=0}^{j-1} f_k p_{\mu_n,k}(x), \quad f_k=\int f(x) p_{\mu_n, k}(x) \mbox{d} \mu_n(x), \quad k=0,\ldots, j-1.$$ Using the Christoffel-Darboux identity (see [@Sze Theorem 3.2.2]) we can deduce $$\label{CDidentity}
S_{f,\mu_n,j}(x)=\int \frac{q_{\mu_n,j}(x)q_{\mu_n,j-1}(t)-q_{\mu_n,j}(t)q_{\mu_n,j-1}(x)}{\left|\left|q_{\mu_n,j-1}\right|\right|_{2,\mu_n}^2(x-t)} f(t) \mbox{d} \mu_n(t).$$ The following result is an extension of [@Sze Theorem 15.2.4 (equality 15.2.7)]

\[mainsec\] Let $\displaystyle (x_{1},\ldots,x_{n})$ be an $\mu$ admissible system of nodes. Given a polynomial $\displaystyle q_{m(n)}$ take the system of weights $\displaystyle (w_{1,n},\ldots,w_{n,n})$ whose elements $w_{j,n}$, $j=1, \ldots,n$, are constructed using (\[lambda\]). Then there always exists a weight $\rho_n$ such that $$\label{identityknown}
\displaystyle \frac{w_{j,n}}{q_{m(n)}(x_{j})}=\frac{\left|\left|q_{\tau_n,n-1}\right|\right|_{2,\tau_n}^2S_{1/\rho_n, \tau_n,n}(x_{j})}{q_{\tau_n,n-1}(x_{j})q_{\tau_n,n}^{\prime}(x_{j})}=-\frac{\left|\left|q_{\tau_n,n}\right|\right|_{2,\tau_n}^2S_{1/\rho_n, \tau_n,n+1}(x_{j})}{q_{\tau_n,n+1}(x_{j})q_{\tau_n,n}^{\prime}(x_{j})},$$ where the measure $\tau_n$ is such that $\displaystyle \frac{\displaystyle \mbox{d} \tau_n}{\mbox{d} \mu}= \frac{\rho_n}{q_{m(n)}}$. Thus $\mbox{sign}\, w_{j,n}= \mbox{sign}\, S_{1/\rho_n, \tau_n,n}(x_{j})=\mbox{sign}\, S_{1/\rho_n, \tau_n,n+1}(x_{j}),$ $j=1,\ldots, n$.

Take an orthogonal weight $\rho_n$ with respect to the system of $n$ nodes $(x_{1},\ldots,x_{n})$ and the measure with differential form $\mbox{d} \mu(x)/q_{m(n)}(x)$. According to (\[lambda\]) and taking into account that $P_n \equiv q_{\mu_n,n}$ where the measure $\tau_n$ has the differential form $\displaystyle \mbox{d} \tau_n(x) =\rho_n(x) \mbox{d} \mu (x)/q_{m(n)}(x)$, we have the following $$w_{j,n}=q_{m(n)}(x_{j})\int \frac{q_{\tau_n,n}(x)}{q_{\tau_n,n}^{\prime}(x_{j})(x-x_{j})} \, \frac{d\mu(x)}{q_{m(n)}(x)}, \quad j=1,\ldots,n.$$ Arranging the above formula and using the identity (\[CDidentity\]) we obtain that $$w_{j,n}=\frac{q_{m(n)}(x_{j})\left|\left|q_{\tau_n,n}\right|\right|_{2,\tau_n}^2}{q_{\tau_n,n+1}(x_{j})q_{\tau_n,n}^{\prime}(x_{j})} \int \frac{q_{\tau_n,n+1}(x_{j})q_{\tau_n,n}(x)}{\left|\left|q_{\tau_n,n}\right|\right|_{2,\tau_n}^2(x-x_{j})} \frac{1}{\rho_n(x)} \frac{\rho_n(x)\, d\mu(x)}{q_{m(n)}(x)}$$ $$=-\frac{q_{m(n)}(x_{j})\left|\left|q_{\tau_n,n}\right|\right|_{2,\tau_n}^2}{q_{\tau_n,n+1}(x_{j})q_{\tau_n,n}^{\prime}(x_{j})} S_{1/\rho_n, \tau_n,n}(x_{j}),$$ which proves the second identity in (\[identityknown\]). Since $q_{\tau_n,n+1}(x_{j})q_{\tau_n,n}^{\prime}(x_{j})<0, \quad j=1,\ldots,n,$ then $\mbox{sign}\,  w_{j,n}= \mbox{sign}\, S_{1/\rho_n, \tau_n,n+1}(x_{j})$. Following the above steps we can prove the first equality in (\[identityknown\]) and $\mbox{sign}\,  w_{j,n}= \mbox{sign}\, S_{1/\rho_n, \tau_n,n}(x_{j})$.

