# A multi-orthogonal polynomial’s approach to bulk queueing theory

U. Fidalgo  Department of Mathematics and Statistics, Case Western Reserve
University, Cleveland, Ohio 43403<EMAIL_ADDRESS>In memoriam to Professors
Elizabeth Meckes, Wojbor A. Woyczynski, and Richard S. Varga

###### Abstract.

We consider a stationary Markov process that models certain queues with a bulk
service of a fixed number $m$ of admitted customers. We find an integral
expression of its transition probability function in terms of certain multi-
orthogonal polynomials with respect to a system of distributions that contain
measures supported on starlike subsets of the complex plane. We give explicit
expressions for such polynomials and distributions in terms of the solution of
an algebraic equation.

###### Key words and phrases:

Bulk queueing models, $M/M(m,m)/1$, Multiple orthogonal polynomials

###### 2000 Mathematics Subject Classification:

Primary 60H10, 60J35; Secondary 41X00

##  1\. Introduction

In [6] N. Bailey considered a queueing model with a bulk service admitting
batches with a maximum size $m\in\mathbb{N}=\\{1,2,3,\ldots\\}$. This model
describes several situations where the customers are admitted in groups, for
example transportation of people in vehicles with a maximum passenger
capacity. In our case we assume that batches have a fixed size $m$ (see [24,
Chapter 4]). We have an example of these situations in some shuttle routes
where a bus only starts its trip if it completes the vehicle’s capacity. We
can find more examples of this behavior in certain bureaucratic processes, for
instance this happens in some public museums where the guides only admit a
fixed number of customers and if this number is not completed they don’t work.

Let us assume that an individual arrives at each epoch of Poisson occurrence
with rate $\lambda>0$, and the service only starts if the queue size reaches
or exceeds $m$. The service for batches is also assumed a Poisson process with
parameter $\mu>0$. In terms of Kendall’s notation [19] used in [24], we say
that our system is $M/M(m,m)/1$.

This system corresponds to a stationary Markov process with path function
$X(t)$ whose transition probability function has form

$P_{i,j}(t)=Pr\left\\{X(t+s)=j|X(s)=i\right\\},\quad(i,j)\in\mathbb{Z}_{+}\times\mathbb{Z}_{+}=\mathbb{Z}_{+}^{2},$

where $\mathbb{Z}_{+}=\mathbb{N}\cup\\{0\\}=\\{0,1,2,\ldots\\}.$

######  Remark 1.1.

Consider the infinite matrix function

$\displaystyle{\bf P}(t)=[P_{i,j}(t)]_{(i,j)\in\mathbb{Z}_{+}^{2}}.$

From [2, pages 5-7] we have that $\displaystyle{\bf
P}=[P_{i,j}]_{(i,j)\in\mathbb{Z}_{+}^{2}}$ is a transition probability
function if it satisfies

  1.   2. (1)

$P_{i,j}(t)\geq 0$, $t\in\mathbb{R}_{+}$, and $P_{i,j}(0)=\delta_{i,j}$ where

$\delta_{i,j}=\left\\{\begin{array}[]{l l l}0&\mbox{if}&i\not=j,\\\ &&\\\
1&\mbox{if}&i=j,\end{array}\right.$

denotes Kroneker’s delta. We denote ${\bf P}(t)\geq 0$. Also ${\bf
P}(0)=\mathbb{I}$. The last equality is the so called identity property.

  3.   4. (2)

$\displaystyle\sum_{j=0}^{\infty}P_{i,j}(t)\equiv 1$, $t\in\mathbb{R}_{+}$,
$n\in\mathbb{Z}_{+}$. This property is called honesty.

  5.   6. (3)

${\bf P}(s+t)={\bf P}(s){\bf P}(t)$, $(s,t)\in\mathbb{R}_{+}^{2}$, which is
the semigroup property (Kolmogorov-Chapman equation).

  7.   8. (4)

$\displaystyle\lim_{t\to 0}P_{i,j}(t)=\delta_{i,j}$,
$(i,j)\in\mathbb{Z}_{+}^{2}$, or $\displaystyle\lim_{t\to 0}{\bf
P}(t)=\mathbb{I}$.

  9.

In our model $M/M(m,m)/1$ the following conditions hold

(1.1) $\begin{array}[]{l}\left\\{\begin{array}[]{rcl}P_{i,i}(\Delta
t)&=&1-\lambda\Delta t+o(\Delta t),\\\ &&\\\ P_{i,i+1}(\Delta
t)&=&\lambda\Delta t+o(\Delta t),\end{array}\quad\mbox{as}\quad\Delta t\to
0,\quad\mbox{if}\quad i<m.\right.\\\ \\\
\left\\{\begin{array}[]{rcl}P_{i,i-m}(\Delta t)&=&\mu\Delta t+o(\Delta t),\\\
&&\\\ P_{i,i}(\Delta t)&=&1-(\lambda+\mu)\Delta t+o(\Delta t),\\\ &&\\\
P_{i,i+1}(\Delta t)&=&\lambda\Delta t+o(\Delta
t).\end{array}\quad\mbox{as}\quad\Delta t\to 0,\quad\mbox{if}\quad i\geq
m.\right.\end{array}$

When $m=1$ we are in a case of a Birth-and-Death process (see [18]). Birth and
Death processes have been used to model any M/M(1,1)/1 system (see [5, Chapter
III]). As in [10, Chapter XVII]), considering (1.1) with the Chapman-
Kolmogorov equations, we obtain the following system with infinitely many
differential equations. Set $P_{0,-1}=0$

(1.2) $P_{i,j}^{\prime}(t)=\left\\{\begin{array}[]{l}-\lambda
P_{i,j}(t)+\lambda
P_{i+1,j}(t),\quad(i,j)\in\left\\{0,1,\ldots,m-1\right\\}\times\mathbb{Z}_{+}=\mathbb{F},\\\
\\\ -(\lambda+\mu)P_{i,j}(t)+\lambda P_{i+1,j}(t)+\mu
P_{i-m,j}(t),\quad(i,j)\in\mathbb{Z}_{+}^{2}\setminus\mathbb{F},\end{array}\right.$

with $t\geq 0$ (we use the notation $t\in\mathbb{R}_{+}$).

Combining condition $(1)$ in Remark 1.1 with the system of differential
equations in (1.2) we have an initial value problem that we write in its
matrix version as follows

(1.3) $\left\\{\begin{array}[]{l l}{\bf P}^{\prime}(t)={\bf A}{\bf
P}(t),&t\geq 0,\\\ &\\\ {\bf P}(0)=\mathbb{I},\end{array}\right.$

where ${\bf A}=[a_{i,j}]_{(i,j)\in\mathbb{Z}_{+}^{2}}$ is such that

$a_{i,j}=\left\\{\begin{array}[]{r c l}\mu&\mbox{if}&j=i-m,\\\ &&\\\
-\lambda&\mbox{if}&j=i<m,\\\ &&\\\ -(\lambda+\mu)&\mbox{if}&j=i\geq m,\\\
&&\\\ \lambda&\mbox{if}&j=i+1,\\\ &&\\\
0&\mbox{if}&j\not\in\\{i+1,i,i-m\\},\end{array}\right.\quad(i,j)\in\mathbb{Z}_{+}^{2},$

which has schematically form

(1.4) ${\bf A}=\left(\begin{array}[]{c c c c c c c c
c}-\lambda&\lambda&0&\cdots&0&0&0&0&\cdots\\\ 0&-\lambda&\lambda&&0&0&0&0&\\\
\vdots&\ddots&\ddots&\ddots&\vdots&&\vdots&\vdots&\\\ &&&&&&&&\\\
0&0&\cdots&0&-\lambda&\lambda&0&0&\cdots\\\
\mu&0&\cdots&0&0&-(\lambda+\mu)&\lambda&0&\ddots\\\
0&\mu&&0&0&0&-(\lambda+\mu)&\lambda&\ddots\\\
\vdots&\ddots&\ddots&&\ddots&&\ddots&\ddots&\ddots\\\ \end{array}\right).$

The system of differential equations (1.2) is deduced from (1.1), which is
independent from the four conditions in Remark 1.1, hence some of its
solutions would not be transition probability functions. In order to find a
matrix function ${\bf P}$ that describes properly a system $M/M(m,m)/1$, we
select from the set of solutions of (1.3) the ones that satisfy such four
conditions. In Section 7 (specifically in subsection 7.1), using techniques of
[30] by W. Sternberg, we are able to prove that there is a unique solution
with analytic entries functions on the complex plane $\mathbb{C}$, being real
valued on $\mathbb{R}_{+}=[0,\infty)$, and so that it satisfies the four
conditions that define what a transition probability function is. The other
possible solutions have non uniform bounded entries on $\mathbb{R}_{+}$. In
case they exist, conditions (1) and (2) in Remark 1.1 would not be satisfied
with them, hence they would not be transition probability functions.

In [18], S. Karlin and J. McGregor arrived at an integral expression for the
entries of ${\bf P}$ in the case of $m=1$, which is a Birth-and-Death process.
They used some aspects of the orthogonal polynomial theory. Using Karlin and
McGregor’s approach, T. S. Chihara and M. E. H. Ismail (see [9]) connected
sequences of orthogonal polynomials with a queueing model. Later on in [15] M.
E. H. Ismail revisited this connection. In this paper we also use Karlin and
McGregor’s method to give an accurate integral expressions for the mentioned
entries of the transition probability function which is a solution of the
initial value problem (1.3), in our case for general $m\geq 1$. To this end we
consider properties of multi-orthogonality relations of polynomials. We use a
family of multi-orthogonal polynomials defined by a three terms of recurrence
relation. Such family is a variation of the ones studied in [3, 4, 21, 22,
23].

Let $\displaystyle\left\\{Q_{n}\right\\}_{n\in\mathbb{Z}_{+}}$ be a sequence
of polynomials generated by the recurrence relation

(1.5) $(\lambda+\mu+x)Q_{n}(x)=\lambda Q_{n+1}(x)+\mu Q_{n-m}(x),\quad n\geq
m,$

with initial conditions

(1.6) $Q_{n}(x)=\frac{1}{\lambda^{n}}(\lambda+x)^{n},\quad n=0,\ldots,m.$

The above conditions (1.6) can be also written as

(1.7) $Q_{0}\equiv 1,\quad Q_{n}(x)=\frac{1}{\lambda}(\lambda+x)Q_{n-1},\quad
n=1,\ldots,m.$

We consider a dual family of vector polynomials ${\bf
q}_{r}=(q_{0,r},\ldots,q_{m-1,r})$ that satisfy

(1.8) $\begin{array}[]{l}\displaystyle{\bf q}_{-1}=(0,\ldots,0),{\bf
q}_{0}=\left(1,0,\ldots,0\right),\\\ \\\ {\bf
q}_{1}=\left(0,1,0,\ldots,0\right),\ldots,{\bf
q}_{m-1}=\left(0,\ldots,,0,1\right),\\\ \\\ (\lambda+x){\bf q}_{r}(x)=\mu{\bf
q}_{r+m}(x)+\lambda{\bf q}_{r-1}(x),\quad r\in\\{0,\ldots,m-1\\},\\\ \\\
(\lambda+\mu+x){\bf q}_{r}(x)=\mu{\bf q}_{r+m}(x)+\lambda{\bf
q}_{r-1}(x),\quad r\in\\{m,m+1,\ldots,\\}.\end{array}$

In the following Theorem 1.2 we state explicit expressions of the polynomials
$Q_{n}$, $n\in\mathbb{Z}_{+}$ and the components of the polynomial vectors
${\bf q}_{r}$, $r\in\mathbb{Z}_{+}$ in terms of the $m+1$ solutions of the
algebraic equation

(1.9) $\lambda\,\omega^{m+1}-(z+\lambda+\mu)\,\omega^{m}+\mu=0.$

In Section 2 we study some the algebraic equations with form

(1.10)
$\alpha\omega^{m+1}-p(z)\omega^{m}+\beta=0,\quad\mbox{with}\quad(\alpha,\beta)\in\mathbb{R}^{2},\quad\alpha\not=0,$

and $p$ is a polynomial with degree exactly 1 ($\deg p=1$), where (1.9) is
contained in such family (1.10). According to Section 2 we denote the solution
functions of (1.9) by $\omega_{0},\omega_{1},\ldots,\omega_{m}$ where for each
$j=1,\ldots,m,$

$\omega_{0}(z)=z+\mathcal{O}(1)\quad\mbox{and}\quad\omega_{j}(z)=\mathcal{O}\left(\frac{1}{z^{1/m}}\right),\quad\mbox{as}\quad
z\to\infty.$

######  Theorem 1.2.

Fix $n\in\mathbb{N}$. The polynomial $Q_{n}$, $n\in\mathbb{N}$ can be
expressed by

(1.11)
$Q_{n}(z)=\sum_{j=0}^{m}\frac{\displaystyle\left(\omega_{j}+\frac{\mu}{\lambda}\left(\frac{1}{\omega_{j}^{m}}-1\right)\right)^{m}-1}{\displaystyle\left(\omega_{j}^{m}-1\right)\left(\omega_{j}^{m}-\frac{m\,\mu}{\lambda\,\omega_{j}}\right)}\,\,\omega_{j}^{m+n}.$

Given $r\in\mathbb{N}$, the vector polynomial ${\bf q}_{r}$ has the form

${\bf q}_{r}=\left(q_{0,r},\,q_{1,r},\ldots,q_{m,r}\right),$

with for each $r\in\\{m,\ldots,2m-1\\}$

(1.12)
$q_{j,r}=\frac{\lambda+x}{\mu}\delta_{j,r-m}-\frac{\lambda}{\mu}\delta_{j,r-m-1},\quad
j=0,\ldots,m-1.$

For $r\geq 2m$

(1.13)
$q_{j,r}=\frac{\lambda}{\mu}\sum_{k=1}^{m}\frac{1}{\omega_{k}^{r-m-j}(\omega_{k}^{m+1}-\frac{m\mu}{\lambda})}$
$\times\left[\left(\omega_{k}^{m+1}+\frac{\mu}{\lambda}\right)\left(\frac{1}{\omega_{k}^{m+1}}+\frac{\lambda}{\mu}\left(1-\omega_{k}\right)-\frac{1}{\omega_{k}}\right)\right.$
$\left.-\frac{\omega_{k}^{m-j}}{\omega_{0}^{m-j}}\left(\omega_{0}^{m+1}+\frac{\mu}{\lambda}\right)\left(\frac{1}{\omega_{0}^{m+1}}+\frac{\lambda}{\mu}\left(1-\omega_{0}\right)-\frac{1}{\omega_{0}}\right)\right],\quad
j=0,\ldots,m-2,$

and

(1.14)
$q_{m-1,r}=\frac{\lambda}{\mu}\sum_{k=1}^{m}\frac{1}{\omega_{k}^{r+1}(\omega_{k}^{m+1}-\frac{m\mu}{\lambda})}$
$\times\left(\frac{1}{\omega_{0}^{m+1}}-\frac{1}{\omega_{0}}+\frac{\lambda}{\mu}\right)\left[\omega_{k}^{m+1}+\frac{\mu}{\lambda}-\frac{\omega_{k}}{\omega_{0}}\left(\omega_{0}^{m+1}+\frac{\mu}{\lambda}\right)\right].$

Given two complex numbers $u$ and $v$ we denote by $[u,\,v]$ the compact
segment of straight line that connects $u$ with $v$. A real valued function
$r$ defined on a certain domain $D\subset\overline{\mathbb{C}}$ of the Riemann
sphere $\overline{\mathbb{C}}=\mathbb{C}\cup\\{\infty\\}$ is said to be a
weight if it does not change sign on $D$. This is
$r:D\subset\overline{\mathbb{C}}\rightarrow\mathbb{R}_{+}$ or
$r:D\subset\overline{\mathbb{C}}\rightarrow\mathbb{R}_{-}=(\mathbb{R}\setminus\mathbb{R}_{+})\cup\\{0\\}$.
Let $\delta_{\zeta}$ denote the measure often called Riemann’s delta with an
only point of mass at $\zeta\in\mathbb{R}$. Fix $j\in\mathbb{Z}_{+}$.
According to [29, Chapter 6] $\delta_{\zeta}^{(j)}$ denotes the $j$th
distribution derivative of $\delta_{\zeta}$, where for each $j$th
differentiable function $f$ at $\zeta$ whose derivatives $f^{(k)}$,
$k=0,1,\ldots,j$ are integrable in $\mathbb{R}$, satisfies that

(1.15) $\int f(z)\,\delta_{\zeta}^{(j)}(z)\,dz=(-1)^{j}f^{(j)}(\zeta).$

We can extend (1.15) for any differentiable function $f$ in an open interval
that contains $\zeta$, replacing $f$ by a proper test function expressed by a
piecewise definition which coincides with $f$ on such interval. Fixed
$j\in\mathbb{N}$ the distribution $\delta_{\zeta}^{(j)}$ is not a measure. The
following result stated in Theorem 1.3 is proved in Section 6.

