Multiplication operator and average characteristic polynomial associated with exceptional Jacobi polynomials

Abstract

Studying the multiplication operator associated with exceptional Jacobi polynomials, the zero distribution of the corresponding average characteristic polynomials is determined. Applying this result, the location of zeros of certain self-inversive polynomials is examined.

1 Introduction

The three term recurrence relation fulfilled by standard orthogonal polynomials implies that the multiplication operator $M: f(x)\rightarrow xf(x)$, acting on the weighted $L^2_w$ space can be represented by a tridiagonal (Jacobi) matrix. Denoting by $\pi _n$ the projection operator to the n-dimensional subspace, the eigenvalues of $\pi _nM\pi _n$ are just the zeros of the orthogonal polynomials in question.

Considering the probability density

$$\begin{aligned} \varrho _{n}(x_1, \dots ,x_n)=c({n})\det |K_n(x_i,x_j)|_{i,j=1}^n\prod _{i=1}^n W(x_i), \end{aligned}$$

where c(n) is a normalization factor, and the average characteristic polynomial is defined as follows:

$$\begin{aligned} \chi _n(z):={\mathbb {E}}\left( \prod _{i=1}^n(z-x_i)\right) , \end{aligned}$$

where the expectation ${\mathbb {E}}$ refers to $\varrho _n$. According to [28, (2.2.10)], the zeros of the average characteristic polynomial are just the zeros of $q_n$ again.

Exceptional orthogonal polynomials are complete systems of polynomials with respect to a positive measure. They are different from the standard or from the classical orthogonal polynomials, since exceptional families have finite codimension in the space of polynomials. Similarly to the classical ones, exceptional polynomials are eigenfunctions of Sturm–Liouville-type differential operators but unlike the classical cases, the coefficients of these operators are rational functions. Exceptional orthogonal polynomials also possess a Bochner-type characterization as each family can be derived from one of the classical families applying finitely many Darboux transformations, see [5].

Exceptional orthogonal polynomials were introduced recently by Gómez-Ullate, Kamran and Milson, cf. e.g. [7, 8] and the references therein. These families of polynomials play a fundamental role for instance in the construction of bound-state solutions to exactly solvable potentials in quantum mechanics. In the last few years, they have seen a great deal of activity in this area both by mathematicians and physicists, cf. e.g. [4, 6, 13, 25], etc. The location of zeros of exceptional orthogonal polynomials is also examined, cf. e.g. [2, 3, 9, 15, 16, 19].

Exceptional orthogonal polynomials fulfil $2L+1$ recurrence formulae with $L\ge 2$. Thus the matrices that represent the multiplication operators associated with exceptional polynomials are $2L+1$ diagonal. Unlike the standard case, the nth average characteristic polynomial does not coincide with the nth orthogonal polynomial; furthermore, these two polynomials are of different degrees.

Below we deal with exceptional Jacobi polynomials. In this case, it is pointed out that the normalized counting measure associated with the modified characteristic polynomial tends to the equilibrium measure of the interval of orthogonality in weak-star sense, see Theorem 1 below, i.e. it coincides asymptotically with the normalized zero-counting measure of the corresponding exceptional Jacobi polynomials, see [2, Theorem 6.5].

Finally, this multiplication operator method is applied to determine the zeros on the unite circle of certain self-inversive polynomials.

2 Preliminaries, notation

2.1 General construction of exceptional orthogonal polynomials

There are two different approaches of derivation exceptional orthogonal polynomials from classical ones. Both methods are based on the quasi-rational eigenfunctions of the classical (Hermite, Laguerre, Jacobi) differential operators. The method, developed first for exceptional Hermite polynomials (see [6]), defines exceptional polynomials by the Wronskian of certain partition of quasi-rational eigenfunctions mentioned above. The other method defines exceptional polynomials recursively, which is by application of finitely many Crum–Darboux transformations to the classical differential operators. A Bochner-type characterization theorem (see [5]) ensures that all exceptional classes can be derived by this method.

Subsequently we follow the construction by Crum-Darboux transformation. According to [5, Propositions 3.5 and 3.6], it is as follows.

Classical orthogonal polynomials $\left\{ P_n^{[0]}\right\} _{n=0}^\infty$ are eigenfunctions of the second-order linear differential operator with polynomial coefficients

$$\begin{aligned} T[y]=py''+qy'+ry, \end{aligned}$$

and its eigenvalues are denoted by $\lambda _n$. T can be decomposed as follows:

$$\begin{aligned} T=BA+\tilde{\lambda }, \text{ with } A[y]=b(y'-wy), B[y]=\hat{b}(y'-\hat{w}y), \end{aligned}$$

(1)

where w is the logarithmic derivative of a quasi-rational eigenfunction of T with eigenvalue $\tilde{\lambda }$, which is w is rational and fulfils the Riccati equation

$$\begin{aligned} p(w'+w^2)+qw+r=\tilde{\lambda }. \end{aligned}$$

(2)

b is a suitable rational function. The coefficients of B can be expressed as follows:

$$\begin{aligned} \hat{b}=\frac{p}{b}, \hat{w}=-w-\frac{q}{p}+\frac{b'}{b}. \end{aligned}$$

(3)

Then the exceptional polynomials are the eigenfunctions of $\hat{T}$ that is the partner operator of T, which is

$$\begin{aligned} \hat{T}[y]=(AB+\tilde{\lambda })[y]=py''+\hat{q}y'+\hat{r}y, \end{aligned}$$

(4)

where

$$\begin{aligned} \hat{q}= & {} q+p'-2\frac{b'}{b}p, \nonumber \\ \hat{r}= & {} r+q'+wp'-\frac{b'}{b}(q+p')+\left( 2\left( \frac{b'}{b}\right) ^2-\frac{b''}{b}+2w'\right) p, \end{aligned}$$

(5)

(2) and (4) imply that

$$\begin{aligned} \hat{T}AP_n^{[0]}=\lambda _nAP_n^{[0]}, \end{aligned}$$

(6)

so exceptional polynomials can be obtained from the classical ones by application of (finitely many) appropriate first-order differential operator(s) to the classical polynomials. This observation motivates the notation below:

$$\begin{aligned} AP_n^{[0]}=b\left( P_n^{[0]}\right) '-bwP_n^{[0]}=:P_n^{[1]}, \end{aligned}$$

(7)

(and recursively $A_sP_n^{[s-1]}=:P_n^{[s]}$ in case of s Darboux transformations.) The degree of $P_n^{[1]}$ is usually greater than n. $\left\{ P_n^{[1]}\right\} _{n=0}^\infty$ is an orthogonal system on I with respect to the weight

$$\begin{aligned} W:=\frac{pw_0}{b^2}, \end{aligned}$$

(8)

where $w_0$ is one of the classical weights.

Remark

Since at the endpoints of I (if there is any) p may possesses zeros, b can be zero here as well, but b does not have zeros inside I, otherwise the moments of W would not be finite.

• As each $P_n^{[1]}$ is a polynomial ($n=0, 1, \dots$), applying the operator A to $P_0^{[0]}$ it can be seen that bw must be a polynomial and to $P_1^{[0]}$ shows that b itself is also a polynomial.

• Let us recall that $r=0$ in the classical differential operators. Considering (2) and comparing degrees, w itself cannot be a polynomial. By the same reasons, pw cannot be a polynomial unless it is of degree one. w is a rational function, and it has no poles in $(-1,1)$.

• Expressing (2) as

  $$\begin{aligned} bBP_n^{[1]}=p\left( P_n^{[1]}\right) '+\left( pw+q-p\frac{b'}{b}\right) P_n^{[1]}=(\lambda _n-\tilde{\lambda })bP_0^{[0]}, \end{aligned}$$

  (9)

  it can be easily seen that if $P_n^{[1]}$ had got a double zero at $x_0\in \mathrm {int}I$, then $P_0^{[0]}(x_0)=0$, and by (7) $\left( P_n^{[0]}\right) '(x_0)=0$, which is impossible, i.e. that zeros of $P_n^{[1]}$ which are in the interior of the interval of orthogonality are simple.

• Let a be (one of) the finite endpoint(s) of the interval of orthogonality. Again by (9), one can derive that if $b(a)\ne 0$, then $P_n^{[1]}(a)\ne 0$ as well.

• If there was an $n\in {\mathbb {N}}$ such that $P_n^{[1]}(a)= 0$, then $b(a)= 0$ and by (7) $(bw)(a)=0$ and so $P_n^{[1]}(a)= 0$ for all n. That is, taking $\tilde{P}_n^{[1]} :=b_1\left( P_n^{[0]}\right) '-b_1wP_0^{[0]}$, where $b_1(x)=\frac{b(x)}{x-a}$, we arrive to the exceptional system orthogonal with respect to $W=\frac{pw_0}{b_1^2}$. Thus, we can assume that $\forall$ n $P_n^{[1]}(a)\ne 0$, and if $b(a)=0$, then $(bw)(a)\ne 0$.