The following two results are consequences of the above Lemma \[mainsec\]

\[theokey1\] An admissible scheme of nodes $\displaystyle {\bf x}=\left\{{\bf x}_n=(x_{1,n},\ldots,x_{n,n})\right\}_{n\in \mathbb{N}}$ is convergent if there exists a family of orthogonal weights $\displaystyle \left\{\rho_n\right\}_{n\in \mathbb{N}}$ with respect to ${\bf x}$ and the sequence of measures $\displaystyle \left\{\mbox{d} \tau_n(x) =\mbox{d} \mu (x) / q_{m(n)}(x)\right\}_{n\in \mathbb{N}}$ satisfying $$\label{keyeq}
    \lim_{n \to \infty} \left|\left|1-\rho_n(x)S_{1/\rho_n, \tau_n,n}(x)  \right|\right|_{[-1,\,1],\infty}=0,$$ where $\displaystyle \left|\left|\cdot\right|\right|_{[-1,1],\infty}$ denotes the supremum norm on $[-1,\,1]$.

Assuming the equality (\[keyeq\]), there exists a number $N>0$ such that for every $n \geq N$ the function $S_{1/\rho_n, \mu_n,n}(x)>0$ on $[-1,\,1]$ particularly at the nodes. According to Lemma \[mainsec\], the coefficients $w_{j,n}$, $j=1, \ldots, n$, are also positive. This completes the proof.

\[lemaclave\] Consider the varying measure $\mbox{d}\, \mu_n(x)=\mbox{d} \, \mu(x)/q_{m(n)}(x)$ and their orthogonal polynomials $\displaystyle q_{\mu_n,n}(x)=\prod_{j=1}^n\left(x-x_{j,n}\right)$ and $\displaystyle q_{\mu_n,n-1}(x)=\prod_{j=1}^{n-1}\left(x-x_{j,n-1}\right)$, $n\in \mathbb{N}$. Let $\displaystyle {\bf y}=\left\{{\bf y}_n=\left(y_{1,n},\ldots, y_{n,n}\right)\right\}_{n\in \mathbb{N}}$ be a scheme of nodes such that for each $n\in \mathbb{N}$ $$\label{entrelazamiento}
-1<y_{1,n}<x_{1,n-1}<y_{2,n}<\cdots < x_{n-1,n-1}<y_{n,n} <1.$$ Assume that the polynomials $\displaystyle P_{n}(x)=\prod_{j=1}^n \left(x-y_{j,n}\right), \quad n\in \mathbb{N}$ satisfy $$\label{extructura}
\lim_{n \to \infty}\frac{1}{\left|\left|q_{\mu_n, n-1}\right|\right|^2_{2, \mu_n}}\left[\frac{(q_{\mu_n,n}-P_n)q_{\mu_n,n-1}}{q_{m(n)}^2}\right]^{\prime}=0, \quad \mbox{on} \quad [-1,\,1].$$ Then ${\bf y}$ is convergent.

From Lemma \[interwendroff\] we ensure the existence of a weight function $\rho_n$ such that the polynomials $q_{\tau_n,n-1}$ and $P_{n}$ belong to the family of orthogonal polynomials corresponding to the measure $\rho_n(x)\mbox{d} \mu (x)/q_{m(n)}(x)$. Let us analyze the function $$\left|1-\frac{\left|\left| P_{n}\right|\right|^2_{2, \rho_n d \mu/q_{m(n)}}}{q_{m(n)}(x) \left|\left|q_{\mu_n, n-1}\right|\right|^2_{2, \mu_n}}S_{1/\rho_n, \rho_n d \mu/q_{m(n),n} }(x)\right|$$ $$=\frac{1}{q_{m(n)}(x)}\left|S_{q_{m(n)}, \mu_n,n}(x) -\frac{\left|\left| \widetilde{P}_{n}\right|\right|^2_{2, \rho_n d \mu/q_{m(n)}}}{\left|\left|q_{\mu_n, n-1}\right|\right|^2_{2, \mu_n}}S_{1/\rho_n, \rho_n d \mu/q_{m(n),n} }(x)\right|$$ We have used that $\displaystyle S_{q_{m(n)}, \mu_n,n}\equiv q_{m(n)}$, hence we need to show that $$\lim_{n\to 0}\frac{1}{q_{m(n)}(x)}\left|S_{1/\rho_n, \mu_n,n}(x)-\frac{\left|\left|\widetilde{P}_{n}\right|\right|^2_{2, \rho_n d \mu/q_{m(n)}}}{\left|\left|q_{\mu_n, n-1}\right|\right|^2_{2, \mu_n}}S_{1/\rho_n, \rho_n d \mu/q_{m(n),n} }(x)\right|=0.$$ Applying (\[CDidentity\]) we observe that $$\frac{1}{q_{m(n)}(x)}\left(S_{1/\rho_n, \mu_n,n}(x)-\frac{\left|\left|\widetilde{P}_{n}\right|\right|^2_{2, \rho_n d \mu/q_{m(n)}}}{\left|\left|q_{\mu_n, n-1}\right|\right|^2_{2, \mu_n}}S_{1/\rho_n, \rho_n d \mu/q_{m(n),n} }(x)\right)=\frac{q_{m(n)}^{-1}(x)}{\left|\left|q_{\mu_n,n-1}\right|\right|_{2,\mu_n}^2}$$ $$\times \int \left( \frac{q_{\mu_n,n}(x)q_{\mu_n,n-1}(t)-q_{\mu_n,n}(t)q_{\mu_n,n-1}(x)}{x-t}-\frac{P_{n}(x)q_{\mu_n,n-1}(t)-P_{n}(t)q_{\mu_n,n-1}(x)}{x-t}\right)$$ $$\times \frac{\mbox{d} \mu(t)}{q_{m(n)}(t)}$$ Let us consider the kernel $$\frac{q_{\mu_n,n}(x)q_{\mu_n,n-1}(t)-q_{\mu_n,n}(t)q_{\mu_n,n-1}(x)}{\left|\left|q_{\mu_n,n-1}\right|\right|_{2,\mu_n}^2q_{m(n)}(x)q_{m(n)}(t)(x-t)}-\frac{P_{n}(x)q_{\mu_n,n-1}(t)-P_{n}(t)q_{\mu_n,n-1}(x)}{\left|\left|q_{\mu_n,n-1}\right|\right|_{2,\mu_n}^2q_{m(n)}(x)q_{m(n)}(t)(x-t)}$$ $$=\frac{(q_{\mu_n,n}-P_n)(x)q_{\mu_n,n-1}(t)-(q_{\mu_n,n}-P_n)(t)q_{\mu_n,n-1}(x)}{\left|\left|q_{\mu_n,n-1}\right|\right|_{2,\mu_n}^2q_{m(n)}(x)q_{m(n)}(t)(x-t)}=\mathcal{K}(x,t).$$ From Taylor’s Theorem we obtain that $$\mathcal{K}(x,t)= \frac{1}{\left|\left|q_{\mu_n, n-1}\right|\right|^2_{2, \mu_n}}\left[\frac{(q_{\mu_n,n}-P_n)q_{\mu_n,n-1}}{q_{m(n)}^2}\right]^{\prime}(s)$$ for some $s$ in between of $x$ and $t$, so the assumption (\[extructura\]) completes the proof.