######  Theorem 1.3.

There exists a system of distribution
$(\sigma_{0},\sigma_{1},\ldots,\sigma_{m-1})$ such that the following bi-
orthogonality relations hold

(1.16) $\delta_{n,r}=\int
Q_{n}(x)\sum_{j=0}^{m-1}q_{j,r}(x)\,d\sigma_{j}(x),\quad(n,r)\in\mathbb{Z}_{+}^{2}.$

Each distribution component $\sigma_{j}$, $j=0,1,\ldots,m-1$, has a type of
Radon Nikodym derivative

(1.17)
$\frac{d\sigma_{j}}{dx}=\rho_{j}(x)\rho(x)+\frac{(-1)^{j-1}}{\lambda^{j}(j-1)!}\left(\delta_{-\lambda}-\delta_{-\lambda-\mu}\right)^{(j-1)},\quad
j=0,1,\ldots,m-1$

with $\rho_{0}\equiv 1$ and $\displaystyle\frac{\delta^{(-1)}}{(-1)!}=0$. The
expression $\rho_{j}(x)\rho(x)$ is a function defined on the starlike set

$\Sigma_{0}=\bigcup_{k=0}^{m}\left[-\lambda-\mu,-\lambda-\mu+\frac{m+1}{m}\left(\frac{\mu\lambda^{m}}{m}\right)^{1/(m+1)}\exp\frac{2\pi
ik}{m+1}\right],$

where fixed $k\in\\{0,1,\ldots,m\\}$ and each point

$x_{k}=-\mu-\lambda+t\exp\frac{2\pi ik}{m+1}\in\Sigma_{0},\quad 0\leq
t\leq\frac{m+1}{m}\left(\frac{\mu\lambda^{m}}{m}\right)^{\frac{1}{m+1}},$

the following symmetry holds

$\rho_{j}\left(x_{k}\right)\rho\left(x_{k}\right)=\exp\left(-\frac{2\pi
ij}{m+1}\right)\rho_{j}\left(t-\mu-\lambda\right)\rho\left(t-\mu-\lambda\right),$

being
$\displaystyle\rho_{j}\left(t-\mu-\lambda\right)\rho\left(t-\mu-\lambda\right)$
a weight.

The functions $\rho,\rho_{1},\ldots,\rho_{m-1}$ are expressed explicitly in
terms of the solutions of the equation (1.9) in Section 6. Let us state our
main result

######  Theorem 1.4.

The matrix function ${\bf P}(t)=[P_{n,r}]_{(n,r)\in\mathbb{Z}_{+}^{2}}$ whose
entries are

(1.18) $P_{n,r}(t)=\int
e^{xt}Q_{n}(x)\sum_{j=0}^{m-1}q_{j,r}(x)\,d\sigma_{j}(x)$

is the unique solution of the initial value problem (1.3), that conform a
transition probability function.

From the equality (1.18) in Theorem 1.4, we show in Subsection 7.3 of Section
7 that any system described by the a queueing model $M/M(m,m)/1$, tends to a
steady state where

(1.19) $\displaystyle\lim_{t\to 0}P_{n,r}(t)=0$

exponentially. According to [2, Theorem 1.6] this behavior is a consequence
that ${\bf P}$ is an irreductible transition probability function. This menas
that all states can be reached from each other.

We prove Theorem 1.2 in Section 3. Theorem 1.3 is proved in Section 6, and
Theorem 1.4, in Section 7 where we show that the functions $P_{n,r}$,
$(n,r)\in\mathbb{Z}_{+}^{2}$ in (1.18) conform a solution corresponding to the
initial value problem in (1.3), and also satisfy the properties corresponding
transition probability functions enumerated in Remark 1.1. To obtain these
results we need to study the solutions of some algebraic equations of the
family (1.9) in Section 2. The work in this publication connects different
math branches. In order to make easier the reading, we show some known results
giving details in their proofs that we need in this context.

##  2\. A family of algebraic equations

This section concerns about the family of algebraic equations in (1.10). After
the change of variable that makes $\displaystyle z=p(\cdot)/\alpha$ and also
denoting $\displaystyle c=\beta/\alpha$, all those equations are reduced to an
easier form which belongs to the present family

(2.1) $\omega^{m+1}-z\omega^{m}+c=0.$

The above equation is studied in [4]. We enumerate some of its results in the
following Proposition 2.1. Previously we introduce some concepts needed to
state it. Let $[0,\infty)\times\exp\left(i\theta\right),$
$\theta\in\mathbb{R}$ denote the ray that starts at the origin with slope
$\tan\theta$. With the expression
$[-\infty,\infty)\times\exp\left(i\theta\right)$ we refer the straight line
with slope $\tan\theta$ containing the origin. Let us declare the following
starlike sets

(2.2) $\begin{array}[]{l}\displaystyle
S_{0}=\bigcup_{k=0}^{m}\left[0,\,a\exp{\frac{2\pi ik}{m+1}}\right],\\\ \\\
\displaystyle
S_{\infty}^{e}=\bigcup_{k=0}^{m}[-\infty,\,\infty)\times\exp\left(\frac{(2k+1)i\pi}{m+1}\right),\\\
\\\ \displaystyle
S_{\infty}^{o}=\bigcup_{k=0}^{m}[0,\infty)\times\exp\left(\frac{2ki\pi}{m+1}\right).\end{array}$

We denote

$\Omega_{0}=\mathbb{C}\setminus
S_{0},\quad\Omega_{j}=\mathbb{C}\setminus\\{S_{\infty}^{o}\cup
S_{\infty}^{e}\\},\quad 1\leq
j<m\quad\mbox{and}\quad\Omega_{m}=\left\\{\begin{array}[]{l
l}\mathbb{C}\setminus S_{\infty}^{e}&\mbox{if }m\mbox{ even},\\\
\mathbb{C}\setminus S_{\infty}^{o}&\mbox{if }m\mbox{ odd},\end{array}\right.$

where

$a=\frac{m+1}{m}\left(\frac{c}{m}\right)^{m/m+1}.$

Let $\pi$ be the projection function that maps a genus zero Riemann surface
$\mathcal{R}$ onto $\mathbb{C}$, with
$\pi^{-1}=\left(\pi_{0}^{-1},\ldots,\pi_{m}^{-1}\right)$ (multivalued)
representing its inverse branches. This Riemann surface $\mathcal{R}$ has form

$\mathcal{R}=\overline{\bigcup_{j=0}^{m}\mathcal{R}_{j}},\quad\mathcal{R}_{0}=\pi_{0}^{-1}(\Omega_{0}\cup\\{\infty\\}),\quad\mathcal{R}_{j}=\pi_{j}^{-1}(\Omega_{j}),\quad
j=1,\ldots,m.$

We also denote $\partial\mathcal{R}_{j,k}$ a closed contour in
$\mathbb{C}^{2}$ which separates $\mathcal{R}_{j}$ and $\mathcal{R}_{k}$. The
contour $\partial\mathcal{R}_{j,k}$ is orientated such that $\mathcal{R}_{j}$
lies on the left, this is positive side. We take

$\begin{array}[]{l
l}\pi(\partial\mathcal{R}_{0,1})&=\partial\Omega_{0}=S_{0+}\cup S_{0-},\\\
&\\\ \pi(\partial\mathcal{R}_{j,j+1})&=\left\\{\begin{array}[]{l
l}S^{e}_{\infty+}\cup S^{e}_{\infty-},&j=2l-1\\\ &\\\ S^{e}_{\infty+}\cup
S^{o}_{\infty-},&j=2l\end{array}\right.\quad j=1,\ldots,m-1.\end{array}$

Hence

$\mathcal{R}=\left\\{\bigcup_{j=0}^{m}\mathcal{R}_{j}\right\\}\bigcup\left\\{\bigcup_{j=0}^{m-1}\partial\mathcal{R}_{j,j+1}\right\\}.$

######  Proposition 2.1.

[4, Proposition 1] There is a multivalued solution $\omega(z)$ of the equation
(2.1) which is an algebraic function of order $m+1$ with global branches of
$\omega(z)$ in (2.1) satisfying that for each $\displaystyle
j\in\left\\{0,1,\ldots,m\right\\}$, $\omega_{j}$ is a holomorphic function on
$\Omega_{j}$ (Notation: $\omega_{j}\in\mathcal{H}(\Omega_{j})$) and

(2.3) $|\omega_{0}|\geq|\omega_{1}|\geq\cdots\geq|\omega_{m}|,\quad\mbox{on
}\,\mathbb{C}.$

In $S_{0}$ the inequalities are strict.

###### Proof.

Consider the polynomial defined in $\overline{\mathbb{C}}^{2}$
$P(z,\omega)=\omega^{m+1}-z\omega^{m}+c$. First we prove that $\mathcal{R}$ is
a apropiate Riemann surface of the multivalued function $\omega$ defined by
the equation $P(z,\omega)=0$ in (2.1). We note that $\omega$ is an algebraic
function of order $m+1$, hence its Riemann surface contains $d=m+1$ sheets
with $m+1$ branches $\omega_{0},\omega_{1},\ldots,\omega_{m}$ that satisfy

$P(z,\omega)=\prod_{j=0}^{m}\left(\omega-\omega_{j}\right)=\omega^{m+1}-z\omega^{m}+c=0,$

that from Vieta’s formulae we have that

(2.4)
$\begin{array}[]{l}\displaystyle\omega_{0}+\omega_{1}+\cdots+\omega_{m}=z,\quad\prod_{j=0}^{m}\omega_{j}=(-1)^{m+1}c,\\\
\\\ \displaystyle\sum_{0\leq j_{1}<j_{2}<\cdots<j_{k}\leq
n}\prod_{j=1}^{k}\omega_{i_{j}}=0,\quad k\in\\{1,\ldots,m-1\\}.\\\ \\\
\end{array}$

Let us find its singular points. Observe

$\frac{\partial
P}{\partial\omega}=(m+1)\,\omega^{m}-z\,m\,\omega^{m-1}\quad\mbox{and}\quad\frac{\partial
P}{\partial z}=\omega^{m}.$

The singular points are the solutions of the systems

$(1)\,\left\\{\begin{array}[]{l}0=\omega^{m},\\\ \\\
0=\omega^{m+1}-z\omega^{m}+c,\end{array}\right.\quad\mbox{and}\quad(2)\,\left\\{\begin{array}[]{l}0=(m+1)\,\omega^{m}-z\,m\,\omega^{m-1},\\\
\\\ 0=\omega^{m+1}-z\omega^{m}+c.\end{array}\right.$

From system (1) we deduce that there is a singular point at $(\infty,0)$.
Consider its inverse rational function

$z(\omega_{j})=\omega_{j}+\frac{c}{\omega_{j}^{m}},\quad j=0,1,\ldots,m,$

hence for each $j\in\\{0,1,\ldots,m\\}$ either

$\omega_{j}(z)=z+\mathcal{O}\left(1\right)\quad\mbox{or}\quad\omega_{j}(z)=\mathcal{O}\left(\frac{1}{z^{1/m}}\right)\quad\mbox{as}\quad
z\to\infty.$

Let $J\subset\\{0,1,\ldots,m\\}$ denote the set such that

$\omega_{j}(z)=z+\mathcal{O}\left(1\right)\quad\mbox{as}\quad z\to\infty,\quad
j\in J.$

Set $\kappa=\\#J\in\mathbb{N}$ the cardinality of $J$. Combining the above
relations and the first identity in (2.4) we obtain that

$\frac{1}{z}\left(\omega_{0}+\omega_{1}+\cdots\omega_{m}\right)=\kappa+\mathcal{O}\left(1/z\right)=1\quad\mbox{as}\quad
z\to\infty.$

This implies $\kappa=1$. We choose the notation:

(2.5)
$\omega_{0}(z)=z+\mathcal{O}(1)\quad\mbox{and}\quad\omega_{j}(z)=\mathcal{O}\left(\frac{1}{z^{1/m}}\right),\quad
j=1,\ldots,m,\quad\mbox{as}\quad z\to\infty.$

We conclude that $\omega$’s branch point $\tau_{\infty}$ at $z=\infty$ has
order $m-1$ or equivalently it has $\eta_{\infty}=m$-cyclic neighborhoods on
its corresponding Riemann surface.