2.2 Exceptional Jacobi polynomials

Let the $n^{th}$ Jacobi polynomial defined by Rodrigues’ formula be:

$$\begin{aligned} (1-x)^{\alpha }(1+x)^{\beta }P_n^{\alpha ,\beta }(x)=\frac{(-1)^n}{2^nn!}\left( (1-x)^{\alpha +n}(1+x)^{\beta +n}\right) ^{(n)}, \end{aligned}$$

where $\alpha , \beta >-1$.

$$\begin{aligned}&p_k:=p_k^{\alpha ,\beta }=\frac{P_k^{\alpha ,\beta }}{\varrho _k^{\alpha ,\beta }}, \nonumber \\&\varrho _k^2:=\left( \varrho _k^{\alpha ,\beta }\right) ^2= \frac{2^{\alpha +\beta +1}\Gamma (k+\alpha +1)\Gamma (k+\beta +1)}{(2k+\alpha +\beta +1)\Gamma (k+1)\Gamma (k+\alpha +\beta +1)}. \end{aligned}$$

(10)

The orthogonal Jacobi polynomials which fulfil the following differential equation (cf. [28, (4.2.1)])

$$\begin{aligned} (1-x^2)y''+(\beta -\alpha -(\alpha +\beta +2)x)y'+n(n+\alpha +\beta +1)y=0. \end{aligned}$$

(11)

With

$$\begin{aligned} P_{n}^{[0]}=p_n^{(\alpha ,\beta )}=p_n \end{aligned}$$

(12)

and

$$\begin{aligned} P_{n}^{[1]}=AP_{n}^{[0]}=b(P_{n}^{[0]})'-bwP_{n}^{[0]}. \end{aligned}$$

(13)

In the following examples $X_m$ stands for a family of exceptional Jacobi polynomials of codimension m which are given by a certain Darboux transformation. Examples of $X_{m}$ can be found in [9]. These are as follows:

$$\begin{aligned} w_0=w^{(\alpha ,\beta )}=(1-x)^\alpha (1+x)^\beta , \end{aligned}$$

(14)

where $\alpha$ and $\beta$ are defined appropriately, see [9, Proposition 5.1].

$$\begin{aligned} T[y]=(1-x^2)y''+(\beta -\alpha -(\alpha +\beta +2)x)y'=BA-(m-\alpha )(m+\beta +1), \end{aligned}$$

where A and B are defined in (2), and

$$\begin{aligned} b(x)=(1-x)P_m^{(-\alpha ,\beta )}(x), w(x)=(\alpha -m)\frac{P_m^{(-\alpha -1,\beta -1)}(x)}{(1-x)P_m^{(-\alpha ,\beta )}(x)}. \end{aligned}$$

So the defined exceptional Jacobi polynomials, $P_n^{[1]}:=AP_n^{[0]}$, are orthogonal with respect to

$$\begin{aligned} W(x)=\frac{(1-x^2)w^{(\alpha ,\beta )}(x)}{(1-x)^2\left( P_m^{(-\alpha ,\beta )}(x)\right) ^2}=\frac{w^{(\alpha -1,\beta +1)}(x)}{\left( P_m^{(-\alpha ,\beta )}(x)\right) ^2}, \end{aligned}$$

and the space spanned by these classes is m-codimensional in the space of polynomials.

We restrict our investigations to the one-step Darboux transformation case. As it is mentioned above, b can be zero at $\pm 1$. Indeed, by [2, Table 1] the quasi-rational eigenfunctions of

$$\begin{aligned} T[y]=(1-x^2)y''+(\beta -\alpha -(\alpha +\beta +2)x)y' \end{aligned}$$

cf. (11), are $p_n^{(\alpha ,\beta )}(x)$, $(1+x)^{-\beta }p_n^{(\alpha ,-\beta )}(x)$, $(1-x)^{-\alpha }p_n^{(-\alpha ,\beta )}(x)$ and $(1+x)^{-\beta }(1-x)^{-\alpha }p_n^{(-\alpha ,-\beta )}(x)$. It gives three kinds of b(x). According to the three types of quasi-rational eigenfunctions, let

$$\begin{aligned} b_0(x)=\left\{ \begin{array}{ll}(1-x)p_n^{(-\alpha ,\beta )}(x) \text{ or }\\ (1+x)p_n^{(\alpha ,-\beta )}(x) \text{ or }\\ (1-x^2)p_n^{(-\alpha ,-\beta )}(x).\end{array}\right. \end{aligned}$$

Thus, b(x) can be taken as

$$\begin{aligned} b(x)=s(x)b_0(x), \end{aligned}$$

(15)

where s is a polynomial such that $s(x)\ne 0$ on $[-1,1]$.

Of course, $\{P_n^{[1]}\}$ is an (exceptional) closed orthogonal system with $s\equiv 1$ in $L^2_{W}$ if and only if $\{sP_n^{[1]}\}$ is orthogonal and closed in $L^2_{\frac{W}{s^2}}$. If s had a zero in $x=\pm 1$, then closedness could fail. So the examples beyond $b_0$ do not give really new classes and can be handled as simple consequences of the $b_0$-type families. Because it does not cause any difficulty, below we use b instead of using $b_0$ first and than taking extension.

Subsequently we denote by $\tilde{b}$ that part of b which appears in the weight function,

$$\begin{aligned} \frac{p}{b}=\frac{\tilde{p}}{\tilde{b}}, \end{aligned}$$

(16)

which is bounded on $[-1,1]$. Introducing the notation

$$\begin{aligned} b(x)=(1-x)^{\frac{1-\varepsilon _1}{2}}(1+x)^{\frac{1-\varepsilon _2}{2}}\tilde{b}(x), \end{aligned}$$

(17)

where $\varepsilon _i=\pm 1$, $i=1,2$, the exceptional Jacobi polynomial system, $\left\{ P_{n}^{[1]}\right\} _{n=0}^{\infty }$, is orthogonal on $(-1,1)$ with respect to

$$\begin{aligned} W=\frac{w^{(\alpha +\varepsilon _1,\beta +\varepsilon _2)}}{\tilde{b}^2}. \end{aligned}$$

(18)

The orthonormal system is denoted by $\left\{ \hat{P}_{n}\right\} _{n=0}^{\infty }.$ The codimension is given by the degree of $\tilde{b}$ cf. [5].

3 Multiplication operator

Below we investigate the location of zeros of the modified characteristic polynomial introduced in [14]. In our case, it is as follows.

Let us denote by $\{\hat{P}_n\}_{n=0}^\infty$ the orthonormal system of exceptional Jacobi polynomials, i.e.

$$\begin{aligned} \hat{P}_n:=\frac{P_n^{[1]}}{\sigma _{n}}, \text{ where } \sigma _{n}:=\Vert P_n^{[1]}\Vert _{W,2}. \end{aligned}$$

(19)

The Christoffel–Darboux kernel is

$$\begin{aligned} K_n(x,y):=\sum _{k=0}^{n-1}\hat{P}_k(x)\hat{P}_k(y). \end{aligned}$$

(20)

The joint probability density generated by an n-point ensemble has the form:

$$\begin{aligned} \rho _{N}(x_1, \dots ,x_N)=c(N)\det |K_N(x_i,x_j)|_{i,j=1}^N\prod _{i=1}^N W(x_i). \end{aligned}$$

(21)

Thus the correlation functions are

$$\begin{aligned} \rho _{N,n}(x_1, \dots ,x_n)=c(n,N)\det |K_N(x_i,x_j)|_{i,j=1}^n\prod _{i=1}^nW(x_i), \end{aligned}$$

where c(N) and c(n, N) are normalization factors. The expectation ${\mathbb {E}}$ refers to (21). The eigenvalue measure becomes a point process with determinantal correlation kernel, cf. [17, Example 2.12], i.e. it is a determinantal point process.

Now we define the modified characteristic polynomial. Let b and $\tilde{b}$ be given by (13) and (16), respectively. Recalling that b is a polynomial, let

$$\begin{aligned} Q(x):= \int ^x \tilde{b}. \end{aligned}$$

(22)

The constant term of Q is chosen to be zero.

Considering Q, the modified average characteristic polynomial is defined as follows:

$$\begin{aligned} \chi _n(z):=\chi _n^{Q}(z)={\mathbb {E}}\left( \prod _{i=1}^n(z-Q(x_i))\right) , \end{aligned}$$

(23)

cf. [14]. Denote by $z_i$ the zeros, of $\chi _n(z)$ and define the not necessarily unique pre-images of $z_i$ by $y_i$, i.e. $z_i=Q(y_i)$, $i=1, \dots , n$. The normalized counting measure based on $y_i$ is

$$\begin{aligned} \tilde{\nu }_n=\frac{1}{n}\sum _{i=1}^n\delta _{y_i} \end{aligned}$$

(24)

that is $\int Q^l(y)d\tilde{\nu }_n=\int z^ld\nu _n(z)$. Note, that since $\tilde{b}>0$ on $[-1,1]$, Q is increasing here and so if $z_i \in Q([-1,1])$, $y_i$ can be chosen from $[-1,1]$ uniquely.