Asymptotic analysis {#asymptotic analysis}
===================

Let us consider the varying measure $\mu_n$ with $\displaystyle \mbox{d} \mu_n(x)/\mbox{d} x= (q_{m(n)}(x)\sqrt{1-x^2})^{-1}$, where $\displaystyle q_{m(n)}(x)= q_{\kappa}^{\displaystyle 2 \frac{1-a}{k} n}(x)=\left(\prod_{k=1}^{\kappa}(x-\zeta_j)\right)^{\displaystyle 2 \frac{1-a}{k} n}, \quad n\in \Lambda.$ Let $\sigma$ be the zero counting measure of $q_k$. This is $\displaystyle \sigma=\frac{1}{\kappa}\sum_{k=1}^{\kappa} \delta_{\zeta_k}.$ Set the analytic logarithmic potential corresponding to the measure $\sigma$: $$\label{externalfield}
g(z,{\sigma})= -\int \log\, (z-\zeta)\, \mbox{d}\, \sigma(\zeta).$$ We take the logarithmic branch such that $g(z,{\sigma})$ is analytic on a domain $D \subset K$ that contains the interval $[-1,\,1]$, and also for every $x \in [-1,\,1]$, $$\label{potentialsigma}
V^{\sigma}(x)= \int \log\, \frac{1}{|x-\zeta|} \mbox{d}\, \sigma(\zeta)=g(x,{\sigma})= -\int \log\, (z-\zeta)\, \mbox{d}\, \sigma(\zeta).$$ Since $\sigma$ is symmetric we $\displaystyle \int \arg (x-\zeta) \mbox{d}\, \sigma(\zeta)=0$. In each compact $K\subset D$ we have that $\displaystyle \frac{1}{2n} \log  \frac{1}{q_{m(n)}(z)}=(1-a) \,g(z,{\sigma})$ on $K.$

\[explicitexp\] Let $\displaystyle \mbox{d} \mu_n(x)/\mbox{d} x= (q_{m(n)}(x)\sqrt{1-x^2})^{-1}$, $n\in \mathbb{N}$ be a sequence of measures as above. Then $$\label{qmunn}
q_{\mu_n,n}= \left(1+\mathcal{O}(e^{-c n})\right) \exp\left\{ -n V^{\overline{\nu}}\right\}K_{1,n}+\mathcal{O}(e^{-c n})\exp\left\{- n V^{\overline{\nu}}\right\} K_{2,n}$$ and $$\label{qmunn-1}
 \frac{d_{n,n-1}}{2^{2 n a}} q_{\mu_n,n-1}=\left(1+\mathcal{O}(e^{-c n})\right) \exp\left\{ -n V^{\overline{\nu}}\right\}K_{2,n}+\mathcal{O}(e^{-c n})\exp\left\{- n V^{\overline{\nu}}\right\} K_{1,n}$$ where $\displaystyle d_{n,n-1}=-\left(2 \pi i \left|\left|q_{\mu_n,n-1}\right|\right|^2_{\mu_n,2}\right)^{-1},$ $$\label{forK1}
K_{1,n}(x)=\displaystyle 2  \cos n\left((1-a)\pi \int^1_x \mbox{d} \widetilde{\sigma}(t) -a \arccos{x} \right),$$ and $$\label{forK2}
K_{2,n}(x)=\displaystyle  \frac{1}{i} \cos n \left((1-a) \pi \int_{x}^1 \mbox{d} \widetilde{\sigma}(t) - (a-1/n) \arccos{x} \right).$$