In system (2) we combine the equations, and considering that $\omega\not=0$ to
obtain that

$m\left(\omega^{m}-z\,\omega^{m-1}\right)+\omega^{m}=0\implies
m\left(\omega^{m+1}-z\,\omega^{m}\right)+\omega^{m+1}=0.$

Taking into account (2.1) we have that $\displaystyle\omega^{m+1}=mc$. This
equations has $m+1$ solutions

(2.6) $\omega_{k}(z_{k})=\sqrt[m+1]{mc}\,\exp\left(\frac{2\pi
ik}{m+1}\right),\quad k=0,\ldots,m,$

with

$z_{k}=\frac{m+1}{m}\sqrt[m+1]{mc}\,\exp\left(\frac{2\pi ik}{m+1}\right),\quad
k=0,\ldots,m.$

For each $\displaystyle k\in\left\\{0,1,\ldots,m\right\\}$ the singular point

$\tau_{k}=\left(\frac{m+1}{m}\sqrt[m+1]{mc}\,\exp\left(\frac{2\pi
ik}{m+1}\right),\,\sqrt[m+1]{mc}\,\exp\left(\frac{2\pi
ik}{m+1}\right)\right),$

with $\eta_{k}=2$-cyclic neighborhoods on its corresponding Riemann surface,
hence $\mathcal{R}$ is appropriate. From Riemann-Hurwitz formula (see [14]),
$\mathcal{R}$ has genus $g=0$

$g=\frac{1}{2}\sum_{k=0}^{m}(\eta_{k}-1)+\frac{1}{2}\left(\eta_{\infty}-1\right)-d+1=\frac{1}{2}\sum_{k=0}^{m}(2-1)+\frac{1}{2}(m-1)-(m+1)+1=0.$

Set the point $\displaystyle a=\frac{m+1}{m}\sqrt[m+1]{mc}$, we have that
$\omega_{0}(z_{0})=\omega_{1}(z_{1})=\sqrt[m+1]{mc}$, then

$\displaystyle\omega_{0}(x)>\sqrt[m+1]{m\,c}>\omega_{1}(x),\quad
x\in\left(a\,,\,+\infty\right).$

When $m$ is even there is a negative branch of $\omega$ on $\mathbb{R}_{+}$,
that we denote by $\omega_{m}$. We label the branches such that we order in
the following way

(2.7)
$|\omega_{0}|\geq|\omega_{1}|>|\omega_{2}|=|\omega_{3}|>\cdots>\left\\{\begin{array}[]{l
l}|\omega_{m-1}|=|\omega_{m}|,&m\,\mbox{odd}\\\ &\\\
-\omega_{m},&m\,\mbox{even}\end{array}\right.\quad\mbox{on }\,\mathbb{R}_{+}.$

Define an analytic continuation of $\omega$ from $\mathbb{R}_{+}$ to
$\mathbb{C}$. Consider $\omega$ a solution of (2.1), then the following
equality holds

$\left(e^{\frac{2\pi i}{m+1}}\omega(z)\right)^{m+1}-ze^{\frac{2\pi
i}{m+1}}\left(e^{\frac{2\pi i}{m+1}}\omega(z)\right)^{m}+\mu\lambda^{m}=0.$

For each fixed $j\in\left\\{0,1,\ldots,m\right\\}$ there exists a choice of
branch $\omega_{k}$, $k\in\left\\{0,1,\ldots,m\right\\}$ satisfying the
symmetric relation:

$\omega_{k}\left(ze^{\frac{2\pi i}{m+1}}\right)=e^{\frac{2\pi
i}{m+1}}\omega_{j}\left(z\right),\quad z\in\mathbb{C}.$

We choose $j=k$, $j=0,\ldots,m.$

(2.8) $\omega_{k}\left(ze^{\frac{2\pi i}{m+1}}\right)=e^{\frac{2\pi
i}{m+1}}\omega_{j}\left(z\right),\quad z\in\mathbb{C}.$

Using monodromy theorem (see [1, Chapter I $\S$ 15]), we extend analytically
the branch $\omega_{0}$ from a neighborhood of $z=\infty$ to the domain
$\Omega_{0}=\mathbb{C}\setminus S_{0}$. We declare
$\omega_{0}\in\mathcal{H}(\Omega_{0})$. Since (2.8) is satisfied near $\infty$
for $j=0$, then

$\omega_{0}\left(ze^{\frac{2\pi i}{m+1}}\right)=e^{\frac{2\pi
i}{m+1}}\omega_{0}\left(z\right),\quad z\in\Omega_{0}.$

Taking into account the behavior at infinity in (2.5), the branch $\omega_{1}$
has an analytic continuation from $\mathbb{R}_{+}\setminus\left[0,\,a\right]$
to the simple connected domain

$\tilde{A}=\left\\{z:|\arg
z|<\frac{\pi}{m+1}\right\\}\setminus\left[0,\,a\right].$

Starting from above set $\widetilde{A}$ and using (2.8) with $j=1$ we extend
analytically $\omega_{1}$ as follows

$\omega_{1}\left(ze^{\frac{2\pi i}{m+1}}\right)=e^{\frac{2\pi
i}{m+1}}\omega_{1}\left(z\right),\quad
z\in\Omega_{1}=\left\\{\begin{array}[]{l l}\Omega_{0}&m=1\\\
\Omega_{0}\setminus S_{\infty}^{e}&m>1\end{array}\right..$

Since $\Im\omega_{1}(x)=0$, $x\in(a,\,+\infty)$ we have that

$\omega_{1}(z)=\overline{\omega_{1}(\overline{z})},\quad z\in\mathbb{C}.$

Combining the above identity with (2.8) we have that $\omega_{1}$ satisfies
the jump condition

$\overline{\omega}_{1+}(z)=e^{\frac{2\pi i}{m+1}}\omega_{1-}(z),\quad
z\in\overset{\circ}{S}_{\infty}^{e},$

where

$\omega_{1\pm}(x)=\lim_{z\to x,z\in S_{\infty\pm}^{e}}\omega_{1}(z),\quad
x\in\overset{\circ}{S}_{\infty}^{e},$

and $S_{\infty}^{e}$ is the closure of $\overset{\circ}{S}_{\infty}^{e}$.
Hence $\omega_{1}\in\mathcal{H}\left(\Omega_{1}\right)$.

We proceed analogously with the remaining branches of $\omega$. Consider the
fundamental sector of $\mathbb{C}$

$A=\left\\{z:|\arg z|<\frac{\pi}{m+1}\right\\}.$

We extend the branches to $\mathbb{C}\setminus A$ using the symmetry in (2.8),

$\omega_{j}\left(ze^{\frac{2\pi i}{m+1}}\right)=e^{\frac{2\pi
i}{m+1}}\omega_{j}\left(z\right),\quad z\in\Omega_{j},\quad j=2,\ldots,m.$

A branch with even index, $\omega_{2k}$, is defined in $A$ as a direct
analytic continuation of the branch $\omega_{2k-1}$ to the domain
$\displaystyle A_{+}=\left\\{z:0<\arg z<\frac{\pi}{m+1}\right\\}$ from the
upper part of the boundary of $A_{+}$, and to the domain $\displaystyle
A_{-}=\left\\{z:-\frac{\pi}{m+1}<\arg z<0\right\\}$ from the lower part of the
boundary of $A_{-}$. While a branch with odd index, $\omega_{2k+1}$, is
defined on $A$ taking the analytic continuation of the branch $\omega_{2k}$ to
$\displaystyle A_{+}$ from the lower part of the boundary of $A_{+}$, and to
$\displaystyle A_{-}$ from the upper part of the boundary of $A_{-}$.

Consider the domain

$B=\left\\{z:0<\arg z<\frac{2\pi}{m+1}\right\\},$

and define the function $\omega_{1,2}\in\mathcal{H}(B)$ as follows

$\omega(z)=\left\\{\begin{array}[]{l l}\omega_{1}(z),&\frac{\pi}{m+1}<\arg
z<\frac{2\pi}{m+1},\\\ &\\\ \omega_{2}(z),&0<\arg z\leq\frac{\pi}{m+1}.\\\
\end{array}\right.$

Let us recall (2.7) and (2.8). If the maximum principle to the holomorphic
function $\omega_{0}/\omega_{1,2}\in\mathcal{H}(B)$ we can extend the
inequality $|\omega_{0}|>|\omega_{1}|$ in
$\displaystyle\left(a,\,+\infty\right)$ to $B$ and therefore to the whole
complex plane $\mathbb{C}$. Analogously we can extend all the relations stated
in (2.7) to $\mathbb{C}$ and obtain (2.3). ∎

######  Lemma 2.2.

There exists a function $\widetilde{\rho}$ defined on $S_{0}$ such that

(2.9) $\widetilde{\rho}(x)=\exp\left({-\frac{2\pi
ik}{m+1}}\right)\widetilde{\rho}(|x|),\quad x\in\left(0,a\exp{\frac{2\pi
ik}{m+1}}\right),$

$k=0,1,\ldots,m$, where $\widetilde{\rho}_{0}(x)$ is a weight in $x\in(0,a))$,
that satisfies

(2.10) $\displaystyle\frac{1}{\omega_{0}(z)}=\frac{1}{2\pi
i}\int\widetilde{\rho}(x)\,\frac{d\,x}{z-x}=\frac{1}{2\pi
i}\int\widetilde{\rho}(|x|)\,\frac{d\,|x|}{z-x},\quad
z\in\overline{\mathbb{C}}\setminus S_{0}.$

###### Proof.

Let us denote

$\omega_{0\pm}(x)=\lim_{z\to
x\in\overset{\circ}{S}_{0\pm}}\omega_{0}(z)\quad\mbox{with}\quad\overset{\circ}{S}_{0}=\bigcup_{k=0}^{m}\left(0,a\exp{\frac{2\pi
ik}{m+1}}\right).$

Consider an arbitrary point $x\in(0,\,a)\subset S_{0}$ in the star’s arm
corresponding to $k=0$. Using (2.3) we observe that the following functions
defined on $\overset{\circ}{S}_{0}$, are real valued and never change sign

$\left[\frac{1}{\omega_{0+}(x)}-\frac{1}{\omega_{0-}(x)}\right]=\frac{1}{\omega_{0}(x)}-\frac{1}{\omega_{1}(x)}=\widetilde{\rho}(x),\quad
x\in(0,\,a).$

With above relation and the indentity (2.8), we obtain that

(2.11) $\widetilde{\rho}(x)=\exp\left({-\frac{2\pi
ik}{m+1}}\right)\widetilde{\rho}(|x|)=\left[\frac{1}{\omega_{0+}(x)}-\frac{1}{\omega_{0-}(x)}\right]=\frac{1}{\omega_{0}(x)}-\frac{1}{\omega_{1}(x)},$

for all

$x\in\left(0,a\exp{\frac{2\pi ik}{m+1}}\right),\quad k\in\\{0,1,\ldots,m\\},$

hence condition (2.9) holds. Since $\omega_{0}\in\mathcal{H}(\Omega_{0})$
never vanishes in $\displaystyle\Omega_{0}$ and taking into account (2.5) we
have that

(2.12)
$\frac{1}{\omega_{0}(z)}\in\mathcal{H}(\Omega_{0})\quad\mbox{and}\quad\frac{1}{\omega_{0}(z)}=\mathcal{O}\left(\frac{1}{z}\right)\quad\mbox{as}\quad
z\to\infty.$

We also recall (2.6) to analyze how the behavior at the endpoints is. Now
using (2.11) and Sokhotski–Plemelj (see [13, Chapter I, equality (4.8)]) we
obtain that

$\displaystyle\frac{1}{\omega_{0}(z)}=\frac{1}{2\pi
i}\int\widetilde{\rho}(x)\frac{d\,x}{z-x},$

that combined to (2.11), completes the proof of (2.10). ∎

We give integral expressions for the functions
$\displaystyle\left(\frac{1}{\omega_{0}}\right)^{j},$ $j=1,\ldots,m$.

######  Proposition 2.3.

There exists a system of weights $r_{j}(x)=x^{j-1}\widetilde{\rho}_{j}(x)$,
$j=1,\ldots,m$, defined on $S_{0}$ that satisfy the following identities

(2.13) $\displaystyle\frac{1}{\omega_{0}^{j}(z)}=\frac{1}{z^{j-1}}\int
x^{j-1}\widetilde{\rho}_{j}(x)\,\widetilde{\rho}(x)\frac{d\,x}{z-x},\quad
j=1,\ldots,m,\quad z\in\overline{\mathbb{C}}\setminus S_{0},$

where $\widetilde{\rho}_{1}\equiv 1$ and $\widetilde{\rho}$ is defined in
Lemma 2.2.

###### Proof.

The case $j=1$ in (2.13) is already proved in Lemma 2.2. For each
$j\in\\{2,\ldots,m\\}$, we have that

$\frac{1}{\omega_{0+}^{j}(x)}-\frac{1}{\omega_{0-}^{j}(x)}=\frac{1}{\omega_{0}^{j}(x)}-\frac{1}{\omega_{1}^{j}(x)}=\widetilde{\rho}(x)\sum_{k=0}^{j-1}\frac{1}{\omega_{0}^{j-1-k}(x)\,\omega_{1}^{k}(x)},\quad
x\in S_{0}.$

Observe that the following functions are weights

$r_{j}(x)=x^{j-1}\widetilde{\rho}_{j}(x)=\sum_{k=0}^{j-1}\frac{x^{j-1}}{\omega_{0}^{j-1-k}(x)\,\omega_{1}^{k}(x)}\in\mathbb{R}_{+},\quad
x\in S_{0},\quad j=1,\ldots,m.$

From (2.12) we obtain that

(2.14)
$\frac{z^{j-1}}{\omega_{0}^{j}(z)}\in\mathcal{H}(\Omega_{0})\quad\mbox{and}\quad\frac{z^{j-1}}{\omega_{0}^{j}(z)}=\mathcal{O}\left(\frac{1}{z}\right)\quad\mbox{as}\quad
z\to\infty,\quad j=2,\ldots,m.$

The equalities in (2.13) are proved combining (2.14) and Sokhotski–Plemelj
([13, Chapter I, equality (4.8)]). The proof is completed. ∎

######  Proposition 2.4.

Let $\widetilde{\rho},\widetilde{\rho}_{1},\ldots,\widetilde{\rho}_{m}$ be the
system of functions in Proposition 2.3. Then

$0=\int x^{\nu}\rho_{j}(x)\,\rho(x)\,dx,\quad\nu=0,\ldots,j-2,\quad
j=2,\ldots,m.$

###### Proof.

From construction we have that

$\widetilde{\rho}_{j}(x)=\sum_{k=0}^{j-1}\frac{1}{\omega_{0}^{j-1-1-k}(x)\omega_{1}^{k}(x)},\quad
x\in S_{0}.$

This means that for each $x\in\mathbb{R}\cap S_{0}$ we have that for each
$\nu\in\\{0,\ldots,j-1\\}$ and $k\in\\{0,1,\ldots,m\\}$

$x^{\nu}\exp\frac{2\pi ik\nu}{m+1}\widetilde{\rho}_{j}\left(x\exp\frac{2\pi
ik}{m+1}\right)=\exp\left(\frac{2\pi
ik(\nu-j+1)}{m+1}\right)x^{\nu}\widetilde{\rho}_{j}(x),$

and

$\int_{S_{0}}x^{\nu}\widetilde{\rho}_{j}(x)\,\widetilde{\rho}(x)\,dx=\sum_{k=0}^{m}\left(\exp\frac{-2\pi
i(j-1-\nu)}{m+1}\right)^{k}\int_{S_{0}\cap\mathbb{R}}x^{\nu}\widetilde{\rho}_{j}(x)\,\widetilde{\rho}(x)\,dx.$

The proof is completed after realization that

$\sum_{k=0}^{m}\left(\exp\frac{-2\pi
i(j-1-\nu)}{m+1}\right)^{k}=0,\quad\nu\in\\{0,\ldots,j-2\\}.$

∎

######  Remark 2.5.

The solutions of the algebraic equations (1.9) can be obtained taking the
change of variable $\displaystyle z=\lambda\zeta-\lambda-\mu$ after solving
the equation

$\omega^{m+1}(\zeta)-\zeta\,\omega^{m}(\zeta)+\frac{\mu}{\lambda}=0,\quad
c=\frac{\mu}{\lambda}.$

Such solutions are the ones used in Section 3.