The next theorem describes zero distribution of the modified average characteristic polynomial. $X_m$ stands for exceptional orthogonal polynomials generated by one Darboux transformation and of codimension m.

Theorem 1

In the exceptional Jacobi case $X_{m}$, where $m\in {\mathbb {N}}$ is arbitrary, if $\alpha +\varepsilon _1, \beta +\varepsilon _2 \ge -\frac{1}{2}$, then

$$\begin{aligned} \tilde{\nu }_n \rightarrow \mu _e \end{aligned}$$

in weak-star sense, where $\mu _e$ is the equilibrium measure of $[-1,1]$.

For this study, our main tool is the multiplication operator $M: f \rightarrow Qf$, which was introduced in [14]. It is defined as follows.

Exceptional orthogonal polynomials fulfil the next recurrence formula with constant coefficient: $QP_n^{[1]}=\sum _{k=-L}^L\tilde{u}_{n,k}P_{n+k}^{[1]}$ (see [24] and [14, (3.4)]). After normalization the recurrence relation above is modified as follows:

$$\begin{aligned} Q\hat{P}_n=\sum _{k=-L}^Lu_{n,k}\hat{P}_{n+k}, \end{aligned}$$

(25)

where $u_{n,k}=\frac{\sigma _{n+k}}{\sigma _{n}}\tilde{u}_{n,k}$, for $\sigma _n$ see (19). That is M can be represented by an infinite matrix, $M_{e}$, in the orthonormal basis $\{\hat{P}_n\}_{n=0}^\infty$. This operator, acting on $L^2_W[-1,1]$ and simultaneously on $l_2$, is denoted by $M_e$. By (25), the matrix of $M_{e}$ is $2L+1$-diagonal:

$$\begin{aligned} M_{e}=\left[ \begin{array}{cccccccc}u_{0,0}&{}u_{0,1}&{}\dots &{}\dots &{}u_{0,L}&{}0&{}0&{}\dots \\ u_{1,-1}&{}u_{1,0}&{}\dots &{}\dots &{}u_{1,L-1}&{}u_{1,L}&{}0&{}\dots \\ \vdots &{}\vdots &{}\dots &{}\ddots &{}\vdots &{}\dots &{}\vdots &{}\dots \\ u_{L,-L}&{}u_{L,-L+1}&{}\dots &{}u_{L,0}&{}\dots &{}\dots &{}u_{L,L}&{}0\\ 0&{}u_{L+1,-L}&{}\dots &{}\vdots &{}\dots &{}\vdots &{}\dots &{}u_{L+1,L}\\ \vdots &{}0&{}\dots &{}u_{L+j,-L}&{}\dots &{}\vdots &{}\dots &{}\dots \end{array}\right] . \end{aligned}$$

(26)

It can be easily seen that $M_e$ is symmetric since

$$\begin{aligned} u_{k,j}=\int _{-1}^1Q\hat{P}_k\hat{P}_{k+j}W^2=\int _{-1}^1Q\hat{P}_{k+j}\hat{P}_{(k+j)-j}W^2=u_{k+j,-j}. \end{aligned}$$

(27)

Let us recall that $\{\hat{P}_n\}_{n=0}^\infty$ is an orthonormal system on $[-1,1]$ with respect to $W=\frac{pw_0}{b^2}$, where $w_0$ is a classical Jacobi weight function. Besides this exceptional orthonormal polynomial system, there is the standard orthonormal polynomial system, $\{q_n\}_{n=0}^\infty$, on $[-1,1]$ with respect to W. These standard orthonormal polynomials fulfil the three-term recurrence relation:

$$\begin{aligned} xq_n=a_{n+1}q_{n+1}+b_nq_n+a_nq_{n-1} \end{aligned}$$

(28)

(see e.g. [28, (3.2.1)]). Since $W>0$ on $(-1,1)$, by [23, Theorem 4.5.7] (see also [26]) in formula (28) the recurrence coefficients fulfil the asymptotics

$$\begin{aligned} \lim _{n\rightarrow \infty }a_n=\frac{1}{2}\text{ and } \lim _{n\rightarrow \infty }b_n=0. \end{aligned}$$

(29)

According to (28) multiplication operator on $L^2_W[-1,1]$, $A: f \mapsto xf$ can be represented in the (Schauder) basis $\{q_n\}$ as an infinite tridiagonal matrix (denoted by A again)

$$\begin{aligned} A=\left[ \begin{array}{ccccc}b_0&{}a_1&{}0&{}0&{}\dots \\ a_1&{}b_1&{}a_2&{}0&{}\dots \\ 0&{}a_2&{}b_2&{}a_3&{}\dots \\ 0&{}0&{}a_3&{}b_3&{}\dots \\ \dots &{}\dots &{}\dots &{}\dots &{}\dots \end{array}\right] , \end{aligned}$$

and so A acts on $L^2_W[-1,1]$ and on $l^2$. Let Q(x) be as above. Then the multiplication operator, $M: f \mapsto Qf$ can be represented as $M=Q(A)$, and in the basis $\{q_n\}$, it has a matrix $M_q=Q(A)$. Let $\Pi _n^q$ and $\Pi _n^e$ be the projections to $\mathrm {span}\{q_0, \dots , q_{n-1}\}$, and to $\mathrm {span}\{\hat{P}_0, \dots , \hat{P}_{n-1}\}$, respectively. Let

$$\begin{aligned} A_{n\times n}:=\Pi _n^qA\Pi _n^q, M_{e,n\times n}:=\Pi _n^eM_e\Pi _n^e, M_{q,n\times n}:=\Pi _n^qM_q\Pi _n^q. \end{aligned}$$

To prove Theorem 1, we compare the trace of the truncated operators in the different bases above.

It is known (see [14] and the references therein) that the characteristic polynomial of the truncated multiplication matrix $M_{e,n\times n}$, coincides with the modified average characteristic polynomial, that is

$$\begin{aligned} \det (zI_n-\pi _n^eM_e\pi _n^e)={\mathbb {E}}\left( \prod _{i=1}^n(z-Q(x_i))\right) , \end{aligned}$$

(30)

where $I_n$ stands for the n-dimensional identity operator. In the standard case, the characteristic polynomial of the truncated multiplication matrix $A_{n\times n}$ coincides with the standard average characteristic polynomial (see [10, 11]) and the eigenvalues are just the zeros of the standard orthonormal polynomials, $q_n$, see [28, (2.2.10)]. Thus, we need the next statement.

Proposition 1

With the notation (18), let us assume that $\alpha +\varepsilon _1, \beta +\varepsilon _2 \ge -\frac{1}{2}$. Then for each $l\in {\mathbb {N}}$

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{1}{n}\left( \mathrm {Tr}((Q(A_{n\times n}))^l)-\mathrm {Tr}((M_{e,n\times n})^l)\right) =0. \end{aligned}$$

Proof of Proposition 1 is postponed to the next section.

Proof

(of Theorem 1) Let $\xi _i=\xi _{i,n}$, $i=1, \dots , n$ the zeros of the standard orthogonal polynomial, $q_n$. Let us recall that $\{\xi _i\}_{i=1}^n$ are the eigenvalues of $A_{n\times n}$. Since $\frac{1}{n}\mathrm {Tr}((Q(A_{n\times n}))^l)=\frac{1}{n}\sum _{i=1}^nQ^l(\xi _i)$ and by (30) $\frac{1}{n}\mathrm {Tr}((M_{e,n\times n})^l)=\frac{1}{n}\sum _{i=1}^nQ^l(y_i)$, according to Proposition 1

$$\begin{aligned} \lim _{n\rightarrow \infty }\left| \frac{1}{n}\sum _{i=1}^nQ^l(\xi _i)- \int Q^ld\tilde{\nu }_n\right| =0. \end{aligned}$$

On the other hand, (29) implies that the normalized counting measure based on the zeros of the orthogonal polynomials, $q_n$, tends to the equilibrium measure of the interval of orthogonality in weak-star sense (see e.g. [18, Proposition 1.1] and the references therein). That is

$$\begin{aligned} \lim _{n\rightarrow \infty }\left| \frac{1}{n}\sum _{i=1}^nQ^l(\xi _i)- \int _{-1}^1 Q^ld\mu _e\right| =0. \end{aligned}$$

Comparing the two limits and taking into consideration that the spectrum of the multiplication operator is the closure the range of Q on $[-1,1]$ (see e.g. [27, Sect. 150]) we have that for all $l\in {\mathbb {N}}$

$$\begin{aligned} \lim _{n\rightarrow \infty }\left( \int _{-1}^{1} Q^l d\tilde{\nu }_n-\int _{-1}^1 Q^l d\mu _{e}\right) =0. \end{aligned}$$

As it is pointed out in the proof of [14, Theorem 4.1], $\mathrm {span}\{Q^l: l\in {\mathbb {N}}\}$ is dense in $C[-1,1]$, which implies the result. $\square$

4 Proof of Proposition 1

First, we need some technical lemmas. Let $C=[c_{ij}]_{i,j=0}^\infty$ be an infinite matrix. $c_{ij}= _i C_j$ stands for the elements of C. Its $n^{th}$ principal minor matrix is $C_{n\times n}:=[c_{ij}]_{i,j=0}^{n-1}$. Its ith row is $_i C$ and jth column is $C_j$. With this notation, we can state the next lemma.