We study a matrix Riemann-Hilbert problem like in [@KMVV Theorem 2.4] whose solution $Y$ is a $2 \times 2$ matrix function satisfying the following conditions:

1.  $Y \in \mathcal{H}(\mathbb{C}\setminus[-1,\,1])$ (all the entries of $Y$ are analytic on $\mathbb{C}\setminus[-1,\,1]$),

2.  $\displaystyle Y_+(x)=Y_-(x)\left(\begin{array}{c c}
    1 & \left(q_{m(n)}(x)\sqrt{1-x^2}\right)^{-1}\\ & \\
    0 & 1
    \end{array}\right)$, $x \in (-1,\,1)$,

3.  $\displaystyle Y(z)\left(\begin{array}{c c}
    z^{-n} & 0\\ & \\
    0 & z^{n}
    \end{array}\right)=\mathbb{I}+\mathcal{O}(1/z)$ as $z \to \infty$, $\mathbb{I}$ is the $2\times2$ identity matrix.

4.  $\displaystyle Y(z)=\mathcal{O}\left(\begin{array}{c c}
    1 & |z\pm 1|^{-1/2} \\ & \\
    1 &  |z\pm 1|^{-1/2}
    \end{array}\right)$ as $z \to \mp 1$.

According to [@KMVV Theorem 2.4] (see also [@Ku]) the $Y$ solution of above matrix Riemann-Hilbert problem (for short Y-RHP) is unique and has the form $$Y(z)=\left(\begin{array}{c c}
q_{\mu_n,n}(z) & \displaystyle -\frac{1}{2 \pi i}\int \frac{q_{\mu_n,n}(x)}{z-x} \, \mbox{d}\mu_n(x) \\ & \\
d_{n,n-1} q_{\mu_n,n-1}(z) & \displaystyle -\frac{d_{n,n-1}}{2 \pi i}\int \frac{q_{\mu_n,n-1}(x)}{z-x} \, \mbox{d}\mu_n(x) 
\end{array}\right).$$

The key of our procedure follows the ideas introduced in [@AV]. We find a relationship between $Y$ and the matrix solution $R: \mathbb{C}\setminus \gamma \rightarrow \mathbb{C}^{2\times 2}$ corresponding to another Riemann-Hilbert problem (R-RHP) for a closed Jordan curve $\gamma$ positively oriented surrounding the interval $[-1,\,1]$:

1.  $R \in \mathcal{H} (\mathbb{C}\setminus \gamma)$,

2.  $R_+(\zeta)=R_-(\zeta)V_n(\zeta),$ $\zeta \in \gamma$, with $V_n\in \mathcal{H}(D)$,

3.  $R(z) \rightarrow \mathbb{I}$ as $z \to \infty$,

where $V_n=\mathbb{I}+\mathcal{O}(\varepsilon^n)$ with $0\leq \varepsilon<1$, uniformly on compact subsets of $K$ as $n \to \infty$. Those conditions imply that $R=\mathbb{I}+\mathcal{O}(\varepsilon^n)$ uniformly on $\mathbb{C}$ as $n \to \infty$. There is a chain of transformations to arrive from $Y$ to $R$, which we represent $Y \rightarrow T \rightarrow S \rightarrow R$. Once we have arrived to $R$, we recover the entries of $Y$ going back from $R$ to $Y$.

From [@BLS2 Corollary 4] we have that the zero counting measures $\nu_n$ defined in (\[zerocountingmeasures\]) corresponding to the monic orthogonal polynomials $\displaystyle q_{\mu_n,n}(z)=\prod_{j=1}^n\left(z-x_{j,n}\right)$ with respect to the varying measures $\mu_n$, satisfy $$\label{equilibriummeasure}
\nu_n \overset{\star}{\rightarrow} \overline{\nu}=(1-a) \widetilde{\sigma} +a \lambda_0 \quad \mbox{as} \quad n \to \infty,$$ where $\widetilde{\sigma}$ denotes the balayage of the measure $\sigma$ out of $\mathbb{C}\setminus [-1, \,1]$ onto $[-1,\,1]$.