##  3\. Proof of Theorem 1.2

We divide the proof in two parts. First we show the expressions for the
elements of the sequence
$\displaystyle\left\\{Q_{n}\right\\}_{n\in\mathbb{Z}_{+}}$ in (1.11). In the
second part we prove the expressions (1.12), (1.13) and (1.14) for the vector
polynomials of the sequence $\displaystyle\left\\{{\bf
q}_{r}\right\\}_{r\in\mathbb{Z}_{+}}$.

###  3.1. Frist part

We follow the steps of [3]. Observe that the following function satisfies the
recurrence relation (1.5)

$Q_{n}=A_{0}\lambda^{n}\,\omega_{0}^{n}+A_{1}\lambda^{n}\,\omega_{1}^{n}+\cdots+A_{m}\lambda^{n}\,\omega_{m}^{n}.$

The coefficients $A_{0}(z),\ldots,A_{m}(z)$ are defined from the initial
conditions in (1.6)

$\left\\{\begin{array}[]{l}A_{0}+A_{1}+A_{2}+\cdots+A_{m}=1\\\ \\\
\displaystyle
A_{0}\lambda\,\omega_{0}+A_{1}\lambda\,\omega_{1}+A_{2}\lambda\,\omega_{2}+\cdots+A_{m}\lambda\,\omega_{m}=z+\lambda\\\
\\\ \displaystyle
A_{0}\lambda^{2}\,\omega_{0}^{2}+A_{1}\lambda^{2}\,\omega_{1}^{2}+A_{2}\lambda^{2}\,\omega_{2}^{2}+\cdots+A_{m}\lambda^{2}\,\omega_{m}^{2}=(z+\lambda)^{2},\\\
\\\ \cdots\cdots\cdots\cdots\cdots\cdots\\\ \displaystyle
A_{0}\lambda^{m}\,\omega_{0}^{m}+A_{1}\lambda^{m}\,\omega_{1}^{m}+A_{2}\lambda^{m}\,\omega_{2}^{m}+\cdots+A_{m}\lambda^{m}\,\omega_{m}^{m}=(z+\lambda)^{m}.\end{array}\right.$

From Cramer’s rule we find

(3.1)
$A_{j}=\frac{\displaystyle\prod_{k=0,k\not=j}^{m}\left(\frac{z+\lambda}{\lambda}-\omega_{k}\right)}{\displaystyle\prod_{k=0,k\not=j}^{m}(\omega_{j}-\omega_{k})},\quad
j=0,\ldots,m.$

Let us denote the polynomial in $\zeta\in\mathbb{C}$:

(3.2)
$P_{z}(\zeta)=\prod_{k=0}^{m}(\zeta-\omega_{k})=\zeta^{m+1}-\frac{z+\lambda+\mu}{\lambda}\zeta^{m}+\frac{\mu}{\lambda}.$

Then

$P_{z}\left(\frac{z+\lambda}{\lambda}\right)=\left(\frac{z+\lambda}{\lambda}\right)^{m+1}-\frac{z+\lambda+\mu}{\lambda}\left(\frac{z+\lambda}{\lambda}\right)^{m}+\frac{\mu}{\lambda},$

and

$P_{z}\left(\frac{z+\lambda}{\lambda}\right)=\frac{\mu}{\lambda}\left[1-\left(\frac{z}{\lambda}+1\right)^{m}\right].$

Since $\omega_{j},$ $j\in\\{0,\ldots,m\\}$, is a solution of (1.9) we have
that

$\frac{z+\lambda}{\lambda}=\omega_{j}+\frac{\mu}{\lambda}\left(\frac{1}{\omega_{j}^{m}}-1\right),\quad
j\in\\{0,\ldots,m\\}$

and so

$P_{z}\left(\frac{z+\lambda}{\lambda}\right)=\frac{\mu}{\lambda}\left[1-\left(\omega_{j}+\frac{\mu}{\lambda}\left(\frac{1}{\omega_{j}^{m}}-1\right)\right)^{m}\right].$

We also observe that

$\frac{z+\lambda}{\lambda}-\omega_{j}=\omega_{j}+\frac{\mu}{\lambda}\left(\frac{1}{\omega_{j}^{m}}-1\right)-\omega_{j}=\frac{\mu}{\lambda}\left(\frac{1}{\omega_{j}^{m}}-1\right).$

This implies that

$\prod_{k=0,k\not=j}^{m}\left(\frac{z+\lambda}{\lambda}-\omega_{k}\right)$
$=\frac{\displaystyle
P_{z}\left(\frac{z+\lambda}{\lambda}\right)}{\displaystyle\frac{z+\lambda}{\lambda}-\omega_{j}}=\frac{\displaystyle\frac{\mu}{\lambda}\left[1-\left(\omega_{j}+\frac{\mu}{\lambda}\left(\frac{1}{\omega_{j}^{m}}-1\right)\right)^{m}\right]}{\displaystyle\frac{\mu}{\lambda}\left(\frac{1}{\omega_{j}^{m}}-1\right)},$

hence

$\prod_{k=0,k\not=j}^{m}\left(\frac{z+\lambda}{\lambda}-\omega_{k}\right)=\frac{\displaystyle
1-\left(\omega_{j}+\frac{\mu}{\lambda}\left(\frac{1}{\omega_{j}^{m}}-1\right)\right)^{m}}{\displaystyle\frac{1}{\omega_{j}^{m}}-1}.$

Now we analyze the denominator of $A_{j}$ in (3.1)

$\prod_{k=0,k\not=j}^{m}(\omega_{j}-\omega_{k})=P_{z}^{\prime}(\omega_{j})=(m+1)\omega_{j}^{m}-\left(\frac{z+\lambda+\mu}{\lambda}\right)\,m\,\omega_{j}^{m-1}$
$\prod_{k=0,k\not=j}^{m}(\omega_{j}-\omega_{k})=\frac{m}{\omega_{j}}\left(\omega_{j}^{m+1}-\left(\frac{z+\lambda+\mu}{\lambda}\right)\,\omega_{j}^{m}\right)+\omega_{j}^{m}.$

Using the identity (1.9) we obtain that

(3.3)
$\prod_{k=0,k\not=j}^{m}(\omega_{j}-\omega_{k})=\omega_{j}^{m}-\frac{m\,\mu}{\lambda\,\omega_{j}}.$

This implies that

$A_{j}=\frac{\displaystyle\left(\omega_{j}+\frac{\mu}{\lambda}\left(\frac{1}{\omega_{j}^{m}}-1\right)\right)^{m}-1}{\displaystyle\left(\omega_{j}^{m}-1\right)\left(\omega_{j}^{m}-\frac{m\,\mu}{\lambda\,\omega_{j}}\right)}\,\,\omega_{j}^{m}.$

This completes the proof of the expression (1.11).

###  3.2. Second Part

Let us prove the expressions in (1.12), (1.13), and (1.14). Consider the dual
vector polynomials satisfying the recurrence relation in (1.8)

${\bf q}_{r}=\left(q_{0,r},\,q_{1,r},\ldots,q_{m-1,r}\right),\quad
r\in\mathbb{Z}_{+}\cup\\{-1\\}.$

From the initial conditions in the recurrence relations (1.8) we compute the
components of the vector polynomials ${\bf q}_{r}$ when $\displaystyle
r\in\left\\{m,\ldots,2m-1\right\\}$

$q_{j,r}=\frac{\lambda+x}{\mu}\delta_{j,r-m}-\frac{\lambda}{\mu}\delta_{j,r-m-1},\quad
j=0,\ldots,m-1.$

Let us analyze the expression (1.13). If we take $q_{j,r}$ as a linear
combination of the functions
$\displaystyle\left\\{\frac{1}{\omega_{0}^{r-m}},\cdots,\frac{1}{\omega_{m}^{r-m}}\right\\}$
of functions with the following form

(3.4)
$q_{j,r}=\frac{B_{j,0}}{\omega_{0}^{r-m}}+\frac{B_{j,1}}{\omega_{1}^{r-m}}+\cdots+\frac{B_{j,m}}{\omega_{m}^{r-m}},\quad
j=0,1,\ldots,m-1,\quad r\geq m.$

as components of ${\bf q}_{r}$, then these vector functions satisfy the
recurrence relation (1.8). If we require the initial conditions in (1.12), we
arrive that they must be vector polynomials. Let us find families of
coefficients $B_{j,0},\ldots,B_{j,m}$, $j=0,1,\ldots,m-1,$ using the initial
conditions with $r=m,\ldots,2\,m-1$

$\frac{B_{j,0}}{\omega_{0}^{r-m}}+\frac{B_{j,1}}{\omega_{1}^{r-m}}+\cdots+\frac{B_{j,m}}{\omega_{m}^{r-m}}=\frac{\lambda+x}{\mu}\delta_{j,r-m}-\frac{\lambda}{\mu}\delta_{j,r-m-1},\quad
j=0,\ldots,m-1.$

This above system of equations has $m$ equations with $m+1$ unknown, hence its
solution is not univocally determined. Since the components functions
$q_{j,r}$ are polynomials and taking into account $\omega_{0}(z)\to\infty$ as
$z\to\infty$ are polynomials we choose $B_{j,0}\equiv 0$.

Let us introduce the Vandermonde matrix ${\bf W}$ and the matrix ${\bf B}$ of
the coefficients $B_{j,i}$,
$\displaystyle(j,i)\in\\{0,1,\ldots,m-1\\}\times\left\\{0,1,\ldots,m\right\\}$
and a $m\times(m+1)$ matrix of ${\bf D}$ with forms

${\bf W}=\left(\begin{array}[]{c c c
c}1&\frac{1}{\omega_{0}}&\cdots&\frac{1}{\omega_{0}^{m}}\\\ &&&\\\
1&\frac{1}{\omega_{1}}&\cdots&\frac{1}{\omega_{1}^{m}}\\\ &&&\\\
\vdots&\vdots&\cdots&\vdots\\\ &&&\\\
1&\frac{1}{\omega_{m}}&\cdots&\frac{1}{\omega_{m}^{m}}\end{array}\right)\qquad{\bf
B}=\left(\begin{array}[]{ c c c c}B_{0,0}&B_{0,1}&\cdots&B_{0,m}\\\ &&\\\
B_{1,0}&B_{1,1}&\cdots&B_{1,m}\\\ &&\\\ \vdots&\vdots&\cdots&\vdots\\\ &&\\\
B_{m-1,0}&B_{m-1,1}&\cdots&B_{m-1,m}\end{array}\right),$

and

${\bf D}(z)=\left(\begin{array}[]{c c c c c c
}\frac{z+\lambda}{\mu}&-\frac{\lambda}{\mu}&0&\cdots&0&D_{0}\\\ &&&&&\\\
0&\frac{z+\lambda}{\mu}&-\frac{\lambda}{\mu}&\cdots&0&D_{1}\\\ &&&&&\\\
0&0&\frac{z+\lambda}{\mu}&\ddots&\vdots&\vdots\\\
\vdots&&\ddots&\ddots&-\frac{\lambda}{\mu}&D_{m-2}\\\ &&&&&\\\
0&0&&0&\frac{z+\lambda}{\mu}&D_{m-1}\end{array}\right),$

where the entries $D_{j}$, $j=0,\ldots,m-1$ guaranty that $B_{j,0}\equiv 0,$
$j=0,\ldots,m-1$.

We solve the matrix equation ${\bf B}{\bf W}={\bf D}$, hence ${\bf B}={\bf
D}{\bf W}^{-1}$. Using [28] we compute an explicit expression for
$\displaystyle W^{-1}_{j,k}$, $\displaystyle(j,k)\in\\{0,\ldots,m\\}^{2}$
which are the entries of inverse matrix of ${\bf W}$ that we denote
$\displaystyle{\bf W}^{-1}$, hence

(3.5)
$\begin{array}[]{l}0=B_{j,0}=\frac{z+\lambda}{\mu}W^{-1}_{j,0}-\frac{\lambda}{\mu}W^{-1}_{j+1,0}+D_{j}\,W^{-1}_{m,0},\quad
j=0,\ldots,m-2,\\\ \\\
0=B_{m-1,0}=\frac{z+\lambda}{\mu}W^{-1}_{m-1,0}+D_{m-1}\,W^{-1}_{m,0},\end{array}$

Consider the polynomials

$p_{k}(\zeta)=\sum_{j=0}^{m}W^{-1}_{j,k}\zeta^{j},\quad k=0,\ldots,m.$

By definition we have that for each $\displaystyle
k\in\left\\{0,\ldots,m\right\\}$

$p_{k}\left(\frac{1}{\omega_{l}}\right)=\sum_{j=0}^{m}W^{-1}_{j,k}\left(\frac{1}{\omega_{l}}\right)^{j}=\delta_{l,k},\quad
l=0,\ldots,m.$

From Lagrange interpolation formula we have that

$p_{k}(\zeta)=\sum_{j=0}^{m}W^{-1}_{j,k}\zeta^{j}=\prod_{l=0,l\not=k}^{m}\frac{\displaystyle\zeta-\frac{1}{\omega_{l}}}{\displaystyle\frac{1}{\omega_{k}}-\frac{1}{\omega_{l}}}.$

This implies that
$\displaystyle\prod_{l=0,j\not=k}^{m}\left(\frac{1}{\omega_{k}}-\frac{1}{\omega_{l}}\right)p_{k}$
is a monic polynomial, and from Vieta’s formulae we obtain that its
coefficients are

(3.6)
$\displaystyle\prod_{l=0,l\not=k}^{m}\left(\frac{1}{\omega_{k}}-\frac{1}{\omega_{l}}\right)W^{-1}_{j,k}=\left(-1\right)^{j+1}\sum_{0\leq\ell_{0}<\cdots<\ell_{m-j-1}\leq
m}\frac{1}{\omega_{\ell_{0}}}\cdots\frac{1}{\omega_{\ell_{m-j-1}}},$

where $\displaystyle\ell_{\tau}\not=k,$ $\tau=0,\ldots,m-j-1$ with
$j=0,\ldots,m-1$, and when $j=m$,

$\displaystyle\prod_{l=0,l\not=k}^{m}\left(\frac{1}{\omega_{k}}-\frac{1}{\omega_{l}}\right)W^{-1}_{m,k}=1,\quad
k=0,\ldots,m.$

Let us consider the factors

$\displaystyle\prod_{l=0,l\not=k}^{m}\left(\frac{1}{\omega_{k}}-\frac{1}{\omega_{l}}\right)=\frac{1}{\displaystyle\omega_{k}^{m}\prod_{l=0,l\not=k}^{m}\omega_{l}}\prod_{l=0,l\not=k}^{m}\left(\omega_{l}-\omega_{k}\right).$

From (3.2) we obtain that

$P_{z}^{\prime}(\omega_{k})=(-1)^{m}\prod_{l=0,l\not=k}^{m}\left(\omega_{l}-\omega_{k}\right),$

and analogously to (3.3) we have that

$\prod_{l=0,l\not=k}^{m}\left(\omega_{l}-\omega_{k}\right)=(-1)^{m}\left(\omega_{k}^{m}-\frac{m\mu}{\lambda\omega_{k}}\right),$

which implies that

$\displaystyle\prod_{l=0,l\not=k}^{m}\left(\frac{1}{\omega_{k}}-\frac{1}{\omega_{l}}\right)=\frac{(-1)^{m}}{\displaystyle\omega_{k}^{m}\prod_{l=0,l\not=k}^{m}\omega_{l}}\left(\omega_{k}^{m}-\frac{m\mu}{\lambda\omega_{k}}\right).$

Let us now consider the other side in (3.6). Fix $j\in\\{0,1,\ldots,m-1\\}$.