Lemma 1

Let $C=[c_{ij}]_{i,j=0}^\infty$ be a $2k+1$-diagonal, infinite matrix. Let $P=\sum _{l=0}^Mp_lx^l$ be a fixed polynomial. If the entries of C are bounded, i.e. there is a K such that $|c_{ij}|<K$ for all $0\le i,j <\infty$, then

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{1}{n}\left( \mathrm {Tr}(P(C_{n\times n}))-\mathrm {Tr}((P(C))_{n\times n})\right) =0. \end{aligned}$$

(31)

Proof

It is obvious that for each $l\in {\mathbb {N}}$ $C^l$ is $2(kl)+1$ diagonal:

Indeed, $c_{ij}=0$ if $|i-j|>k$. Assuming that $(C^l)_{ij}=:C^l_{ij}=0$ if $|i-j|>kl$ we compute $C^{l+1}_{ij}=\langle _iC^l, C_j\rangle =\sum _{\begin{array}{c} {|i-p|\le kl} {|p-j|\le k} \end{array}} C^l_{ip}C_{pj}$. That is $C^{l+1}_{ij}=0$ $|i-j|>(l+1)k$.

By induction on l, we show that the elements of the principal diagonal are coincided in $C_{n\times n}^l$ and $(C^l)_{n\times n}$ except at most the last $k(l-1)$ ones.

Let us assume that $(C_{n\times n}^l)_{ij}=((C^l)_{n\times n})_{ij}$ if $i\le n-(l-1)k$ or $j\le n-(l-1)k$. (For $l=1$, it is obviously fulfilled.) Let $i\le n-kl$. By the assumption and by $2kl+1$-diagonality $_i((C_{n\times n})^l)=_i(C^l)$, and similarly if $j\le n-k$, then $(C_{n\times n})_j=C_j$. Thus, $((C_{n\times n})^{l+1})_{ij}=C^{l+1}_{ij}$ if $i\le n-kl$ and $j\le n-k$. If $i\le n-kl$, then $C^l_{ip}=0$ if $p>n$ and if $j>n-k$ then $C_j-(C_{n\times n})_j$ starts with n zeros; thus, if $i\le n-kl$ and $j> n-k$, $_i((C_{n\times n})^{l+1})_{j}=C^{l+1})_{ij}$ again. For $j\le n-kl$, we have the same by symmetry.

Finally considering that the rows and columns contain finitely many non-zero elements, these imply that $\frac{1}{n}\left( \mathrm {Tr}((C_{n\times n})^l)-\mathrm {Tr}((C^l)_{n\times n})\right) \le \frac{1}{n}k(l-1)(2kl+1)^lK^l$, that is

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{1}{n}\left( \mathrm {Tr}((C_{n\times n})^l)-\mathrm {Tr}(((C^l))_{n\times n})\right) =0. \end{aligned}$$

(32)

Since $\mathrm {Tr}(P(C_{n\times n}))=\sum _{l=0}^Mp_l\mathrm {Tr}((C_{n\times n})^l)$ and $\mathrm {Tr}((P(C))_{n\times n})=\sum _{l=0}^Mp_l\mathrm {Tr}((C^l)_{n\times n})$, (32) implies (31). $\square$

To ensure the boundedness of the entries of the matrices in question, we recall the asymptotic behaviour of recurrence coefficients. First, we compute the norm of $P_n^{[1]}$.

Lemma 2

With the notation (19), if $\alpha +\frac{\varepsilon _1}{2}, \beta +\frac{\varepsilon _2}{2}> -\frac{1}{2}$,

$$\begin{aligned} \sigma _k=\sqrt{k(k+\alpha +\beta +1)+\tilde{\lambda }}. \end{aligned}$$

(33)

Proof

With the notation $p_k^{(\alpha ,\beta )}=p_k$, cf. (12)

$$\begin{aligned} \sigma _k^2\hat{P}_k^2W= & {} (bp_k'-bwp_k)^2\frac{pw^{(\alpha ,\beta )}}{b^2} \\ \quad= & {} (p_k')^2w^{(\alpha +1,\beta +1)}+ (bw)^2p_k^2\frac{pw^{(\alpha ,\beta )}}{b^2}-2b^2wp_k'p_k\frac{pw^{(\alpha ,\beta )}}{b^2}. \end{aligned}$$

By (2) $pw^2=\tilde{\lambda }-qw-pw'$ and considering (11) $(w^{(\alpha +1,\beta +1)})'=qw^{(\alpha ,\beta )}$

$$\begin{aligned} \sigma _k^2\hat{P}_k^2W=(p_k')^2w^{(\alpha +1,\beta +1)}+\tilde{\lambda }p_k^2w^{(\alpha ,\beta )}-\left( p_k^2ww^{(\alpha +1,\beta +1)}\right) '. \end{aligned}$$

Thus,

$$\begin{aligned} \sigma _k^2=\int _{-1}^1(p_k')^2w^{(\alpha +1,\beta +1)}+\tilde{\lambda }p_k^2w^{(\alpha ,\beta )}-\left( p_k^2ww^{(\alpha +1,\beta +1)}\right) '=I_1+I_2+I_3. \end{aligned}$$

According to [28, (4.21.7)]

$$\begin{aligned}&p_k'=\frac{k+\alpha +\beta +1}{2}p_{k-1}^{(\alpha +1,\beta +1)}\frac{\varrho _{k-1}^{\alpha +1,\beta +1}}{\varrho _{k}^{\alpha ,\beta }}\\&I_1=\left( \frac{k+\alpha +\beta +1}{2}\frac{\varrho _{k-1}^{\alpha +1,\beta +1}}{\varrho _{k}^{\alpha ,\beta }}\right) ^2. \end{aligned}$$

$I_2=\tilde{\lambda }$. By the assumption on $\alpha$ and $\beta$, one can see that $I_3=0$. Substituting the values of the corresponding $\varrho _k^{\alpha ,\beta }$, (33) is proved. $\square$

Analogous to (29), the coefficients in (25) fulfil a symmetric limit relation

$$\begin{aligned} \lim _{n\rightarrow \infty }u_{n,j}=:U_{|j|}, \end{aligned}$$

(34)

where $U_{|j|}$ depends on the polynomial $\tilde{b}$ (cf. (2), (16)) as follows. Let

$$\begin{aligned} \tilde{b}(x)=\sum _{k=0}^{L-1} d_kx^k. \end{aligned}$$

(35)

Since Lemma 2 ensures that the asymptotics of $\tilde{u}_{n,j}$ and $u_{n,j}$ coincide, according to [14, Proposition 1]

$$\begin{aligned} U_{|j|}=\left\{ \begin{array}{ll}\sum _{p=\max \{l,1\}}^{\left[ \frac{L}{2}\right] }\frac{d_{2p-1}}{2p}\left( {\begin{array}{c}2p\\ p-l\end{array}}\right) \frac{1}{2^{2p}}, \text{ if }|j|=2l\\ \sum _{p=l}^{\left[ \frac{L-1}{2}\right] }\frac{d_{2p}}{2p+1}\left( {\begin{array}{c}2p+1\\ p-l\end{array}}\right) \frac{1}{2^{2p+1}}, \text{ if }|j|=2l+1. \end{array}\right. \end{aligned}$$

(36)

Let us recall that $\frac{p}{b}$ is bounded on $[-1,1]$ and $W(x)=\frac{(1-x)^{\alpha +\varepsilon _1}(1+x)^{\beta +\varepsilon _2}}{\tilde{b}^2}$ ($\varepsilon _i=\pm 1$), where $b(x)=\tilde{b}(x)(1-x)^{\frac{1-\varepsilon _1}{2}}(1+x)^{\frac{1-\varepsilon _2}{2}}$ and $0<c<\tilde{b}<C$ on $[-1,1]$.

Subsequently, the next lemma proved by Badkov (see [1]) is useful. Here, we cite the formulation given in [22].

Lemma A

[22, Lemma 2.E] Let $\{q_k\}_{k=0}^\infty$ be the standard orthonormal system with respect to W. For each $j\ge 0$ integer

$$\begin{aligned} \left| q_k^{(j)}(x)\right|&\le c \left( \frac{k}{\sqrt{1-x^2}+\frac{1}{k}}\right) ^j\nonumber \\&\frac{1}{\left( \sqrt{1-x}+\frac{1}{k}\right) ^{\alpha +\varepsilon _1}\left( \sqrt{1+x}+\frac{1}{k}\right) ^{\beta +\varepsilon _2}\sqrt{\sqrt{1-x^2}+\frac{1}{k}}}. \end{aligned}$$

(37)

We need the following estimations on norms of exceptional Jacobi polynomials.