The measure $\overline{\nu}$ is the so called (see [@ST Theorem I.1.3]) equilibrium measure under the influence of the external field $ (1-a) V^{\sigma}(z)$. From (\[equilibriummeasure\]) we have the following equilibrium condition $$\label{equilibriumcondition}
V^{\overline{\nu}}(t)-(1-a) V^{\sigma}(t)=a V^{\lambda_0}(t)=a \log 2, \quad t\in [-1,\,1].$$

Observe the conditions $(3)$ in both Riemann Hilbert problems. $Y$ requires a normalization at infinity to get to $R$’s behavior at infinity. We modify $Y$ to obtain a Riemann-Hilbert problem whose solution is defined on the same set as $Y$, which approaches $\mathbb{I}$ as $n \rightarrow \infty$. Let us introduce the function $g(z,\overline{\nu})$, which is the analytic potential corresponding to the measure $\overline{\nu}$ described in (\[equilibriummeasure\]) $$\label{defgn}
g(z,\overline{\nu})=-\int \log\, (z-t)\, \mbox{d} \overline{\nu}(t)=V^{\overline{\nu}}(z)-\,i\,\int \arg (z-t) \, \mbox{d}\overline{\nu}(t),$$ with $\arg$ denoting the principal argument $g(z,\overline{\nu}) \in \mathcal{H}(K\setminus (-\infty,\,1])$. Substituting $g(z,\overline{\nu})$ in (\[equilibriumcondition\]) we obtain $$\label{roi}
g_+(x,\overline{\nu})+g_-(x,\overline{\nu})-2a \log 2-2(1-a)g(x,\sigma)=0, \quad x \in [-1,\,1].$$ and $$\label{casta}
g_-(x,\overline{\nu})-g_+(x,\overline{\nu})=\left\{\begin{array}{l l l}
0 & \mbox{if} & x\geq 1 \\
2\pi i &\mbox{if} & x\leq 1 \\
2 \wp(x) &\mbox{if} & x\in (-1,\,1),
\end{array}
\right.$$ with $$\label{california}
\wp(x)=\displaystyle \pi i \int^1_{x} \mbox{d} \overline{\nu}(t)=\pi i \left[(1-a) \displaystyle  \int^1_{x} \mbox{d} \widetilde{\sigma}(t)-\frac{a}{\pi}\arccos(x)  \right].$$

Consider the matrices $\displaystyle G(z)=\left(\begin{array}{c c}
e^{n g(z,\overline{\nu})} & 0 \\ 
0 & e^{-n g(z,\overline{\nu})} 
\end{array}
\right)$ and $\displaystyle L=\left(\begin{array}{c c}
2^{n a} & 0 \\ 
0 & 2^{-n a} 
\end{array}
\right).$ We define the matrix function $T=LYGL^{-1}$. So $T$ is the unique solution of the following Riemann-Hilbert problem (T-RHP)

1.  $T \in \mathcal{H}(\mathbb{C}\setminus [-1,\,1])$,

2.  $T_+(x)=T_-(x)M(x)$, $x\in (-1,\,1),$

3.  $T(z)=\mathbb{I}+\mathcal{O}\left(1/z\right)$ as $z \to \infty$,

4.  $\displaystyle T(z)=\mathcal{O}\left(\begin{array}{c c}
    1 &|z\pm 1|^{-1/2} \\
    1 & |z\pm 1|^{-1/2} 
    \end{array}\right)$ as $z \to \mp 1$,

where according to (\[roi\]) and (\[casta\]) the jump matrix $\displaystyle M(x)=\left(\begin{array}{c c}
e^{-2 n \wp(x)} & (1-x^2)^{-1/2}  \\
0 & e^{2 n\wp(x)} 
\end{array}
\right),$ $x\in (-1,\,1).$

According to [@DKMc Theorem 1.34] there exists a domain $D$ containing the interval $[-1,\,1]$ where the function $\wp$ in (\[california\]) admits an analytic extension on $D\setminus [-\infty,\,1]$ as $$\label{defwp}
\mathcal{A}(z)=\pi i \int_z^1 \, \mbox{d}\overline{\nu}(\zeta)=\pi i \int_z^1 \, \overline{\nu}^{\prime}(\zeta) \mbox{d} \zeta,$$ where $\displaystyle \overline{\nu}^{\prime}(\zeta)=\frac{\psi(\zeta)}{\sqrt{1-\zeta^2}},$ with $\psi \in \mathcal{H}(D)$ and $\psi(x)>0,$ $x\in (-1,1).$ Observe that $\mathcal{A}_+(x)=\wp(x)=-\mathcal{A}_-(x)$, then we write $\displaystyle M(x)=\left(\begin{array}{c c}
e^{-2 n \mathcal{A}_+(x)} & (1-x^2)^{-1/2}\\
0 & e^{-2 n\mathcal{A}_-(x)} 
\end{array}
\right)$.

We now seek jump conditions as we have in R-RHP. Consider a closed Jordan curve $\gamma \in D$ surrounding $[-1,\,1]$ as we have in R-RHP. Let $\Omega$ denote the bounded connected component of $\mathbb{C}\setminus\gamma$. We consider the function $\sqrt{z^2-1}\in \mathcal{H} (\mathbb{C}\setminus [-1,\, 1]$ with $\sqrt{x^2-1}_\pm=\pm i \sqrt{1-x^2}$, $x \in (-1,\,1)$. We introduce the matrix function $S$ as follows $$S(z)=\left\{\begin{array}{c l l}
T(z) & \mbox{when} & z\in \mathbb{C} \setminus (\gamma \cup \Omega) \\ 
T(z) \left(\begin{array}{c c}
1 & 0 \\ 
-i\sqrt{z^2-1}\, e^{-2 n \mathcal{A}(z)}& 1
\end{array} \right) & \mbox{when} & z\in \Omega
\end{array} 
\right..$$ The matrix function $S$ is the solution of the following Riemann-Hilbert problem (S-RHP):