$\frac{(-1)^{m}}{\displaystyle\omega_{k}^{m}\prod_{l=0,l\not=k}^{m}\omega_{l}}\left(\omega_{k}^{m}-\frac{m\mu}{\lambda\omega_{k}}\right)W^{-1}_{j,k}=\frac{\left(-1\right)^{j+1}}{\displaystyle\prod_{l=0,l\not=k}^{m}\omega_{l}}\sum_{0\leq\ell_{0}<\cdots<\ell_{j}\leq
m}\omega_{\ell_{0}}\,\cdots\omega_{\ell_{j}},$

$\displaystyle\ell_{\tau}\not=k,$ $\tau=0,\ldots,j$ with $j=0,\ldots,m-1$.
Hence

(3.7)
$(-1)^{m}\left(1-\frac{m\mu}{\lambda\omega_{k}^{m+1}}\right)W^{-1}_{j,k}=\left(-1\right)^{j+1}\sum_{0\leq\ell_{0}<\cdots<\ell_{j}\leq
m}\omega_{\ell_{0}}\,\cdots\omega_{\ell_{j}}.$

When $j=m$ we obtain that

$\frac{(-1)^{m}}{\displaystyle\omega_{k}^{m}\prod_{l=0,l\not=k}^{m}\omega_{l}}\left(\omega_{k}^{m}-\frac{m\mu}{\lambda\omega_{k}}\right)W^{-1}_{m,k}=1,$

which gives

(3.8)
$W^{-1}_{m,k}=\frac{\mu}{\lambda}\frac{\omega_{k}^{m}}{\displaystyle\omega_{k}^{m+1}-\frac{m\mu}{\lambda}},\quad
k=0,1,\ldots,m.$

For the case $0\leq j\leq m-1$, the equality (3.7) and Vieta’s formulae yield
the following chain of equalities

$(-1)^{j+1}\sum_{0\leq\ell_{0}<\cdots<\ell_{j}\leq
m}\omega_{\ell_{0}}\,\omega_{\ell_{1}}\cdots\omega_{\ell_{j}}$
$=(-1)^{j+2}\,\omega_{k}\sum_{0\leq\ell_{0}<\cdots<\ell_{j-1}\leq
m}\omega_{\ell_{0}}\,\omega_{\ell_{1}}\cdots\omega_{\ell_{j-1}}$
$=(-1)^{j+3}\,\omega_{k}^{2}\sum_{0\leq\ell_{0}<\cdots<\ell_{j-2}\leq
m}\omega_{\ell_{0}}\,\omega_{\ell_{1}}\cdots\omega_{\ell_{j-2}}$
$=\cdots=(-1)^{2j+1}\omega_{k}^{j}\sum_{l=0,l\not=k}\omega_{l}=\omega_{k}^{j}\,\frac{z+\lambda+\mu}{\lambda}=\frac{\omega_{k}^{m+1}+\frac{\mu}{\lambda}}{\omega_{k}^{m-j}}.$

Hence

(3.9)
$W^{-1}_{j,k}=\frac{\mu}{\lambda}\frac{\omega_{k}^{j}\left(\omega_{k}^{m+1}+\frac{\mu}{\lambda}\right)}{\displaystyle\omega_{k}^{m+1}-\frac{m\mu}{\lambda}},\quad\left(j,k\right)\in\\{0,1,\ldots,m-1\\}\times\\{0,1,\ldots,m\\}.$

Combining (3.5), (3.8) and (3.9) we obtain that

$\begin{array}[]{l}\displaystyle
D_{j}=\frac{1}{\omega_{0}^{m-j}}\left[\frac{1}{\omega_{0}}-\frac{\lambda}{\mu}\left(1-\omega_{0}\right)-\frac{1}{\omega_{0}^{m+1}}\right]\left(\omega_{0}^{m+1}+\frac{\mu}{\lambda}\right),\\\
\\\ \hfill j=0,\ldots,m-2,\\\ \\\ \mbox{and when }j=m-1\\\ \\\ \displaystyle
D_{m-1}=\frac{1}{\omega_{0}}\left(\frac{1}{\omega_{0}}-\frac{\lambda}{\mu}-\frac{1}{\omega_{0}^{m+1}}\right)\left(\omega_{0}^{m+1}+\frac{\mu}{\lambda}\right).\end{array}$

We are now ready to find the entries of the matrix ${\bf B}$ for
$(j,k)\in\\{0,1,\ldots,m-2\\}\times\\{1,\ldots,m\\}$

$B_{j,k}=\frac{z+\lambda}{\mu}W_{j,k}^{-1}-\frac{\lambda}{\mu}W_{j+1,k}^{-1}+D_{j}W_{m,k}^{-1}$
$=\frac{\mu}{\lambda}\frac{\omega_{k}^{j}}{\omega_{k}^{m+1}-\frac{m\mu}{\lambda}}\left[\left(\omega_{k}^{m+1}+\frac{\mu}{\lambda}\right)\left(\frac{1}{\omega_{k}^{m+1}}+\frac{\lambda}{\mu}\left(1-\omega_{k}\right)-\frac{1}{\omega_{k}}\right)\right.$
$\left.-\frac{\omega_{k}^{m-j}}{\omega_{0}^{m-j}}\left(\omega_{0}^{m+1}+\frac{\mu}{\lambda}\right)\left(\frac{1}{\omega_{0}^{m+1}}+\frac{\lambda}{\mu}\left(1-\omega_{0}\right)-\frac{1}{\omega_{0}}\right)\right],$

and

$B_{m-1,k}=\frac{z+\lambda}{\mu}W^{-1}_{m-1,0}+D_{m-1}\,W^{-1}_{m,0}$
$=\frac{\mu}{\lambda}\frac{\omega_{k}^{m-1}}{\omega_{k}^{m+1}-\frac{m\mu}{\lambda}}\left(\frac{1}{\omega_{0}^{m+1}}-\frac{1}{\omega_{0}}+\frac{\lambda}{\mu}\right)\left[\omega_{k}^{m+1}+\frac{\mu}{\lambda}-\frac{\omega_{k}}{\omega_{0}}\left(\omega_{0}^{m+1}+\frac{\mu}{\lambda}\right)\right]$

From (3.4) we have that for each $r\geq 2\,m$

$q_{j,r}=\frac{\lambda}{\mu}\sum_{k=1}^{m}\frac{1}{\omega_{k}^{r-m-j}(\omega_{k}^{m+1}-\frac{m\mu}{\lambda})}\left[\left(\omega_{k}^{m+1}+\frac{\mu}{\lambda}\right)\left(\frac{1}{\omega_{k}^{m+1}}+\frac{\lambda}{\mu}\left(1-\omega_{k}\right)-\frac{1}{\omega_{k}}\right)\right.$
$\left.-\frac{\omega_{k}^{m-j}}{\omega_{0}^{m-j}}\left(\omega_{0}^{m+1}+\frac{\mu}{\lambda}\right)\left(\frac{1}{\omega_{0}^{m+1}}+\frac{\lambda}{\mu}\left(1-\omega_{0}\right)-\frac{1}{\omega_{0}}\right)\right],\quad
j=0,\ldots,m-2,$

and

$q_{m-1,r}=\frac{\lambda}{\mu}\sum_{k=1}^{m}\frac{1}{\omega_{k}^{r+1}(\omega_{k}^{m+1}-\frac{m\mu}{\lambda})}\left(\frac{1}{\omega_{0}^{m+1}}-\frac{1}{\omega_{0}}+\frac{\lambda}{\mu}\right)\left[\omega_{k}^{m+1}+\frac{\mu}{\lambda}-\frac{\omega_{k}}{\omega_{0}}\left(\omega_{0}^{m+1}+\frac{\mu}{\lambda}\right)\right],$

which shows (1.13) and (1.14). This completes the proof of Theorem 1.2.

##  4\. Bi-orthogonality and resolvent functions

In this Section we follow ideas from [3, 4] to arrive at necessary results to
prove Theorem 1.3 and Theorem 1.4 in Section 6 and Section 7, respectively.
Set $\eta>0$, $\iota>0$ $\xi\in\mathbb{R}_{+}$, $\gamma\in\mathbb{R}$, and
consider the sequence of polynomials
$\displaystyle\left\\{H_{n}\right\\}_{n\in\mathbb{Z}_{+}}$ whose elements
satisfy the recurrence relation

(4.1) $(\xi+x)H_{n}(x)=\iota H_{n+1}(x)+\eta H_{n-m}(x),\quad n\geq m,$

with initial conditions

$H_{n}(x)=\frac{1}{\iota^{n}}(\gamma+x)^{n},\quad n=0,\ldots,m.$

Let us consider the associated linear operator ${\bf H}$ acting on the space
of sequences $\ell^{2}$, which is determined by a infinite non-symmetric
matrix $\displaystyle{\bf H}=[h_{i,j}]_{(i,j)\in\mathbb{Z}_{+}^{2}}$ with
entries satisfying

$h_{i,j}=\left\\{\begin{array}[]{r c l}\eta&\mbox{if}&j=i-m,\\\ &&\\\
-\gamma&\mbox{if}&j=i<m,\\\ &&\\\ -\xi&\mbox{if}&j=i\geq m,\\\ &&\\\
\iota&\mbox{if}&j=i+1,\\\ &&\\\
0&\mbox{if}&j\not\in\\{i+1,i,i-m\\},\end{array}\right.\quad(i,j)\in\mathbb{Z}_{+}^{2},$

with schematic form

(4.2) ${\bf H}=\left(\begin{array}[]{c c c c c c c c
c}-\gamma&\iota&0&\cdots&0&0&0&0&\cdots\\\ 0&-\gamma&\iota&&0&0&0&0&\\\
\vdots&\ddots&\ddots&\ddots&\vdots&&\vdots&\vdots&\\\ &&&&&&&&\\\
0&0&\cdots&0&-\gamma&\iota&0&0&\cdots\\\
\eta&0&\cdots&0&0&-\xi&\iota&0&\ddots\\\ 0&\eta&&0&0&0&-\xi&\iota&\ddots\\\
\vdots&\ddots&\ddots&&\ddots&&\ddots&\ddots&\ddots\\\ \end{array}\right).$

Set $\\{{\bf e}_{j}\\}_{j\in\mathbb{Z}_{+}}$ the standard basis of $\ell^{2}$,
then the operator ${\bf H}$ is defined as follows

(4.3) ${\bf H}^{\top}{\bf e}_{j}=\left\\{\begin{array}[]{l l}\iota{\bf
e}_{j+1}-\gamma{\bf e}_{j},&j\in\\{0,1,2,\ldots,m-1\\},\\\ &\\\ \iota{\bf
e}_{j+1}-\xi{\bf e}_{j}+\eta{\bf e}_{j-m},&j\geq m,\end{array}\right.$

where $\top$ denotes the operator of matrix transposition.

Let $p(x)=b_{\kappa}\,x^{\kappa}+\cdots+b_{1}\,x+b_{0}$ be a polynomial with
an arbitrary degree $\kappa$. Then $p({\bf H})$ denotes the operator

$p({\bf H}^{\top}){\bf u}=\left(b_{\kappa}{\bf
H}^{\top\kappa}+\cdots+b_{1}\,{\bf H}^{\top}+b_{0}\right){\bf
u}=b_{\kappa}{\bf H}^{\top\kappa}{\bf u}+\cdots+b_{1}\,{\bf H}^{\top}{\bf
u}+b_{0}{\bf u},\quad{\bf u}\in\ell^{2}.$

######  Lemma 4.1.

Consider the sequence of polynomials
$\displaystyle\left\\{H_{n}\right\\}_{n\in\mathbb{Z}_{+}}$ satisfying the
recurrence relation (4.1). Then

(4.4) ${\bf e}_{n}=H_{n}({\bf H}^{\top})\,{\bf e}_{0},\quad
n\in\mathbb{Z}_{+}.$

###### Proof.

For each $\displaystyle n\in\left\\{0,1,\ldots,m-1\right\\}$ we have that
$\displaystyle H_{n}(x)=\frac{1}{\iota^{n}}(\gamma+x)^{n}$, then
$\displaystyle H_{n}({\bf H}^{\top})=\frac{1}{\iota^{n}}({\bf
H}^{\top}+\gamma)^{n}$. This means that

$\begin{array}[]{r c l}H_{0}({\bf H}^{\top})\,{\bf
e}_{0}&=&\displaystyle\frac{1}{\iota^{0}}\left({\bf
H}^{\top}+\gamma\right)^{0}{\bf e}_{0}={\bf e}_{0},\\\ &&\\\ H_{1}({\bf
H}^{\top})\,{\bf e}_{0}&=&\displaystyle\frac{1}{\iota}\left({\bf
H}^{\top}+\gamma\right){\bf e}_{0}=\frac{1}{\iota}\left(\iota{\bf
e}_{1}-\gamma{\bf e}_{0}+\gamma{\bf e}_{0}\right)={\bf e}_{1},\\\ &&\\\
H_{2}({\bf H}^{\top})\,{\bf e}_{0}&=&\displaystyle\frac{1}{\iota^{2}}({\bf
H}^{\top}+\gamma)^{2}{\bf e}_{0}=\frac{1}{\iota}\left({\bf
H}^{\top}+\gamma\right){\bf e}_{1}=\frac{1}{\iota}\left(\iota{\bf
e}_{2}-\gamma{\bf e}_{1}+\gamma{\bf e}_{1}\right)={\bf e}_{2},\\\ &&\\\
&\vdots&\\\ &&\\\ H_{m-1}({\bf H}^{\top})\,{\bf
e}_{0}&=&\displaystyle\frac{1}{\iota^{m-1}}({\bf H}^{\top}+\gamma)^{m-1}{\bf
e}_{0}\\\ &&\\\ &=&\displaystyle\frac{1}{\iota^{m-2}}({\bf
H}^{\top}+\gamma)^{m-2}{\bf e}_{1}=\cdots=\frac{1}{\iota}\left({\bf
H}^{\top}+\gamma\right){\bf e}_{m-2}={\bf e}_{m-1}.\end{array}$

We now consider $n\geq m-1$. For each $k=1,\ldots,n$, assume that $H_{k}({\bf
H}^{\top})\,{\bf e}_{0}={\bf e}_{k}$. By the recurrence relation (4.1), we
have that

$H_{n+1}({\bf H}^{\top}){\bf e}_{0}=\frac{1}{\iota}\left[({\bf
H}^{\top}+\xi)H_{n}({\bf H}^{\top})-\eta H_{n-m}({\bf H}^{\top})\right]{\bf
e}_{0}.$

hence

$H_{n+1}({\bf H}^{\top})=\frac{1}{\iota}\left[\iota{\bf e}_{n+1}-\xi{\bf
e}_{n}+\xi{\bf e}_{n}+\eta{\bf e}_{j-m}-\eta{\bf e}_{n-m}\right]={\bf
e}_{n+1}.$

This completes the proof by induction. ∎

Let us recall the family of polynomials
$\displaystyle\left\\{Q_{n}\right\\}_{n\in\mathbb{Z}_{+}}$ defined by the
recurrence relation (1.5), which is associated to the infinite non-symmetric
matrix ${\bf A}$ in (1.4), determining an operator ${\bf A}$ in the space of
sequences $\ell^{2}$ such as