Lemma 3

If $\alpha +\varepsilon _1, \beta +\varepsilon _2\ge -\frac{1}{2}$, and $0\le \delta \le \min \left\{ \frac{1}{4},\frac{\alpha +\varepsilon _1}{2}+\frac{1}{4}, \frac{\beta +\varepsilon _2}{2}+\frac{1}{4}\right\}$, then

$$\begin{aligned} \left\| \hat{P}_k(x)\sqrt{W(x)}(1-x^2)^{\frac{1}{4}-\delta }\right\| _\infty \le c k^{\max \{-\varepsilon _1,-\varepsilon _2\}-1+2\delta }, \end{aligned}$$

(38)

where c is a constant (independent of k).

Proof

We have

$$\begin{aligned}&\left| \hat{P}_k(x)\sqrt{W(x)}(1-x^2)^{\frac{1}{4}}\right| \\&\quad \le \frac{c}{\sigma _k}\left( |b(x)p_k'(x)|+|(bw)(x)p_k(x)|\right) (1-x)^{\frac{\alpha +\varepsilon _1}{2}+\frac{1}{4}}(1+x)^{\frac{\beta +\varepsilon _2}{2}+\frac{1}{4}}\\&\quad =K_1+K_2. \\&K_1\le \left| p_{k-1}^{(\alpha +1,\beta +1)}\right| (1-x)^{\frac{\alpha }{2}+\frac{3}{4}}(1+x)^{\frac{\beta }{2}+\frac{3}{4}}, \end{aligned}$$

which is bounded, see e.g. [28, (8.21.10], or (37). Since bw is bounded on $[-1,1]$ (it is a polynomial), by (37)

$$\begin{aligned}&K_2\le \frac{c}{k}|p_k(x)|(1-x)^{\frac{\alpha +\varepsilon _1}{2}+\frac{1}{4}-\delta }(1+x)^{\frac{\beta +\varepsilon _2}{2}+\frac{1}{4}-\delta } \\&\quad \le \frac{c}{k}\frac{1}{(\sqrt{1-x}+\frac{1}{k})^\alpha (\sqrt{1+x}+\frac{1}{k})^\beta \sqrt{\sqrt{1-x^2}+\frac{1}{k}}} \\&\quad \le \frac{c\sqrt{k}}{k}\frac{1}{(\sqrt{1-x}+\frac{1}{k})^{-\varepsilon _1-\frac{1}{2}+2\delta }(\sqrt{1+x}+\frac{1}{k})^{-\varepsilon _2-\frac{1}{2}+2\delta }}\le c k^{\max \{-\varepsilon _1,-\varepsilon _2\}-1+2\delta }. \end{aligned}$$

To prove Proposition 1, we introduce the operator O which changes the basis in the Hilbert space in question:

$$\begin{aligned} O^{-1}M_qO=M_e, \end{aligned}$$

(39)

where

$$\begin{aligned} O=[o_{ij}]_{i,j=0}^\infty , \end{aligned}$$

and

$$\begin{aligned} \hat{P}_j=\sum _{i=0}^{j+m}o_{ij}q_i. \end{aligned}$$

(40)

$\square$

Proof

(of Proposition 1) Notice that Lemma 1 can be applied to A and $M_e$. Indeed, A is tridiagonal, and $M_e$ is $2L+1$-diagonal. Furthermore, the entries of A and $M_e$ are bounded (see (29) and (34) respectively). Thus it is enough to show that

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{1}{n}\left( \mathrm {Tr}(\pi _n^qQ^l(A)\pi _n^q)-\mathrm {Tr}(\pi _n^e(M_e)^l\pi _n^e)\right) =0. \end{aligned}$$

(41)

Considering (39), it is enough to prove that

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{1}{n}\left( \mathrm {Tr}(\pi _n^qQ^l(A)\pi _n^q)-\mathrm {Tr}(\pi _n^eO^{-1}Q^l(A)O\pi _n^e)\right) =0. \end{aligned}$$

(42)

To prove (42), we compute the diagonal elements of $\pi _n^eO^{-1}Q^l(A)O\pi _n^e$. According to (40), for $0\le i\le n-1$

$$\begin{aligned} _i(O^{-1}Q^l(A)O)_i=\sum _{k=\max \{0,i-L\}}^{\infty }\ _i(O^{-1}Q^l(A))_k \ _kO_i \end{aligned}$$

and considering the $2lL+1$-diagonality of $Q^l(A)$

$$\begin{aligned} _i(O^{-1}Q^l(A))_k= & {} \sum _{j=0}^{i+L}\ _iO^{-1}_j \ _j(Q^l(A))_k\\= & {} \sum _{\max \{0,k-lL\}\le j \le \min \{i+L, k+lL\}} \ _jO_i \ _j(Q^l(A))_k. \end{aligned}$$

Thus,

$$\begin{aligned}&\sum _{i=0}^{n-1} \ _i(O^{-1}Q^l(A)O)_i=\sum _{i=0}^{n-1}\sum _{k=0}^{i+L}\sum _{\max \{0,k-lL\}\le j \le \min \{i+L, k+lL\}} o_{ji} o_{ki} \ _j(Q^l(A))_k \nonumber \\&\quad =\sum _{i=0}^{n-1}\sum _{k=0}^{i+L}o_{ki}^2\ _k(Q^l(A))_k + \sum _{i=0}^{n-1}\sum _{k=0}^{i+L}\sum _{\begin{array}{c} {\max \{0,k-lL\}\le j \le \min \{i+L, k+lL\}} \\ {j\ne k} \end{array}} o_{ji} o_{ki} \ _j(Q^l(A))_k\nonumber \\&\quad =\tilde{M}_n+H_n. \end{aligned}$$

(43)

Reversing the order of summation, we get

$$\begin{aligned} \tilde{M}_n= & {} \sum _{k=0}^L\sum _{i=0}^{n-1}o_{ki}^2\ _k(Q^l(A))_k+\sum _{k=L+1}^{n-1}\ _k(Q^l(A))_k\sum _{i=k-L}^{n-1}o_{ki}^2\\&+\sum _{k=n}^{n-1+L}\ _k(Q^l(A))_k\sum _{i=k-L}^{n-1}o_{ki}^2 \\= & {} E_{1,n}+M_n+E_{2,n}. \end{aligned}$$

Thus, the expression after the limit in (42) becomes

$$\begin{aligned}&\frac{1}{n}\sum _{k=0}^{n-1} \left( \ _k(Q^l(A))_k - \ _k(O^{-1}Q^l(A)O)_k\right) \nonumber \\&\quad =\frac{1}{n}\left( \sum _{k=0}^L \ _k(Q^l(A))_k +E_{1,n}+E_{2,n}\right) \nonumber \\&\quad \quad +\frac{1}{n}\sum _{k=L+1}^{n-1} \ _k(Q^l(A))_k\left( 1-\sum _{i=k-L}^{n-1}o_{ki}^2\right) + \frac{1}{n}H_n. \end{aligned}$$

(44)

According to (36), the entries $_i(Q^l(A))_k$ are bounded, and by orthonormality, we have

$$\begin{aligned} \sum _{i=k-L}^{n-1}o_{ki}^2=1. \end{aligned}$$

(45)

Thus,

$$\begin{aligned} \frac{1}{n}\left( \sum _{k=0}^L \ _k(Q^l(A))_k +E_{1,n}+E_{2,n}\right) \le \frac{K}{n}, \end{aligned}$$

where K depends on L and on the supremum of $|_i(Q^l(A))_k|$, but it is independent of n, that is the first term tends to zero, when n tends to infinity.

We can start the estimation of the second term of (44) in the same way. By the boundedness of the entries of $Q^l(A)$, and by (45), we have

$$\begin{aligned} \frac{1}{n}\sum _{k=L+1}^{n-1} \ _k(Q^l(A))_k\left( 1-\sum _{i=k-L}^{n-1}o_{ki}^2\right) \le C\frac{1}{n}\sum _{k=0}^{n-1}\sum _{i=n}^\infty o_{ki}^2. \end{aligned}$$

(46)

Let $i>k$. Referring to (40), we have

$$\begin{aligned} o_{ki}=\int _{-1}^1\hat{P}_iq_kW=\frac{1}{\lambda _i}\int _{-1}^1\left( p\hat{P}''_i+ \hat{q}\hat{P}'_i+\hat{r}\hat{P}_i\right) q_kW, \end{aligned}$$

(47)

where the differential equation of $\hat{P}_i$ was taken into consideration, cf. (4) and (6). By the assumption on $\alpha , \beta$ and (16), integrating by parts the first term, we have

$$\begin{aligned} \int _{-1}^1p\hat{P}''_iq_kW=-\int _{-1}^1\hat{P}'_i(pq_kW)'=-\int _{-1}^1\hat{P}'_i(\hat{q}q_k+pq'_k)W, \end{aligned}$$

(48)

because $pW'=(\hat{q}-p')W$, cf. (5).