1.  $S \in \mathcal{H}(\mathbb{C} \setminus (\gamma \cup [-1,\,1]))$,

2.  $\displaystyle S_+(x)=S_-(x) \left(\begin{array}{c c}
    0 & \displaystyle (1-x^2)^{-1/2} \\ 
    - (1-x^2)^{1/2} &0 
    \end{array}
    \right)$, when $x \in (-1,\,1)$ and

3.  $\displaystyle S_+(\zeta)=S_-(\zeta) \left(\begin{array}{c c}
    1 & 0 \\ 
    -i\sqrt{z^2-1}\, e^{-2 n \mathcal{A}(z)}& 1
    \end{array} \right),$ when $\zeta \in \gamma.$

4.  $S(z)=\mathbb{I}+\mathcal{O}\left(1/z\right)$ as $z \to \infty$,

5.  $\displaystyle S(z)=\mathcal{O}\left(\begin{array}{c c}
    1 & |z\pm 1|^{-1/2} \\
    1 & |z\pm 1|^{-1/2}
    \end{array}\right)$ as $z \to \mp 1$.

The jump matrix on $\gamma$ approaches uniformly the identity matrix $I$. However it does not happen in $[-1,\, 1]$. We fix this problem in the interval following the steps in [@Ku]. Consider the matrix $$\label{Ndef}
N(z)=\left(\begin{array}{l l}
\displaystyle \frac{\displaystyle a(z)+a^{-1}(z)}{\displaystyle 2} \frac{D(\infty)}{D(z)}& \displaystyle \frac{\displaystyle a(z)-a^{-1}(z)}{\displaystyle 2i} D(\infty) D(z) \\ 
\displaystyle \frac{\displaystyle a(z)-a^{-1}(z)}{\displaystyle -2i}\frac{1}{D(\infty)D(z)} & \displaystyle \frac{\displaystyle a(z)+a^{-1}(z)}{\displaystyle 2}\frac{D(z)}{D(\infty)} 
\end{array}
\right),$$ where $\displaystyle D(z)=\left(\frac{z}{\sqrt{z^2-1}}+1\right)^{1/2}, \quad D(\infty)=\sqrt{2}$ and $\displaystyle a(z)= \frac{(z-1)^{1/4}}{(z+1)^{1/4}}.$ Hence $N$ is the solution of the following Riemann-Hilbert problem

1.  $N \in \mathcal{H}(\mathbb{C}\setminus [-1,\,1])$,

2.  $\displaystyle N_+(x)=N_-(x)\left(\begin{array}{c c}
    0 & \displaystyle (1-x^2)^{-1/2} \\
    -(1-x^2)^{1/2} &0 
    \end{array}
    \right)$, $x\in (-1,\,1),$

3.  $N(z)=\mathbb{I}+\mathcal{O}\left(1/z\right)$ as $z \to \infty$,

4.  $\displaystyle N(z)=\mathcal{O}\left(\begin{array}{ c c }
    1 & \left| z\pm 1\right|^{-1/2} \\
    1 & \left| z \pm 1\right|^{-1/2} 
    \end{array}\right)$ as $z \to \mp 1$.

Introduce the matrix function $R(z)=S(z)N^{-1}$. Taking into account that $R$ and $S$ satisfy the same jump conditions across $(-1,\,1)$ we have that $R_+(x)=R_-(x)$. So $R \in \mathcal{H}(\mathbb{C} \setminus (\gamma \cup \{-1,1\}))$. Since $\det N=1$ and from (\[Ndef\]) we have that $$N^{-1}(z)=\mathcal{O}\left(\begin{array}{ c c }
\left| z\pm 1\right|^{-1/2} & \left| z\pm 1\right|^{-1/2} \\
1 & 1 
\end{array}\right) \quad \mbox{as} \quad z \to \mp 1.$$ Thus, when $z \to \mp 1$ $$R(z)=\mathcal{O}\left(\begin{array}{c c}
1 & \left| z\pm 1\right|^{-1/2}  \\
1 & \left| z\pm 1\right|^{-1/2}  \end{array}\right)\mathcal{O}\left(\begin{array}{ c c }
\left| z\pm 1\right|^{-1/2}  & \left| z\pm 1\right|^{-1/2}  \\
1 & 1 
\end{array}\right).$$ This implies $$R(z)= \mathcal{O}\left(\begin{array}{ c c }
\left| z\pm 1\right|^{-1/2} & \left| z\pm 1\right|^{-1/2} \\
\left| z\pm 1\right|^{-1/2} & \left| z \pm 1\right|^{-1/2} 
\end{array}\right),$$ which means that each entry of $R$ has isolated singularities at $z=-1$ and $z=1$ with $R(z)=\mathcal{O}|z\pm 1|^{-1/2}$ as $z \to \mp 1$, and they are removable. So $R$ satisfies the following Riemann-Hilbert conditions:

1.  $R \in \mathcal{H}(\mathbb{C} \setminus \gamma)$,

2.  $\displaystyle R_+(\zeta)=R_-(\zeta) \left(\begin{array}{c c}
    1 & 0 \\
    e^{-2 n \mathcal{A}(\zeta)} & 1 
    \end{array}
    \right),$ when $\zeta \in \gamma.$

3.  $R(z)=\mathbb{I}+\mathcal{O}\left(1/z\right)$ as $z \to \infty$.