${\bf A}^{\top}{\bf e}_{j}=\lambda{\bf e}_{j+1}-\lambda{\bf e}_{j}\quad
j\in\\{0,1,2,\ldots,m-1\\}$

and

${\bf A}^{\top}{\bf e}_{j}=\lambda{\bf e}_{j+1}-(\lambda+\mu){\bf
e}_{j}+\mu{\bf e}_{j-m},\quad j\geq m.$

We also introduce two sequences of polynomials
$\displaystyle\left\\{L_{n}\right\\}_{n\in\mathbb{Z}_{+}}$ and
$\displaystyle\left\\{T_{n}\right\\}_{n\in\mathbb{Z}_{+}}$ that we use in
Section 6. The sequence
$\displaystyle\left\\{L_{n}\right\\}_{n\in\mathbb{Z}_{+}}$ is obtained from
$\displaystyle\left\\{Q_{n}\right\\}_{n\in\mathbb{Z}_{+}}$ and
$\displaystyle\left\\{T_{n}\right\\}_{n\in\mathbb{Z}_{+}}$ is a variation of
$\displaystyle\left\\{L_{n}\right\\}_{n\in\mathbb{Z}_{+}}$. Consider the
change of variables $z=x+\lambda+\mu$ and define

$L_{n}(z)=\lambda^{n}Q_{n}(z-\lambda-\mu),\quad n\in\mathbb{Z}_{+}.$

From (1.5) and (1.6) we deduce the recurrence relation

(4.5) $z\,L_{n}(z)=L_{n+1}(z)+\mu\lambda^{m}L_{n-m},\quad n\geq m,$

with initial conditions

(4.6) $L_{n}(z)=(z-\mu)^{n},\quad n=0,\ldots,m-1.$

The sequence $\displaystyle\left\\{T_{n}\right\\}_{n\in\mathbb{Z}_{+}}$
satisfies the recurrence relation (4.5)

(4.7) $z\,T_{n}(z)=T_{n+1}(z)+\mu\lambda^{m}T_{n-m},\quad n\geq m,$

with different initial conditions

(4.8) $T_{n}(z)=z^{n},\quad n=0,\ldots,m-1.$

Sequences of polynomials with recurrence relations of
$\displaystyle\\{T_{n}\\}_{n\in\mathbb{Z}_{+}}$ are analyzed in several
publications such as [3, 4, 21, 22, 23]. Section 5 deals specifically about
these polynomials.

We now introduce the two linear operators ${\bf L}$ and ${\bf T}$
corresponding to the recurrence relations of
$\displaystyle\\{L_{n}\\}_{n\in\mathbb{Z}_{+}}$ and
$\displaystyle\\{T_{n}\\}_{n\in\mathbb{Z}_{+}}$, respectively:

(4.9) ${\bf L}^{\top}{\bf e}_{j}={\bf e}_{j+1}-\mu{\bf e}_{j},\quad 0\leq
j\leq m-1,\quad{\bf L}^{\top}{\bf e}_{j}={\bf e}_{j+1}+\mu\lambda^{m}{\bf
e}_{j-m},\quad j\geq m,$

and

${\bf T}^{\top}{\bf e}_{j}={\bf e}_{j+1},\quad 0\leq j\leq m-1,\quad{\bf
T}^{\top}{\bf e}_{j}={\bf e}_{j+1}+\mu\lambda^{m}{\bf e}_{j-m},\quad j\geq m.$

The following statements are particular cases of Lemma 4.1

######  Lemma 4.2.

Consider the sequences of polynomials
$\displaystyle\left\\{Q_{n}\right\\}_{n\in\mathbb{Z}_{+}}$,
$\displaystyle\left\\{L_{n}\right\\}_{n\in\mathbb{Z}_{+}}$, and
$\displaystyle\left\\{T_{n}\right\\}_{n\in\mathbb{Z}_{+}}$ satisfying the
recurrence relations (1.5), (4.5), and (4.7), respectively, and their
corresponding initial conditions (1.6), (4.6), and (4.8). Then

(4.10) ${\bf e}_{n}=Q_{n}({\bf A}^{\top})\,{\bf e}_{0}=L_{n}({\bf
L}^{\top})\,{\bf e}_{0}=H_{n}({\bf H}^{\top})\,{\bf e}_{0},\quad
n\in\mathbb{Z}_{+}.$

###### Proof.

Proving Lemma 4.2 deduces immediately after substituting change of variables
in Lemma 4.1

  1.   2. (1)

for $\displaystyle\\{Q_{n}\\}_{n\in\mathbb{Z}_{+}}$, $\gamma=\iota=\lambda$,
$\eta=\mu$, and $\xi=\lambda+\mu$,

  3.   4. (2)

for $\displaystyle\\{L_{n}\\}_{n\in\mathbb{Z}_{+}}$, $\gamma=-\mu$, $\iota=1$,
$\eta=\mu\lambda^{m}$, and $\xi=0$,

  5.   6. (3)

For $\displaystyle\\{T_{n}\\}_{n\in\mathbb{Z}_{+}}$, $\gamma=\xi=0$,
$\iota=1$, and $\eta=\mu\lambda^{m}$.

  7.

∎

######  Remark 4.3.

The matrix expressions of the operators ${\bf A}$ and ${\bf L}$ are related as
follows

(4.11) ${\bf A}={\bf\Lambda}^{-1}\left[{\bf
L}-(\lambda+\mu)\mathbb{I}\right]{\bf\Lambda},$

where

${\bf\Lambda}=\left(\begin{array}[]{c c c c}1&0&0&\cdots\\\ 0&\lambda&0&\\\
0&0&\lambda^{2}&\ddots\\\ \vdots&&\ddots&\ddots\end{array}\right).$

This relationship between ${\bf L}$ and ${\bf A}$ is used in Section 6.

Recall the sequence of polynomial vectors $\displaystyle\left\\{{\bf
q}_{r}=\left(q_{0,r},\ldots,q_{m-1,r}\right)\right\\}_{r\in\mathbb{Z}_{+}\cup\\{-1\\}}$
which is defined by the recurrence relation (1.8). Set an arbitrary vector
polynomial $\displaystyle{\bf p}=\left(p_{0},\ldots,p_{m-1}\right)$, let us
denote for every system ${\bf U}=\left({\bf u}_{0},\ldots,{\bf
u}_{m-1}\right)$ with ${\bf u}_{j}\in\ell^{2}$, $j=0,\ldots,m-1$ the ”dot”
product

${\bf p}({\bf A})\bigodot{\bf U}=\sum_{j=0}^{m-1}p_{j}({\bf A}){\bf u}_{j}.$

######  Lemma 4.4.

Let $\displaystyle{\bf E}=\left({\bf e}_{0},\ldots,{\bf e}_{m-1}\right)$
denote. Then

${\bf e}_{r}={\bf q}_{r}({\bf A})\bigodot{\bf E},\quad
r\in\mathbb{Z}_{+}=\\{0,1,2,\ldots\\}.$

###### Proof.

We observe that for each $\displaystyle r\in\left\\{0,1,\ldots,m-1\right\\}$

${\bf q}_{r}({\bf A})\bigodot{\bf
E}=(\overset{r+1}{\overbrace{0,\ldots,0}},1,0,\ldots,0)\bigodot({\bf
e}_{0},\ldots,{\bf e}_{m-1})={\bf e}_{r}.$

Let us consider $2m-1\geq r\geq m-1$. For each $k=0,\ldots,r-1$, let us assume
that $\displaystyle{\bf q}_{k}({\bf A})\bigodot{\bf E}={\bf e}_{k}$. By the
recurrence relation (1.8), we have that

${\bf q}_{r}({\bf A})\bigodot{\bf E}=\frac{1}{\mu}\left[\left(\lambda+{\bf
A}\right){\bf q}_{r-m}({\bf A})\bigodot{\bf E}-\lambda{\bf q}_{r-m-1}({\bf
A})\bigodot{\bf E}\right]$ $=\frac{1}{\mu}\left[\lambda{\bf q}_{r-m}({\bf
A})\bigodot{\bf E}+{\bf A}{\bf q}_{r-m}({\bf A})\bigodot{\bf E}-\lambda{\bf
q}_{r-m-1}({\bf A})\bigodot{\bf E}\right]$ $=\frac{1}{\mu}\left[\lambda{\bf
e}_{r-m}+{\bf A}{\bf e}_{r-m}-\lambda{\bf e}_{r-m-1}\right]$
$=\frac{1}{\mu}\left[\lambda{\bf e}_{r-m}-\lambda{\bf e}_{r-m}+\lambda{\bf
e}_{r-m-1}+\mu{\bf e}_{r}-\lambda{\bf e}_{r-m-1}\right]={\bf e}_{r}.$

We now consider $r\geq 2m-1$

${\bf q}_{r}({\bf A})\bigodot{\bf E}=\frac{1}{\mu}\left[\left(\lambda+\mu+{\bf
A}\right){\bf q}_{r-m}({\bf A})\bigodot{\bf E}-\lambda{\bf q}_{r-m-1}({\bf
A})\bigodot{\bf E}\right]$ $=\frac{1}{\mu}\left[(\lambda+\mu){\bf
q}_{r-m}({\bf A})\bigodot{\bf E}+{\bf A}{\bf q}_{r-m}({\bf A})\bigodot{\bf
E}-\lambda{\bf q}_{r-m-1}({\bf A})\bigodot{\bf E}\right]$
$=\frac{1}{\mu}\left[(\mu+\lambda){\bf e}_{r-m}+{\bf A}{\bf
e}_{r-m}-\lambda{\bf e}_{r-m-1}\right]$ $=\frac{1}{\mu}\left[(\mu+\lambda){\bf
e}_{r-m}-(\mu+\lambda){\bf e}_{r-m}+\iota{\bf e}_{r-m-1}+\mu{\bf
e}_{r}-\lambda{\bf e}_{r-m-1}\right]={\bf e}_{r}.$

The proof is completed by induction. ∎

In $\ell^{2}$ we consider the regular inner product. This means that given two
sequences ${\bf u}=\left\\{u_{j}\in\mathbb{R}\right\\}_{j\in\mathbb{Z}_{+}}$
and ${\bf v}=\left\\{v_{j}\in\mathbb{R}\right\\}_{j\in\mathbb{Z}_{+}}$ of
$\ell^{2}$

${\bf u}\cdot{\bf v}=\sum_{j=0}^{\infty}u_{j}v_{j}={\bf u}^{\top}{\bf v}.$

We think the sequences as infinite column vectors.

######  Proposition 4.5.

Set ${\bf E}=({\bf e}_{0},\ldots,{\bf e}_{m-1})$. Given an order pair
$(n,r)\in\mathbb{Z}_{+}^{2}$, the polynomial $Q_{n}$ in (1.5) and ${\bf
q}_{r}$ in (1.8) satisfy the orthogonality relations:

(4.12) ${\bf q}_{r}({\bf A})\bigodot{\bf E}\cdot Q_{n}({\bf A}^{\top}){\bf
e}_{0}=\delta_{n,r}.$

###### Proof.

From Lemma 4.1 and Lemma 4.4 we have immediately

${\bf q}_{r}({\bf A})\bigodot{\bf E}\cdot Q_{n}({\bf A}^{\top}){\bf
e}_{0}={\bf e}_{r}\cdot{\bf e}_{n}=\delta_{r,n}.$

This proves (4.12). ∎

The following results Lemma 4.6 and Lemma 4.7 refers the cases of ${\bf L}$
and ${\bf T}$. This means that ${\bf H}$ is such that $\eta=\mu\lambda^{m}$,
$\xi=0$, and $\iota=1$, while $\gamma$ takes the values either $\gamma=\mu$
for ${\bf L}$ or $\gamma=0$ for ${\bf T}$.

######  Lemma 4.6.

Set $\displaystyle j\in\left\\{1,\ldots,m\right\\}$. For each
$\displaystyle\nu\in\\{0,1,\ldots,j-1\\}$

(4.13) $\displaystyle({\bf H}+\gamma\mathbb{I})^{\nu}{\bf
e}_{j-1}=\sum_{k=0}^{\nu}{\nu\choose k}\left(\mu\lambda^{m}\right)^{\nu-k}{\bf
e}_{(\nu-k)m+j-1-k}.$

###### Proof.

We proceed by induction. In case $\nu=0$ according to (4.3) we have that

$({\bf H}+\gamma\mathbb{I})^{0}{\bf e}_{j-1}={\bf e}_{j-1}.$

Then assuming that the statement (4.13) is true for a $\nu\in\\{0,1,\ldots
m-2\\}$ we prove it for $\nu+1$.

$\displaystyle({\bf H}+\gamma\mathbb{I})^{\nu+1}{\bf e}_{j-1}=({\bf
H}+\gamma\mathbb{I})({\bf H}+\mu\mathbb{I})^{\nu}{\bf e}_{j-1}$ $=({\bf
H}+\gamma\mathbb{I})\sum_{k=0}^{\nu}{\nu\choose
k}\left(\lambda\mu^{m}\right)^{\nu-k}{\bf e}_{(\nu-k)m+j-1-k}$
$=\sum_{k=0}^{\nu}{\nu\choose k}\left(\mu\lambda^{m}\right)^{\nu-k}({\bf
H}+\iota\mathbb{I})\,{\bf e}_{(\nu-k)m+j-1-k}$
$\displaystyle=\sum_{k=0}^{\nu}{\nu\choose
k}\left(\mu\lambda^{m}\right)^{\nu-k}\left({\bf
e}_{(\nu-k)m+j-1-k-1}+\mu\lambda^{m}{\bf e}_{(\nu-k+1)m+j-1-k}\right)$
$\displaystyle=\sum_{k=0}^{\nu}{\nu\choose
k}\left(\mu\lambda^{m}\right)^{\nu-k}{\bf
e}_{(\nu-k)m+j-1-k-1}+\sum_{k=0}^{\nu}{\nu\choose
k}\left(\mu\lambda^{m}\right)^{\nu-k+1}{\bf e}_{(\nu-k+1)m+j-1-k}$
$\displaystyle=\sum_{k=1}^{\nu+1}{\nu\choose
k-1}\left(\mu\lambda^{m}\right)^{\nu-k+1}{\bf e}_{(\nu-k+1)m+j-1-k}$
$+\sum_{k=0}^{\nu}{\nu\choose k}\left(\mu\lambda^{m}\right)^{\nu-k+1}{\bf
e}_{(\nu-k+1)m+j-1-k}$ $\displaystyle={\bf
e}_{j-1-\nu-1}+\sum_{k=1}^{\nu}\left({\nu\choose k-1}+{\nu\choose
k}\right)\left(\lambda\mu^{m}\right)^{\nu-k+1}{\bf e}_{(\nu-k+1)m+j-1-k}$
$+\left(\lambda\mu^{m}\right)^{\nu+1}{\bf e}_{j-1-\nu-1}$