So (47) and (48) imply that

$$\begin{aligned} o_{ki}=\frac{1}{\lambda _i}\int _{-1}^1 -p\hat{P}'_iq'_kW+\hat{r}\hat{P}_iq_kW=I_{ki}+II_{ki}. \end{aligned}$$

(49)

Considering (9), we get

$$\begin{aligned} p\hat{P}'_i=\frac{\lambda _i-\tilde{\lambda }}{\sigma _i}bp_i-\left( pw+q-p\frac{b'}{b}\right) \hat{P}_i. \end{aligned}$$

(50)

By (50), we have

$$\begin{aligned} I_{ki}= & {} -\frac{\lambda _i-\tilde{\lambda }}{\sigma _i \lambda _i}\int _{-1}^1 bp_iq'_kW+\frac{1}{\lambda _i}\int _{-1}^1\left( pw+q-p\frac{b'}{b}\right) \hat{P}_iq'_kW\nonumber \\= & {} I_{ki}^{(1)}+I_{ki}^{(2)}. \end{aligned}$$

(51)

By (16) $\left| pw+q-p\frac{b'}{b}\right|$ is bounded on $[-1,1]$. Indeed, only the first term needs some investigation; recalling that bw is a polynomial, notice that $pw=\frac{p}{b}bw$.

Thus, according to (37) and (38)

$$\begin{aligned}&|I_{ki}^{(2)}|\le \frac{ck}{i^2}\left\| \hat{P}_k(x)\sqrt{W(x)}(1-x^2)^{\frac{1}{4}}\right\| _\infty \\&\qquad \quad \times \int _{-1}^1\frac{\sqrt{1-x}^{\alpha +\varepsilon _1}\sqrt{1+x}^{\beta +\varepsilon _2}}{(1-x^2)^{\frac{1}{4}}\left( \sqrt{1-x^2}+\frac{1}{k}\right) ^\frac{3}{2}\left( \sqrt{1-x}+\frac{1}{k}\right) ^{\alpha +\varepsilon _1}\left( \sqrt{1+x}+\frac{1}{k}\right) ^{\beta +\varepsilon _2}}dx \\&\quad \le c\frac{k}{i^2}J_k. \\&J_k\le c \int _{-1}^0\frac{\left( \sqrt{1+x}\right) ^{\beta +\varepsilon _2+\frac{3}{2}-2\delta }}{\left( \sqrt{1+x}+\frac{1}{k}\right) ^{\beta +\varepsilon _2+\frac{3}{2}-2\delta +2\delta }(1+x)^{1-\delta }} \\&\qquad \quad +c\int _0^1\frac{\left( \sqrt{1-x}\right) ^{\alpha +\varepsilon _1+\frac{3}{2}-2\delta }}{\left( \sqrt{1-x}+\frac{1}{k}\right) ^{\alpha +\varepsilon _1+\frac{3}{2}-2\delta +2\delta }(1-x)^{1-\delta }}\le c \frac{k^{2\delta }}{\delta }, \end{aligned}$$

where the last inequality fulfils if $\alpha +\varepsilon _1+\frac{3}{2}-2\delta \ge 0$ and $\beta +\varepsilon _2+\frac{3}{2}-2\delta \ge 0$. Let $\delta =ck^{-\frac{1}{4}}$. Then,

$$\begin{aligned} |I_{ki}^{(2)}|\le c\frac{k^{\frac{5}{4}}}{i^2}, \end{aligned}$$

(52)

provided that $\alpha +\varepsilon _1,\beta +\varepsilon _2> - \frac{3}{2}$. For sake of simplicity, let us denote by $c(i):=-\frac{\lambda _i-\tilde{\lambda }}{\sigma _i \lambda _i}$. After simplification, we arrive at

$$\begin{aligned} I_{ki}^{(1)}= c(i)\int _{-1}^1\frac{\tilde{p}}{\tilde{b}}q'_kp_iw^{(\alpha ,\beta )}. \end{aligned}$$

Recalling that $\frac{1}{\tilde{b}}$ is bounded on $[-1,1]$, let $S_{i-k-2}$ be its uniformly best approximating polynomial on $[-1,1]$ of degree $i-k-2$. Then, as its degree is less than i, it follows that $S_{i-k-2}\tilde{p}q'_k$ is orthogonal to $p_i$ with respect to $w^{(\alpha ,\beta )}$. Thus,

$$\begin{aligned} I_{ki}^{(1)}= c(i)\int _{-1}^1\left( \frac{1}{\tilde{b}}-S_{i-k-2}\right) \tilde{p}q'_kp_iw^{(\alpha ,\beta )}. \end{aligned}$$

Taking into account, (11), (33) and (2), we get $c(i) \le c \frac{1}{i}$. Since $\left( \frac{1}{\tilde{b}}\right) '$ is bounded on $[-1,1]$ too, according to the classical Jackson’s theorem, we have

$$\begin{aligned} |I_{ki}^{(1)}|&\le \frac{c}{i(i-k)}\int _{-1}^1\tilde{p}|q'_k||p_i|w^{(\alpha ,\beta )} \\&\le \frac{c}{i(i-k)}\left\| q'_k(x)\sqrt{W(x)}(1-x^2)^{\frac{3}{4}}\right\| _\infty \int _{-1}^1|p_i(x)|\sqrt{W(x)}\frac{b(x)}{(1-x^2)^{\frac{3}{4}}}dx \\&\le c\frac{k}{i(i-k)}\int _{-1}^1|p_i(x)|(1-x)^{\frac{\alpha }{2}-\frac{1}{4}}(1+x)^{\frac{\beta }{2}-\frac{1}{4}}, \end{aligned}$$

where the norm of $q'_k$ is estimated by (37). Finally by [28, (7.34.1)], the last integral can be estimated by a constant independent of i, that is

$$\begin{aligned} |I_{ki}^{(1)}|\le c\frac{k}{i(i-k)}. \end{aligned}$$

(53)

To estimate $II_{ki}$, first we remark that $\hat{r}$ is bounded on $[-1,1]$. Indeed, it is clear that we have to deal with only the endpoints. Recalling that $\frac{p}{b}$ is bounded on $[-1,1]$ and considering (5), we see that $\hat{q}$ is bounded there too. As it is mentioned in the Remark of section 2, $\hat{P}_n(1)\ne 0$ that is due to (4) and (6) $\hat{r}$ must be bounded at 1. By the same chain of ideas, we can show that $\hat{r}$ is bounded at $-1$. Thus, by the Cauchy-Schwarz inequality

$$\begin{aligned} |II_{ki}|\le \frac{c}{i^2}\int _{-1}^1|\hat{P}_i|\sqrt{W}|q_k|\sqrt{W}\le \frac{c}{i^2}. \end{aligned}$$

(54)

Since $\sum _{i=n}^\infty o_{ki}^2\le 1$, according to (52), (53) and (54), we have

$$\begin{aligned}&\frac{1}{n}\sum _{k=0}^{n-1}\sum _{i=n}^\infty o_{ki}^2 \le c \frac{1}{n}\sum _{k=n^{\frac{3}{4}}}^{n-n^{\frac{3}{4}}}\sum _{i=n}^\infty o_{ki}^2+ O\left( \frac{1}{n^{\frac{1}{4}}}\right) \nonumber \\&\quad \le \frac{c}{n}\sum _{k=n^{\frac{3}{4}}}^{n-n^{\frac{3}{4}}}\sum _{i=n}^\infty \left( \frac{k^{\frac{5}{2}}}{i^4}+ \frac{k^2}{i^2n^{\frac{3}{2}}}\right) +O\left( \frac{1}{n^{\frac{1}{4}}}\right) =O\left( \frac{1}{\sqrt{n}}\right) + O\left( \frac{1}{n^{\frac{1}{4}}}\right) , \end{aligned}$$

(55)

so the expression in (46) tends to zero.