From (\[defwp\]) $2\mathcal{A}\in \mathcal{H}(D\setminus[-\infty,\,1])$ and $\mathcal{R}e(2\mathcal{A}_{\pm}(x))=0$, $x \in [-1,\,1]$. Using the fact $\displaystyle 2\mathcal{A}_{\pm}^{\prime}(x)=\pm 2 i \overline{\nu}^{\prime}(x)=\mp 2 \pi i\frac{\psi(x)}{\sqrt{1-x^2}}$, $x \in (-1,\,1)$ and the Cauchy-Riemann conditions we have that $\displaystyle \frac{\partial \mathcal{R}e(2\mathcal{A}_\pm)}{\partial y}(x)>0$, $x \in [-1,\,1]$. Since $\mathcal{R}e(2\mathcal{A})$ is a harmonic function on $D\setminus [-1,\,1]$ we have $\mathcal{R}e(2\mathcal{A}(z))>0,$ $z\in D\setminus [-1,\, 1]$. So given an arbitrary compact set $K \subset D\setminus [-1,\,1]$ there exists a constant $c(K)>0$ and an $N\in \mathbb{N}$ large enough such that for every $n\geq N$ the function $\mathcal{R}e(2\mathcal{A}(z))(z))>c(K)$, $z \in K$ and $n\geq N$. Note also that $\phi_n \to 0$ as $n \to \infty$. So according [@AV] we arrive at $R(z)=\mathbb{I}+\mathcal{O}(e^{-c n})$ uniformly as $n \to \infty$ for each compact set $K \subset \mathbb{C} \setminus [-1,\,1]$. Take $z \in \mbox{Int}(\gamma)$. Going back now from $R$ to $Y$, and considering just the first column, we have that: $$e^{n g(z, \overline{\nu})} \left(\begin{array}{l}
q_{\mu_n,n}(z) \\
2^{-2 n a} d_{n,n-1} q_{\mu_n,n-1}(z) 
\end{array}
\right)=\left(\mathbb{I}+\mathcal{O}(e^{-c n}) \right)$$ $$\times\left(\begin{array}{l l}
\displaystyle \frac{\displaystyle a(z)+a^{-1}(z)}{\displaystyle 2} \frac{D(\infty)}{D(z)}& \displaystyle \frac{\displaystyle a(z)-a^{-1}(z)}{\displaystyle 2i} D(\infty) D(z) \\
\displaystyle \frac{\displaystyle a(z)-a^{-1}(z)}{\displaystyle -2i}\frac{1}{D(\infty)D(z)} & \displaystyle \frac{\displaystyle a(z)+a^{-1}(z)}{\displaystyle 2}\frac{D(z)}{D(\infty)} 
\end{array}
\right)  \left(\begin{array}{c}
1 \\ \\
(1-z^2)^{1/2}e^{-2 n \mathcal{A} (z)} 
\end{array} \right).$$

Take the $+$ boundary values of all quantities involved when $z \to x \in (-1,1)$. Using the following identities from [@Ku] or [@KMVV] $$\frac{a_+(x)\pm a_+(x)}{2}=\frac{1}{\sqrt{2}(1-x^2)^{1/4}} \exp \left(\displaystyle \pm \frac{i}{2} \arccos x \mp i \frac{\pi}{4} \right),$$ we have $\displaystyle 
\exp\left\{ n V^{\overline{\nu}}(x)\right\} \left(\begin{array}{l}
q_{\mu_n,n}(x) \\
2^{-2 n a} d_{n,n-1} q_{\mu_n,n-1}(x) 
\end{array}
\right)=\left(\mathbb{I}+\mathcal{O}(e^{-c n}) \right) \left(\begin{array}{c}
    K_{1,n} (x)\\
    K_{2,n} (x)
\end{array}\right),$ where $\displaystyle 
K_{2,n}(x)=\displaystyle \frac{1}{i} \cos n \left((1-a) \pi \int_{x}^1 \mbox{d} \widetilde{\sigma}(t) - (a-1/n) \arccos{x} \right)$ and

$\displaystyle K_{1,n}(x)=\displaystyle 2 \cos n\left((1-a)\pi \int^1_x \mbox{d} \widetilde{\sigma}(t) -a \arccos{x}\right)$.

Finally we obtain $$q_{\mu_n,n}(x)= \left(1+\mathcal{O}(e^{-c n})\right) e^{ -n V^{\overline{\nu}}(x)}K_{1,n}(x)+\mathcal{O}(e^{-c n})e^{- n V^{\overline{\nu}}(x)} K_{2,n}(x)$$ and $$\frac{d_{n,n-1}}{2^{2 n a}} q_{\mu_n,n-1}(x)=\left(1+\mathcal{O}(e^{-c n})\right) e^{ -n V^{\overline{\nu}}(x)}K_{2,n}(x)
+\mathcal{O}(e^{-c n})e^{- n V^{\overline{\nu}}(x)} K_{1,n}(x),$$ which are exactly the equalities stated in (\[qmunn\]) and (\[qmunn-1\]).