Now using Pascal’s identity we obtain that

$\displaystyle({\bf H}+\gamma\mathbb{I})^{\nu+1}{\bf e}_{j-1}=$
$\displaystyle={\bf e}_{j-1-\nu-1}+\sum_{k=1}^{\nu}{\nu+1\choose
k}\left(\mu\lambda^{m}\right)^{\nu-k+1}{\bf
e}_{(\nu-k+1)m+j-1-k}+\left(\mu\lambda^{m}\right)^{\nu+1}{\bf e}_{j-1-\nu-1}$
$=\sum_{k=0}^{\nu+1}{\nu+1\choose k}\left(\mu\lambda^{m}\right)^{\nu+1-k}{\bf
e}_{(\nu+1-k)m+j-1-k}.$

This completes the proof by induction. ∎

######  Lemma 4.7.

The linear operator is ${\bf H}$ is bounded in $\ell^{2}$, and its support
satisfies that

$\mbox{supp}({\bf H})\subset\left\\{z\in\mathbb{C}:|z|\leq
1+\mu\lambda^{m}\right\\}\cup\left\\{z\in\mathbb{C}:|z+\gamma|\leq
1+\mu\lambda^{m}\right\\}.$

###### Proof.

Set ${\bf H}_{n}=(H_{0},\ldots,H_{n-1})^{\top}$, the vector ${\bf
e}_{n}=(0,0\ldots,0,1)^{\top}$ with $n$ components, and the $n\times n$ matrix
block $\displaystyle{\bf H}_{n}=[h_{i,j}]_{(i,j)\in\\{0,1,\ldots,n-1\\}^{2}}$
of the infinite matrix form corresponding to ${\bf H}$, where for each
$i\in\\{1,\ldots,n-1\\}$

$h_{i,i+1}=1,\quad h_{i+m,i}=\mu\lambda^{m},\quad h_{i,i}=-\gamma,\quad
i\in\\{0,\ldots,m\\}\quad\mbox{otherwise}\quad h_{i,j}=0,$

schematically

${\bf H}_{n}^{\top}=\left(\begin{array}[]{c c c c c c c c c c c c
}-\gamma&0&0&\cdots&0&\mu\lambda^{m}&0&0&\cdots&0&0\\\
1&-\gamma&0&\cdots&0&0&\mu\lambda^{m}&0&\cdots&0&0\\\
0&1&-\gamma&\cdots&0&0&0&\mu\lambda^{m}&\ddots&0&0\\\
0&0&1&\cdots&0&0&0&0&&0&0\\\ \vdots&&\ddots&&\vdots&&\vdots&&\ddots&\ddots&\\\
0&0&\cdots&1&-\gamma&0&0&0&&0&\mu\lambda^{m}\\\
0&0&\cdots&0&1&0&0&0&\cdots&0&0&\\\ 0&0&\cdots&0&0&1&0&0&\cdots&0&0\\\
\vdots&&&\vdots&\ddots&\ddots&\ddots&\ddots&&\vdots&\vdots\\\
0&0&\cdots&0&\cdots&0&0&1&0&0&0\\\ 0&0&\cdots&0&\cdots&0&0&0&1&0&0\\\
0&0&\cdots&0&\cdots&0&0&0&0&1&0\\\ \end{array}\right).$

From the recurrence relations in (4.5) and (4.6) we observe that $\deg
H_{n}=n$ then $H_{n}$ has $n$ zeros. Let $x_{k}$ denote the zeros of $H_{n}$.
The relations (4.5) and (4.6) can be rewrite as follows

$z{\bf H}_{n}(z)={\bf H}_{n}\,(z)+H_{n}(z){\bf e}_{n}\implies x_{k}{\bf
H}_{n}(h_{k})={\bf H}_{n}\,(x_{k}),$

which implies that $x_{k}$, $k=1,\ldots,n,$ are the $n$ eigenvalues of ${\bf
H}_{n}$. Now using Gershgorin Disc Theorem (see [31]) we obtain that
$|x_{j}+\gamma|\leq 1+\mu\lambda^{m}$ or $|x_{j}|\leq 1+\mu\lambda^{m}$,
$j=1,\ldots,n$. Then the operator ${\bf H}$ has a bounded support
$\mbox{supp}({\bf H})\subset\left\\{z\in\mathbb{C}:|z|\leq
1+\mu\lambda^{m}\right\\}\cup\left\\{z\in\mathbb{C}:|z+\gamma|\leq
1+\mu\lambda^{m}\right\\}$. This completes the proof. ∎

Let us consider the resolvent functions (see for instance [16, 25]) on the
largest domain where they can be defined
($\overline{\mathbb{C}}\setminus\mbox{supp}({\bf H})$)

(4.14) $f_{j}(z)=\left(z\mathbb{I}-{\bf H}\right)^{-1}{\bf e}_{j-1}\cdot{\bf
e}_{0},\quad j=1,\ldots,m.$

######  Lemma 4.8.

Fix $j\in\\{1,\ldots,m\\}$. The resolvent function $f_{j}$ is holomorphic on
$\overline{\mathbb{C}}\setminus\mbox{supp}\left({\bf H}\right)$, where
$\overline{\mathbb{C}}=\mathbb{C}\cup\\{\infty\\}$ is the Riemann sphere.
Denote $\displaystyle
f_{j}\in\mathcal{H}\left(\overline{\mathbb{C}}\setminus\mbox{supp}\left({\bf
H}\right)\right)$. The function $f_{j}$ also behave close to infinity as

(4.15) $f_{j}(z)=\mathcal{O}\left(\frac{1}{z^{j}}\right)\quad\mbox{as}\quad
z\to\infty.$

###### Proof.

By construction $\displaystyle
f_{j}\in\mathcal{H}\left(\overline{\mathbb{C}}\setminus\mbox{supp}\left({\bf
H}\right)\right)$. According to Lemma 4.7 we have that
$\displaystyle\mbox{supp}({\bf H})\subset\left\\{z\in\mathbb{C}:|z|\leq
1+\lambda\mu^{m}\right\\}\cup\left\\{z\in\mathbb{C}:|z+\gamma|\leq
1+\lambda\mu^{m}\right\\}$ is bounded. We then apply Neumann’s series
inversion (see for instance [27])

$\displaystyle\left(z\mathbb{I}-{\bf
H}\right)^{-1}=\left(\left(z+\gamma\right)\mathbb{I}-\left({\bf
H}+\gamma\mathbb{I}\right)\right)^{-1}$

to obtain the following expression that holds uniformly in any compact set
contained in the widest open disk of $\overline{\mathbb{C}}$ centered at
$\infty$ leaving outside $\mbox{supp}\left({\bf H}\right)$

$f_{j}(z)=\sum_{\nu=0}^{\infty}\frac{({\bf H}+\gamma\mathbb{I})^{\nu}{\bf
e}_{j-1}\cdot{\bf e}_{0}}{(z+\gamma)^{\nu+1}}.$

From (4.13) in Lemma 4.6 we have that for each $\nu\in\\{0,\ldots,j-2\\}$

$\displaystyle({\bf H}+\gamma\mathbb{I})^{\nu}{\bf e}_{j-1}\cdot{\bf
e}_{0}=\left(\sum_{k=0}^{\nu}{\nu\choose
k}\left(\lambda\mu^{m}\right)^{\nu-k}{\bf e}_{(\nu-k)m+j-1-k}\right)\cdot{\bf
e}_{0}=0,$

hence

$f_{j}(z)=\sum_{\nu=j-1}^{\infty}\frac{({\bf H}+\gamma\mathbb{I})^{\nu}{\bf
e}_{j-1}\cdot{\bf
e}_{0}}{(z+\gamma)^{\nu+1}}=\frac{1}{(z+\gamma)^{j-1}}\sum_{\nu=0}^{\infty}\frac{({\bf
H}+\gamma\mathbb{I})^{\nu+j-1}{\bf e}_{j-1}\cdot{\bf
e}_{0}}{(z+\gamma)^{\nu+1}},$

which proves the statement. ∎

We define the sequences of moments
$\displaystyle\left\\{c_{\nu,j}\right\\}_{\nu\in\mathbb{Z}_{+}}$,
$j=1,\ldots,m$, corresponding to then operator ${\bf H}$ as follows

(4.16) $c_{\nu,j}=\left({\bf H}\right)^{\nu}{\bf e}_{j-1}\cdot{\bf
e}_{0},\quad\nu\in\mathbb{Z}_{+},\quad j=1,\ldots,m.$

######  Lemma 4.9.

Given $j\in\\{1,\ldots,m\\}$

$c_{\nu,j}=\left({\bf H}\right)^{\nu}{\bf e}_{j-1}\cdot{\bf
e}_{0}=0,\quad\nu=0,\ldots,j-1,\quad j=1,\ldots,m.$

###### Proof.

Fix $j\in\\{1,\ldots,m\\}$. Since $\displaystyle\mbox{supp}({\bf H})$ is
bounded we apply Neumann’s series inversion to
$\displaystyle\left(z\mathbb{I}-{\bf H}\right)^{-1}$ such that at any compact
set contained in the widest open disk of $\overline{\mathbb{C}}$ centered at
$\infty$ leaving outside $\mbox{supp}\left({\bf H}\right)$, the following
equality holds uniformly

$f_{j}(z)=\sum_{\nu=0}^{\infty}\frac{{\bf H}^{\nu}{\bf e}_{j-1}\cdot{\bf
e}_{0}}{z^{\nu+1}}=\sum_{\nu=0}^{\infty}\frac{c_{\nu}}{z^{\nu+1}}.$

Taking now into account (4.15) we obtain that the first $j-1$th coefficients
$c_{\nu}$ must vanish. This completes the proof. ∎

##  5\. Chebishev’s star sequence
$\displaystyle\left\\{T_{n}\right\\}_{n\in\mathbb{Z}_{+}}$

In this Section we reproduce several of the result stated in [3, 4, 21, 22,
23]. We accommodate them to our notation, and give detailed proofs which are
comfortable to understand some results in coming sections. Consider the
sequence of polynomials
$\displaystyle\left\\{T_{n}\right\\}_{n\in\mathbb{Z}_{+}}$ satisfying the
recurrence relation (4.7) with initial conditions (4.8).

Let us introduce the so called second kind polynomials

(5.1) $T_{n,j}(z)=\left(z\mathbb{I}-{\bf T}\right)^{-1}{\bf
e}_{j-1}\cdot\left(T_{n}(z\mathbb{I})-T_{n}({\bf H}^{\top})\right)\,{\bf
e}_{0},\quad j=1,\ldots,m.$

######  Lemma 5.1.

Let $n\in\mathbb{Z}_{+}$ and $j\in\\{1,\ldots,m\\}$ be given, with
$n=\kappa\,m+s$, $(\kappa,s)\in\mathbb{Z}_{+}\times\\{0,1,\ldots,m-1\\}$. The
resolvent function in (4.14) and the polynomial (5.1) satisfy the identity

(5.2) $T_{n}(z)f_{j}(z)-T_{n,j}(z)=\left(z\mathbb{I}-{\bf T}\right)^{-1}{\bf
e}_{j-1}\cdot T_{n}({\bf T}^{\top})\,{\bf e}_{0},$

where

(5.3) $\left(z\mathbb{I}-{\bf T}\right)^{-1}{\bf e}_{j-1}\cdot T_{n}({\bf
T}^{\top})\,{\bf
e}_{0}\in\mathcal{H}\left(\overline{\mathbb{C}}\setminus\mbox{supp}({\bf
T})\right),$

and

(5.4) $\left(z\mathbb{I}-{\bf T}\right)^{-1}{\bf e}_{j-1}\cdot T_{n}({\bf
T}^{\top})\,{\bf
e}_{0}=\mathcal{O}\left(\frac{1}{z^{\tau(n,j)}}\right)\quad\mbox{as}\quad
z\to\infty$

with

$\tau(n,j)=\left\\{\begin{array}[]{l l
l}\kappa+m+j-s+1&\mbox{if}&j\in\\{1,\ldots,s\\},\\\ &&\\\
\kappa+j-s+1&\mbox{if}&j\in\\{s+1,\ldots,m\\}.\end{array}\right.$

###### Proof.

Fix $j\in\\{1,2,\ldots,m\\}$. Expression (5.1) yields immediately identity
(5.2). From Lemma 4.8 we have that the resolvent function $\displaystyle
f_{j}\in\mathcal{H}\left(\overline{\mathbb{C}}\setminus\mbox{supp}({\bf
T})\right)$. So taking into account (5.2) we show (5.3).

Since $\mbox{supp}({\bf T})$ is bounded we apply again Neumann’s series for
operator inversion to $\displaystyle\left(z\mathbb{I}-{\bf T}\right)^{-1}$, in
the sense of uniform convergence on any compact set of the maximum open disk
in $\overline{\mathbb{C}}$ centered at $\infty$ which leaves outside
$\mbox{supp}\left({\bf T}\right)$. Hence

$\left(z\mathbb{I}-{\bf T}\right)^{-1}{\bf e}_{j-1}\cdot T_{n}({\bf
T}^{\top})\,{\bf e}_{0}=\sum_{\nu=0}^{\infty}\frac{{\bf T}^{\nu}{\bf
e}_{j-1}\cdot T_{n}({\bf T}^{\top})\,{\bf e}_{0}}{z^{\nu+1}},\quad
j=1,\ldots,m.$

Using the identity in (4.10) for $T_{n}$, we obtain that

$\left(z\mathbb{I}-{\bf T}\right)^{-1}{\bf e}_{j-1}\cdot T_{n}({\bf
T}^{\top})\,{\bf e}_{0}=\sum_{\nu=0}^{\infty}\frac{{\bf T}^{\nu}{\bf
e}_{j-1}\cdot{\bf e}_{n}}{z^{\nu+1}}.$

Recall (4.13) in Lemma 4.6

$\left(z\mathbb{I}-{\bf T}\right)^{-1}{\bf e}_{j-1}\cdot T_{n}({\bf
T}^{\top})\,{\bf
e}_{0}=\sum_{\nu=0}^{\infty}\frac{1}{z^{\nu+1}}\sum_{k=0}^{\nu}{\nu\choose
k}\left(\mu\lambda^{m}\right)^{\nu-k}{\bf e}_{(\nu-k)m+j-1-k}\cdot{\bf e}_{n}$
$=\sum_{\nu=0}^{\infty}\frac{1}{z^{\nu+1}}\sum_{k=0}^{\nu}{\nu\choose
k}\left(\mu\lambda^{m}\right)^{\nu-k}{\bf e}_{(\nu-k)m+j-1-k}\cdot{\bf
e}_{\kappa m+s}$
$=\sum_{\nu=0}^{\infty}\frac{1}{z^{\nu+1}}\sum_{k=0}^{\nu}{\nu\choose
k}\left(\mu\lambda^{m}\right)^{\nu-k}\delta_{(\nu-k)m+j-1-k,\kappa m+s}.$

Let us start by analyzing the cases $j\in\\{1,\ldots,s\\}$. In above summation
the first term which does not vanish occurs when $\nu-k-1=s$ and $m+j-1-k=s$,
hence $\nu=\kappa+m+j-s$. This implies that

$\left(z\mathbb{I}-{\bf T}\right)^{-1}{\bf e}_{j-1}\cdot T_{n}({\bf
T}^{\top})\,{\bf
e}_{0}=\mathcal{O}\left(\frac{1}{z^{\kappa+m+j-s+1}}\right)\quad\mbox{as}\quad
z\to\infty.$