Finally, we estimate the error term $\frac{1}{n}H_n$, cf. (43) and (44). Using the properties of $|_j(Q^l(A))_k |$ and $o_{ki}$ again, it can be estimated as follows:

$$\begin{aligned}&\frac{1}{n}|H_n|\le c\frac{1}{n}\sum _{i=0}^{(l-1)L}\sum _{k=0}^{i+L}\sum _{\begin{array}{c} {\max \{0,k-lL\}\le j \le i+L} \\ {j\ne k} \end{array}} |o_{ji}|| o_{ki}| \\&\quad +\frac{1}{n}\left| \sum _{i=(l-1)L+1}^{n-1}\sum _{k=0}^{i+L}\sum _{\begin{array}{c} {\max \{0,k-lL\}\le j \le k+lL} \\ {j\ne k} \end{array}} o_{ji} o_{ki} \ _j(Q^l(A))_k \right| =\Sigma _1(n)+\Sigma _2(n). \end{aligned}$$

Certainly, $\lim _{n\rightarrow \infty }\Sigma _1(n)=0$. Splitting $\Sigma _2(n)$ to two parts and changing the ordering of summation, we get

$$\begin{aligned} \Sigma _2(n)&\le \frac{1}{n}\left| \sum _{i=(l-1)L+1}^{n-1}\sum _{k=0}^{lL-1}\sum _{\begin{array}{c} {0\le j \le k+lL} \\ {j\ne k} \end{array}} o_{ji} o_{ki} \ _j(Q^l(A))_k \right| \\&\quad +\frac{1}{n}\left| \sum _{i=(l-1)L+1}^{n-1}\sum _{k=lL}^{i+L}\sum _{\begin{array}{c} {k-lL\le j \le k+lL} \\ {j\ne k} \end{array}} o_{ji} o_{ki} \ _j(Q^l(A))_k \right| \\&\le \frac{1}{n}\sum _{k=0}^{lL-1}\sum _{\begin{array}{c} {0\le j \le k+lL} \\ {j\ne k} \end{array}}\left| _j(Q^l(A))_k\right| \sum _{i=(l-1)L+1}^{n-1}|o_{ji} o_{ki}| \\&\quad +\left| \frac{1}{n}\sum _{k=lL}^{n-1+L}\sum _{\begin{array}{c} {k-L\le j \le k+lL} \\ {j\ne k} \end{array}} \ _j(Q^l(A))_k\sum _{i=(l-1)L+1}^{n-1}o_{ji} o_{ki}\right| \\&=\Sigma _{21}(n)+\Sigma _{22}(n). \end{aligned}$$

Again, $\lim _{n\rightarrow \infty }\Sigma _{21}(n)=0$. According to (40) $o_{ki}=0$ if $i\le k-L$. Thus, by orthogonality $\sum _{i=(l-1)L+1}^{n-1}o_{ji} o_{ki}=\sum _{i=n}^{\infty }o_{ji} o_{ki}$. That is

$$\begin{aligned} \Sigma _{22}(n)\le c\frac{1}{n}\sum _{k=lL}^{n-1+L}\sum _{i=n}^{\infty }|o_{ji} o_{ki}|\le c\frac{1}{n}\sum _{k=lL}^{n-1+L}\max _{k-L\le j \le k+lL}\sum _{i=n}^{\infty }o_{ji}^2. \end{aligned}$$

We can proceed as in (55) again, and so $\lim _{n\rightarrow \infty }\Sigma _{22}(n)=0$, which finishes the estimation of $\frac{1}{n}|H_n|$ and the proof of Proposition 1 as well. $\square$

5 Certain self-inversive polynomials

In this section, we use the multiplication operator and the infinite matrix $M_e$ introduced in Sect. 3 to investigate certain self-inversive polynomials.

Let b be defined by (15), and $\tilde{b}$ by (16). Investigation of Sect. 3 was independent of the constant term of the polynomial $Q=\int ^x \tilde{b}$. At first, as above, take $Q_0(x):= \int ^x \tilde{b}$, with zero constant term. Recalling (25), we have $Q_0\hat{P}_n=\sum _{k=-L}^Lu_{n,k}\hat{P}_{n+k}$, and by (34) we get $\lim _{n\rightarrow \infty }u_{n,0}=U_0$, where $U_0$ depends on $\tilde{b}$, cf. (35). Rearranging the equation above, we have $(Q_0-U_0)\hat{P}_n=\sum _{\begin{array}{c} {-L\le k\le L} \\ {k\ne 0} \end{array}}u_{n,k}\hat{P}_{n+k}+ (u_{n,0}-U_0)\hat{P}_n$; thus, in this section, we define

$$\begin{aligned} Q(x):=\int ^x \tilde{b}-U_0, \end{aligned}$$

where $\int ^x \tilde{b}$ means the primitive function without any constant term. The operator $M_e$ refers to this Q, that is, denoting by $M^0_e$ the operator defined by $Q_0$, $M_e=M^0_e-U_0I$. Since the asymptotics of the elements of $M_e$ can be described by (36) $M_e$ can be decomposed to a bounded symmetric and a compact symmetric part again (cf. (27)), that is

$$\begin{aligned} M_{e}=M_{e,s}+M_{e,c}, \end{aligned}$$

(56)

where

$$\begin{aligned} M_{e}=\left[ \begin{array}{cccccccc}u_{0,0}-U_0&{}u_{0,1}&{}\dots &{}\dots &{}u_{0,L}&{}0&{}0&{}\dots \\ u_{1,-1}&{}u_{1,0}-U_0&{}\dots &{}\dots &{}u_{1,L-1}&{}u_{1,L}&{}0&{}\dots \\ \vdots &{}\vdots &{}\dots &{}\ddots &{}\vdots &{}\dots &{}\vdots &{}\dots \\ u_{L,-L}&{}u_{L,-L+1}&{}\dots &{}u_{L,0}-U_0&{}\dots &{}\dots &{}u_{L,L}&{}0\\ 0&{}u_{L+1,-L}&{}\dots &{}\vdots &{}\dots &{}\vdots &{}\dots &{}u_{L+1,L}\\ \vdots &{}0&{}\dots &{}u_{L+j,-L}&{} \dots &{}\vdots &{}\dots &{}\dots \end{array}\right] \end{aligned}$$

and

$$\begin{aligned} M_{e,s}=\left[ \begin{array}{ccccccc}0&{}U_1&{}\dots &{}U_L&{}0&{}0&{}\dots \\ U_1&{}0&{}\dots &{}U_{L-1}&{}U_L&{}\dots \\ \vdots &{}\vdots &{}\ddots &{}\vdots &{}\dots &{}\vdots &{}\dots \\ U_{L}&{}\dots &{}0&{}\dots &{}\dots &{}U_{L}&{}\dots \\ 0&{}U_{L}&{}\vdots &{}\dots &{}\vdots &{}\dots &{}U_{L}\\ \vdots &{}\dots &{}U_{L}&{} \dots &{}\vdots &{}\dots &{}\dots \end{array}\right] . \end{aligned}$$

(57)

Investigation of $M_{e,s}$ leads to the so-called self-inversive or palindrome polynomials, cf. [21, Lemma 12]. A polynomial P of degree n with real coefficients is self-inversive if $z^nP\left( \frac{1}{z}\right) =P(z)$. The location of zeros of self-inversive polynomials has been extensively studied, see e.g. [20, 29], etc. Of course, the zeros of a self-inversive polynomial are symmetric with respect to the unit circle. One of the statements on the location of zeros is the next one (see eg. [29] ): if $P_{2m}(z)=\sum _{k=0}^{2m}a_kz^k$ is self-inversive and $|a_m|>\sum _{\begin{array}{c} {0\le k \le m} \\ {k\ne m} \end{array}} |a_k|$, then $P_{2m}$ has no zeros on the unite circle. In our special case we get something similar.

Let us recall that $\tilde{b}(x)=\sum _{k=0}^md_kx^k$, cf. (35). Now define

$$\begin{aligned} P_{2L,\lambda }(z)=\sum _{k=1}^LU_k\left( z^{L+k}+z^{L-k}\right) -\lambda z^L, \end{aligned}$$

where $U_k$ depends on b see (36), and

$$\begin{aligned} \tilde{P}_{2L}(z)=P_{2L,\lambda }(z)+\lambda z^L. \end{aligned}$$

Theorem 2

$P_{2L,\lambda }(z)$ has no zeros on the unit circle if and only if

$$\begin{aligned} \lambda \not \in \left[ 2\sum _{k=1}^L(-1)^kU_k,2\sum _{k=1}^LU_k\right] =\left[ (-1)^L\tilde{P}_{2L}(-1),(-1)^L\tilde{P}_{2L}(1)\right] . \end{aligned}$$

Let us consider $M_{e,s}$ as an operator on $l^2$ and on the Hardy space

$$\begin{aligned}&H^2:=\left\{ f(z)=\sum _{k=0}^\infty c_kz^k : \text{ is } \text{ holomorphic } \text{ on } |z|<1,\right. \\&\quad \left. \lim _{r\rightarrow 1} f(re^{i\varphi })=f(e^{i\varphi }) \text{ a.e. } \varphi \in (0,2\pi ]\right\} . \end{aligned}$$

It is a Hilbert space under the norm $\Vert f\Vert ^2=\sum _{k=0}^\infty |c_k|^2=\frac{1}{2\pi }\int _0^{2\pi }|f(e^{i\varphi })|^2d\varphi$.

Lemma 4

$M_{e,s}-\lambda I$ has a bounded inverse if and only if $P_{2L,\lambda }(z)$ has no zeros on the unit circle.