Proof of Theorem \[main\] {#proofmaintheorem}
=========================

We combine Lemma \[explicitexp\] and Lemma \[extructura\]. First we choose a special scheme of nodes $\displaystyle {\bf y}=\left\{{\bf y}_n=\left(y_{1,n},\ldots,y_{n,n}\right)\right\}_{n \in \Lambda}$ which satisfies (\[maincond\]). The corresponding polynomials have the following form $$\label{drpn}
P_n(x)=\prod_{j=1}^n \left(x-y_{j,n}\right)=\Phi_n(x) \cos n \left((1-a)\pi \int^1_x \mbox{d} \widetilde{\sigma}(t)-a \arccos x \right),$$ Where $\Phi_n$ is a real valued function on $[-1,\,1]$ that never vanishes. Let us rewrite the relation (\[qmunn\]) as follows $$q_{\mu_n,n}(x)= 2\exp\left\{-n V^{\overline{\nu}}(x)\right\} \left(\cos  n \left((1-a)\pi \int^1_x \mbox{d} \widetilde{\sigma}(t)-a \arccos x \right) +\mathcal{O}(e^{-cn}) \right)$$ $$= 2\exp\left\{-n V^{\overline{\nu}}(x)\right\}\cos  n \left((1-a)\pi \int^1_x \mbox{d} \widetilde{\sigma}(t)-a \arccos x + \mathcal{O}(e^{-cn}) \right).$$ Combining the above equality with (\[drpn\]) we obtain that $\displaystyle \int_{x_{j,n}}^{y_{j,n}} \, \mbox{d} \overline{\nu}(t)=\mathcal{O}(e^{-cn})$ and $x_{j,n}-y_{j,n}=\mathcal{O}(e^{-cn}).$ This implies that $\displaystyle \limsup_{n\to \infty}|q_{\mu_n,n}-P_n|^{1/n}(x)=\exp(-c-V^{\overline{\nu}}(x))$ on $ [-1,\,1].$ Hence $$\limsup_{n\to \infty} \left(\frac{1}{\left|\left|q_{\mu_n, n-1}\right|\right|^2_{2, \mu_n}}\left[\frac{(q_{\mu_n,n}-P_n)q_{\mu_n,n-1}}{q_{m(n)}^2}\right]^{\prime}\right)^{1/n}(x)=\exp (-c+V^{\sigma}(x))<1.$$ Here we have taken into account that $\displaystyle \mathrm{dist}\left(\left\{\zeta_1,\ldots,\zeta_{\kappa}\right\},[-1,\,1]\right)>1$, which yields $V^{\sigma}(x)<0$, $x \in [-1, \, 1]$. Then we see that condition (\[extructura\]) in Lemma \[lemaclave\] is satisfied. We now prove that condition (\[entrelazamiento\]) holds.

Taking into account the equality (\[qmunn-1\]) we have that the zeros of the polynomials $q_{\mu_n,n-1}$ satisfy that for each $j=1, \ldots, n-1$, $n\in \mathbb{N}$ $$\label{qn-1zeros}
(1-a) \pi \int_{x_{j,n-1}}^1 \mbox{d} \widetilde{\sigma}(t)-(a-1/n) \arccos x + \mathcal{O}(e^{-cn})= \frac{2j-1}{2 n} \pi.$$ For each $j=1,\ldots,n-1$, we subtract the above equality (\[qn-1zeros\]) to (\[maincond\]), and we obtain that $$\int_{y_{j,n}}^{x_{j,n-1}} \mbox{d} \overline{\nu}(t) = \frac{1}{n}\left(1 + o(1)\right) \quad \mbox{as} \quad n \to \infty.$$ This means that for $n$ large enough $y_{j,n} < x_{j,n-1}$, $j=1, \ldots, n-1$. Considering now the $j$th equality in (\[maincond\]) and the $j+1$th in (\[qn-1zeros\]) we have that $$\int_{x_{j,n-1}}^{y_{j+1,n}} \mbox{d} \overline{\nu}(t)=\frac{1}{n}\left(\pi-1+o(1)\right) \quad \mbox{as} \quad n\to \infty,$$ which implies that $x_{j,n-1}<y_{j+1,n}$. So condition (\[entrelazamiento\]) holds. This proves that the scheme ${\bf y}$ is convergent.

Once we know that ${\bf y}$ is convergent, we can construct another convergent scheme $\displaystyle {\bf x}=\left\{{\bf x}_n=\left(x_{1,n},\ldots,x_{n,n}\right)\right\}_{n \in \Lambda}$ taking $$x_{j,n}-y_{j,n}\leq A e^{-\ell n}, \quad j=1, \ldots,n, \quad n\in \Lambda,$$ and follow the previous process. This completes the proof.