Proof

With this interpretation if $f \in H^2$,

$$\begin{aligned} (M_{e,s}f)(z)=\sum _{k=1}^LU_k(z^k+z^{-k})f(z)-\sum _{k=1}^LU_k\sum _{j=0}^{k-1}\frac{f^{(j)}(0)}{j!}z^{j-k}. \end{aligned}$$

Let $g\in H^2$ be arbitrary. Then considering the equation

$$\begin{aligned} (M_{e,s}-\lambda I)f=g \end{aligned}$$

we have

$$\begin{aligned} f(z)&=\frac{z^Lg(z)+\sum _{k=1}^LU_k\sum _{j=0}^{k-1}\frac{f^{(j)}(0)}{j!}z^{L+j-k}}{P_{2L,\lambda }(z)} \nonumber \\&=\frac{z^Lg(z)+\sum _{j=0}^{L-1}\frac{f^{(j)}(0)}{j!}\sum _{k=j}^{L-1}U_{L-k+j}z^k}{P_{2L, \lambda }(z)}. \end{aligned}$$

(58)

By symmetry, counting with multiplicity, $P_{2L,\lambda }$ has l zeros in the unit disc and $2(L-l)$ on the unit circle. We can determine $f\in H^2$ if and only if these zeros can be compensated. That is if $\xi _m$ are the zeros of the denominator (in the closed unit disc) of multiplicity $k_m$ with $\sum _m k_m=2L-l$, then it means a system of linear equations in the numerator. According to (58) this system looks like

$$\begin{aligned} \left[ \begin{array}{cccc}C_0(\xi _1)&{}C_1(\xi _1)&{}\dots &{}C_{L-1}(\xi _1)\\ \vdots &{}\vdots &{}\dots &{}\vdots \\ C_0(\xi _2)&{}C_1(\xi _2)&{}\dots &{}C_{L-1}(\xi _2)\\ C'_0(\xi _2)&{}C'_1(\xi _2)&{}\dots &{}C'_{L-1}(\xi _2)\\ \vdots &{}\vdots &{}\dots &{}\vdots \\ C_0^{(k_2-1)}(\xi _2)&{}C_1^{(k_2-1)}(\xi _2)&{}\dots &{}C_{L-1}^{(k_2-1)}(\xi _2)\\ \vdots &{}\vdots &{}\dots &{}\vdots \\ c_{2L-l,0}&{}\dots &{}\dots &{}c_{2L-1,L-1}\end{array}\right] \left[ \begin{array}{c}f(0)\\ f'(0)\\ f''(0)\\ \vdots \\ \vdots \\ f^{(L-1)}(0)\end{array}\right] =\left[ \begin{array}{c}h(\xi 1)\\ \vdots \\ h(\xi _2)\\ h'(\xi _2)\\ \vdots \\ h^{(k_2-1)}(\xi _2) \\ \vdots \end{array}\right] , \end{aligned}$$

where $C_j(x)=\frac{1}{j!}\sum _{k=j}^{L-1}U_{L-k+j}x^k$, $j=0, \dots ,L-1$; $h(x)=-x^Lg(x)$, the coefficient matrix is of $2L-l \times L$. We deal with the rank of the coefficient matrix. Let us denote by $O_j$, the jth column of the coefficient matrix $(j = 0; 1;\ldots ;L - 1)$. The first step is

$$\begin{aligned} O_{j} - {{\frac{(L-1)!}{j!}}{\frac{U_{j+1}}{U_{L}}}}{O_{L-1}}, \quad {\text {for}}\quad j=0, 1, \dots ,L-2. \end{aligned}$$

After the first step, the sums in the entries of the new columns $O_{j}^{(1)}$, are shorter than they were in $O_{j}$ for $j=0, 1, \dots ,L-2$. Repeating this procedure we finally arrive at a coefficient matrix which inherits the structure of the original one. Its entries can be calculated by the new functions $\tilde{C}_j(x)={a_{j}}{x^j}$.

Thus, the rank of the coefficient matrix coincides with its less size. That is to get a unique solution, it must be of $L\times L$ which means that $2L-l=L$, i.e. $l=L$ which means that $P_{2L,\lambda }$ has L zeros inside the unite disc. The unique solution of this linear system is equivalent to the existence of a (well-defined) bounded inverse of the operator. Indeed, after determination of the first L coefficients of f, the remainder ones can be uniquely determined. $\square$

Lemma 5

The spectrum of $M_{e,s}$ is $Q([-1,1])$.

Proof

Let us recall the information on $M_{e,s}$. According to Weyl’s theorem (see e.g. [27, Sect. 134]), the essential spectrum of $M_{e,s}$ agrees with the essential spectrum of $M_{e}$. Recalling that the spectrum of $M_{e}$ is the closure of $Q([-1,1])$, the essential spectrum and the spectrum of $M_{e}$ are the same. As the spectrum of $M_{e,s}$ does not contains any isolated points as well, see [12], it also coincides with $Q([-1,1])$.

The fact that the essential spectrum and the spectrum of $M_{e,s}$ coincide in this setup can be proved as follows.

Let us consider $P_{2L, \lambda }(r,\varphi )$ ($z=(r\cos \varphi ,r\sin \varphi )$) as a function from ${\mathbb {R}}^{1+2}$ to ${\mathbb {R}}^{2}$. If there was an isolated point of $\sigma (M_{e,s})$, then there would be a $(\lambda _0, r_0, \varphi _0)$, $r_0=1$, such that $P_{2L, \lambda _0}(r_0,\varphi _0)=0$ and there would be a neighbourhood of $\lambda _0$ such that for any $\lambda$ from this neighbourhood, the zeros of $P_{2L, \lambda }$ are not on the unit circle.

So $P_{2L, \lambda }(r,\varphi )=(p_1(\lambda ,r,\varphi ),p_2(\lambda ,r,\varphi )$. Let us consider $\partial _2P_{2L, \lambda }=\left[ \begin{array}{cc}a_{11}&{}a_{12} \\ a_{21}&{}a_{22}\end{array}\right]$, where $a_{i1}=\frac{\partial }{\partial r}p_i$, $a_{i2}=\frac{\partial }{\partial \varphi }p_i$, $i=1,2$. As $\det \partial _2P_{2L, \lambda }=\frac{1}{r}(a_{22}^2+a_{12}^2)$, we can apply the implicit function theorem at $(\lambda _0,r_0,\varphi _0)$ that is there is a neighbourhood of $\lambda _0$ denoted by U and an arc g in U such that if $\lambda \in U$ then $P_{2L}(\lambda , g(\lambda ))=P_{2L, \lambda }(r(\lambda ),\varphi (\lambda ))=0$. If $\lambda \in U$ $g'(\lambda )=\left[ \begin{array}{c}r'(\lambda ) \varphi '(\lambda )\end{array}\right]$, where $r'(\lambda )=\frac{r^{2L+1}}{a_{22}^2+a_{12}^2}\sum _{k=1}^LkU_k\cos k\varphi (r^k-r^{-k})$ (that is $r'(\lambda _0)=0$), and $\varphi '(\lambda )=\frac{-r^{2L}}{a_{22}^2+a_{12}^2}\sum _{k=1}^LkU_k\sin k\varphi (r^k+r^{-k})$. Thus, the slope of the tangent line at $\lambda _0$ to g is $-\cot (\varphi (\lambda _0))$, which is just the the slope of the tangent line to the unit circle at the same point, and the curvature of g at $\lambda _0$ is 1. Considering the symmetry of the zeros with respect to the unite circle g has to coincide locally with the unit circle, which means that $\lambda _0$ cannot be isolated. $\square$

Proof

(of Theorem 2) Notice, that $b\ge 0$ on $[-1,1]$. Thus Q is increasing here and $Q(-1)<Q(1)$. That is to prove the statement it is enough to compute these two values. In view of (36)

$$\begin{aligned}&2\sum _{k=1}^L(\pm 1)^kU_k \\&\quad =2\sum _{l=1}^{\left[ \frac{L}{2}\right] }\sum _{p=l}^{\left[ \frac{L}{2}\right] }\frac{d_{2p-1}}{2p}\left( {\begin{array}{c}2p\\ p-l\end{array}}\right) \frac{1}{2^{2p}}\pm 2\sum _{l=0}^{\left[ \frac{L-1}{2}\right] }\sum _{p=l}^{\left[ \frac{L-1}{2}\right] }\frac{d_{2p}}{2p+1}\left( {\begin{array}{c}2p+1\\ p-l\end{array}}\right) \frac{1}{2^{2p+1}}=:S. \end{aligned}$$

Changing the order of summation and by the definition of Q

$$\begin{aligned} S=\sum _{p=l}^{\left[ \frac{L}{2}\right] }\frac{d_{2p-1}}{2p}\left( 1-\frac{1}{2^{2p}}\left( {\begin{array}{c}2p\\ p\end{array}}\right) \right) \pm \sum _{p=l}^{\left[ \frac{L-1}{2}\right] }\frac{d_{2p}}{2p+1}=\sum _{k=1}^L(\pm 1)^k\frac{d_{k-1}}{k}-U_0=Q(\pm 1). \end{aligned}$$

Thus, the spectrum of the operator $[M]_{e,s}$ is $\left[ \sum _{k=1}^L(-1)^kU_k,2\sum _{k=1}^LU_k\right]$, which together with Lemma 4 proves the statement. $\square$