When $j\in\\{s+1,\ldots,m\\}$, the first term that is different from zero,
corresponds to $\nu-k-1=\kappa$ and $j-1-k$, so $\nu=\kappa+j-s$ and

$\left(z\mathbb{I}-{\bf T}\right)^{-1}{\bf e}_{j-1}\cdot T_{n}({\bf
T}^{\top})\,{\bf e}_{0}=\sum_{\nu=0}^{\infty}\frac{{\bf T}^{\nu}{\bf
e}_{j-1}\cdot{\bf
e}_{n}}{z^{\nu+1}}=\mathcal{O}\left(\frac{1}{z^{\kappa+j-s+1}}\right),\,\,\mbox{as}\,z\to\infty.$

These equalities complete the proof. ∎

######  Lemma 5.2.

Fix $\displaystyle j\in\left\\{1,\ldots,m\right\\}$. The sequence of
polynomials $\displaystyle\left\\{T_{n,j}\right\\}_{n\in\mathbb{Z}_{+}}$ in
(5.1) satisfy the following recurrence relations

(5.5) $zT_{n,j}(z)=T_{n+1,j}(z)+\mu\lambda^{m}T_{n-m,j}(z),$

with the initial conditions

(5.6) $T_{n,j}\equiv 0,\quad n=0,1,\ldots,j-1\quad\mbox{and}\quad
T_{n,j}(z)=z^{n-j},\quad n=j,\ldots,m.$

###### Proof.

For each $\displaystyle j\in\\{1,\ldots,m\\}$, consider (5.2)

$T_{n,j}(z)=T_{n}(z)f_{j}(z)-\left(z\mathbb{I}-{\bf T}\right)^{-1}{\bf
e}_{j-1}\cdot T_{n}({\bf T}^{\top})\,{\bf e}_{0}.$

Let us recall Lemma 4.8. We have that $\displaystyle
f_{j}\in\mathcal{H}\left(\overline{\mathbb{C}}\setminus\mbox{supp}\left({\bf
T}\right)\right)$ is analytic in a neighborhood of $\infty$ where it vanishes
with a zero of order $j$ as in (4.15). Combining this fact with (5.3) and
(5.4) in Lemma 5.1, we conclude that for any $n\in\\{0,\ldots,j-1\\}$ the
polynomial

$T_{n,j}(z)=T_{n}(z)f_{j}(z)-\left(z\mathbb{I}-{\bf T}\right)^{-1}{\bf
e}_{j-1}\cdot T_{n}({\bf T}^{\top})\,{\bf e}_{0}$
$=\mathcal{O}\left(\frac{1}{z}\right)+\mathcal{O}\left(\frac{1}{z^{\tau(n,j)}}\right)=\mathcal{O}\left(\frac{1}{z}\right)\quad\mbox{as}\quad
z\to\infty,$

which yields $\displaystyle T_{n,j}\equiv 0,$ $n=0,\ldots,j-1$. When $n=j$ we
have that the polynomial $T_{n,j}$ is bounded in $\overline{\mathbb{C}}$,
hence its a constant. From (4.13) in Lemma 4.6 we obtain that

$\lim_{z\to\infty}T_{n,j}(z)=\lim_{z\to\infty}T_{n}(z)f_{j}(z)=\lim_{z\to\infty}z^{j}\frac{1}{z^{j}}{\bf
e}_{j-1}\cdot\left({\bf T}^{\top}\right)^{j}{\bf e}_{0}=1.$

We now consider $n\in\\{j,\ldots,m-1\\}$ and with (4.6)

$zT_{n,j}(z)=z\left(z\mathbb{I}-{\bf T}\right)^{-1}{\bf
e}_{j-1}\cdot\left(T_{n}(z\mathbb{I})-T_{n}({\bf T}^{\top})\right)\,{\bf
e}_{0}$ $=\left(z\mathbb{I}-{\bf T}\right)^{-1}{\bf
e}_{j-1}\cdot\left(zT_{n}(z\mathbb{I})-zT_{n}({\bf T}^{\top})\right)\,{\bf
e}_{0}$ $=\left(z\mathbb{I}-{\bf T}\right)^{-1}{\bf
e}_{j-1}\cdot\left(zT_{n}(z\mathbb{I})-(z\mathbb{I}-{\bf
T}^{\top})\,T_{n}({\bf T}^{\top})-({\bf T}^{\top}+\iota\mathbb{I})T_{n}({\bf
T}^{\top})\right)\,{\bf e}_{0}$ $=\left(z\mathbb{I}-{\bf T}\right)^{-1}{\bf
e}_{j-1}\cdot\left(z\mathbb{I}T_{n}(z\mathbb{I})-(z\mathbb{I}-{\bf
T}^{\top})\,T_{n}({\bf T}^{\top})-{\bf T}^{\top}\,T_{n}({\bf
T}^{\top})\right)\,{\bf e}_{0}$ $=\left(z\mathbb{I}-{\bf T}\right)^{-1}{\bf
e}_{j-1}\cdot\left(T_{n+1}(z\mathbb{I})-T_{n+1}({\bf T}^{\top})\right)\,{\bf
e}_{0}$ $-\left(z\mathbb{I}-{\bf T}\right)^{-1}{\bf
e}_{j-1}\cdot\left((z\mathbb{I}-{\bf T}^{\top})\,T_{n}({\bf
T}^{\top})\right)\,{\bf e}_{0}$ $=T_{n+1,j}(z)-\left(z\mathbb{I}-{\bf
T}\right)^{-1}{\bf e}_{j-1}\cdot\left((z\mathbb{I}-{\bf T}^{\top})\,T_{n}({\bf
T}^{\top})\right)\,{\bf e}_{0}$ $=T_{n+1,j}(z)-{\bf
e}_{j-1}^{\top}\,\left(z\mathbb{I}-{\bf T}^{\top}\right)^{-1}(z\mathbb{I}-{\bf
T}^{\top})\,T_{n}({\bf T}^{\top})\,{\bf e}_{0}=T_{j,n+1}(z)-{\bf
e}_{j-1}^{\top}\,{\bf e}_{n}.$

Since $n\geq j$ we obtain (5.6).

Let us prove (5.5). Set $n\geq m$ and recall that (4.5)

$z\,T_{n,j}(z)=z\left(z\mathbb{I}-{\bf T}\right)^{-1}{\bf
e}_{j-1}\cdot\left(T_{n}(z\mathbb{I})-T_{n}({\bf T}^{\top})\right)\,{\bf
e}_{0}$ $=\left(z\mathbb{I}-{\bf T}\right)^{-1}{\bf
e}_{j-1}\cdot\left(z\mathbb{I}T_{n}(z\mathbb{I})-z\mathbb{I}T_{n}({\bf
T}^{\top})\right)\,{\bf e}_{0}$ $=\left(z\mathbb{I}-{\bf T}\right)^{-1}{\bf
e}_{j-1}\cdot\left(z\mathbb{I}T_{n}(z\mathbb{I})-(z\mathbb{I}-{\bf
T}^{\top})T_{n}({\bf T}^{\top})-{\bf T}^{\top}T_{n}({\bf
T}^{\top})\right)\,{\bf e}_{0}$ $=\left(z\mathbb{I}-{\bf T}\right)^{-1}{\bf
e}_{j-1}\cdot\left(T_{n+1}(z\mathbb{I})+\mu\lambda^{m}\,T_{n-m}(z\mathbb{I})-T_{n+1}({\bf
T}^{\top})-\mu\lambda^{m}T_{n-m}({\bf T})\right)\,{\bf e}_{0}$
$-\left(z\mathbb{I}-{\bf T}\right)^{-1}{\bf
e}_{j-1}\cdot\left((z\mathbb{I}-{\bf T}^{\top})\,T_{n}({\bf
T}^{\top})\right)\,{\bf e}_{0}=T_{n+1,j}(z)+\mu\lambda^{m}T_{n-m,j}(z).$

This proves (5.5). ∎

Let us consider the second kind polynomials corresponding to $T_{n}$ defined
in (5.1) denote by $T_{n,j}$ $j=1,\ldots,m$. We include in the notation
$T_{n}=T_{0,n}$.

######  Lemma 5.3.

Given $n\in\mathbb{Z}_{+}$

(5.7) $T_{n}\equiv T_{n,0}\equiv T_{n+1,1}\equiv\cdots\equiv T_{n+m,m}.$

###### Proof.

Fix $j\in\\{0,1,\ldots,m\\}$. The recurrence relations in Lemma 5.2 can be
written as

(5.8) $T_{n,j}\equiv 0,\quad n=j-m,j-m+1,\ldots,j-1\quad\mbox{and}\quad
T_{j,i}(z)=1,$

and

(5.9) $zT_{n,j}(z)=T_{n+1,j}(z)+\mu\lambda^{m}T_{n-m,j}(z),\quad n\geq j.$

The initial conditions in (5.8) coincide with the ones in (5.6) for every
$n\in\\{0,1,\ldots,j\\}$. We now consider the cases
$n\in\\{j,j+1,\ldots,m-1\\}$. Taking into account that $T_{n-m,j}\equiv 0$ we
see that the recurrence relation (5.9) becomes

$T_{n,j}(z)=zT_{n-1,j}(z)=z\,z^{n-1}=z^{n}.$

Since the recurrence relations (5.5) and (5.9) coincide we have that for each
$j\in\\{0,\ldots,m\\}$ the sequence $\displaystyle\\{T_{n,j}:n\geq j-m\\}$ is
well defined. This proves (5.7). ∎

######  Lemma 5.4.

Set $n\in\mathbb{Z}_{+}$ with $n=d(m+1)+r$ where $d\in\mathbb{Z}_{+}$ and
$r\in\\{0,1,\ldots,m\\}$. There exists a polynomial $h_{n}$ with degree $d$
such that $T_{n}(z)=z^{r}h_{n}(z^{m+1})$. The polynomial elements of the
sequence $\displaystyle\left\\{h_{n}\right\\}_{n\in\mathbb{Z}_{+}}$ satisfy
the following initial conditions and recurrence relations, respectively

(5.10) $h_{0}=h_{1}=\cdots=h_{m}=1$

and when $d\geq 1$

(5.11) $h_{n+1}(z)=\left\\{\begin{array}[]{l c
l}h_{n}(z)-\mu\lambda^{m}h_{n-m}(z)&\mbox{if}&r\in\\{0,1,\ldots,m-1\\},\\\
&&\\\ z\,h_{n}(z)-\mu\lambda^{m}h_{n-m}(z)&\mbox{if}&r=m.\end{array}\right.$

###### Proof.

The initial conditions (4.8) make evident of above statement for each
$n\in\\{0,1,\ldots,m\\}$

$T_{n}(z)=z^{n}\implies h_{n}\equiv 1\quad\mbox{and}\quad r=n,$

which proves (5.10). Let us use the induction method. Given $n=d(m+1)+r\geq m$
we assume that the statement holds for every
$\widetilde{n}\in\\{n-m,n-m+1,\ldots,n\\}$. Observe that

$n-m=d(m+1)+r-m=d(m+1)-(m+1)+r+1=(d-1)(m+1)+r+1.$

We divide the proof in two cases

  1.   2. (1)

Case $r=0,1,\ldots,m-1$. We have that $n+1=d(m+1)+r+1$, with
$d\in\mathbb{Z}_{+}$, $r+1\in\\{1,\ldots,m\\}$.

$T_{n+1}(z)=zT_{n}(z)-\mu\lambda^{m}T_{n-m}=zz^{r}h_{n}(z^{m+1})-\mu\lambda^{m}z^{r+1}h_{n-m}(z^{m+1}),$

then

$T_{n+1}(z)=z^{r+1}h_{n}(z^{m+1})-\mu\lambda^{m}z^{r+1}h_{n-m}(z^{m+1})$
$=z^{r+1}\left(h_{n}(z^{m+1})-\mu\lambda^{m}h_{n-m}(z^{m+1})\right)=z^{r+1}h_{n+1}(z^{m+1}).$

We see that $\deg h_{n+1}=\deg h_{n}$. We also observe that the equlity

$z^{r+1}\left(h_{n}(z^{m+1})-\mu\lambda^{m}h_{n-m}(z^{m+1})\right)=z^{r+1}h_{n+1}(z^{m+1})$

implies that

$h_{n+1}(z)=h_{n}(z)-\mu\lambda^{m}h_{n-m}(z).$

This completes this part of the proof for cases $r=0,1,\ldots,m-1$.

  3.   4. (2)

Case $r=m$. Since $n=d(m+1)+m$ then

$n+1=d(m+1)+m+1=(d+1)(m+1)$

and

$n-m=d(m+1)+m-m=d(m+1).$

Hence

$T_{n+1}(z)=zT_{n}(z)-\mu\lambda^{m}T_{n-m}=zz^{m}h_{n}(z^{m+1})-\mu\lambda^{m}z^{m+1}h_{n-m}(z^{m+1})$
$=z^{m+1}h_{n}(z^{m+1})-\mu\lambda^{m}h_{n-m}(z^{m+1}))=h_{n+1}(z^{m+1}).$

This implies that $\deg h_{n+1}=\deg h_{n}+1=d+1$, and the expression

$z^{m+1}\left(h_{n}(z^{m+1}-\mu\lambda^{m}h_{n-m}(z^{m+1})\right)=h_{n+1}(z^{m+1})$

yields

$h_{n+1}(z)=z\,h_{n}(z)-\mu\lambda^{m}h_{n-m}(z).$

This completes the proof.

  5.

∎

We now study the distribution of zeros the polynomials $h_{n}$,
$n\in\mathbb{Z}_{+}$.

######  Lemma 5.5.

Consider the sequence $\displaystyle\\{h_{n}\\}_{n\in\mathbb{Z}_{+}}$ as in
Lemma 5.4. Their elements satisfy the following statements

  1.   2. (1)

$h_{n}(0)\not=0$, $n\in\mathbb{Z}_{+}$.

  3.   4. (2)

Set $n=d(m+1)+r$ with $d\in\mathbb{Z}_{+}$ and $r\in\\{0,1,\ldots,m\\}$. Then
the $d$ zeros of $h_{n}$ are real and simple.

  5.   6. (3)

Fix $d\in\mathbb{Z}_{+}$, for each $r\in\\{0,1,\ldots,m\\}$ and denote
$x_{n,1}<x_{n,2}<\cdots<x_{n,d}$ the zeros of $h_{n}$ with $n=d(m+1)+r$. Set
$d\geq 1$ and $n=(d-1)(m+1)+m$ ($r=m$). Then for each $j\in\\{1,\ldots,d\\}$

$x_{n+1,j}<\cdots<x_{n+m+1,j}<x_{n,j}<x_{n+1,j+1}<\cdots<x_{n+m+1,j+1}<x_{n,j+1}.$

  7.

###### Proof.

We use induction method. When $n\in\\{0,\ldots,m\\},$ $d=0$ and from (5.10)
the first two statements (1) and (2) are trivially satisfied. Consider $d=1$.
From recurrence relations in (5.11) we have that

$\begin{array}[]{lcl}h_{m+1}(z)&=&z\,h_{m}-\mu\lambda^{m}\,h_{0}=z-\mu\lambda^{m},\\\
&&\\\ h_{m+2}(z)&=&h_{m+1}-\mu\lambda^{m}\,h_{1}=z-2\mu\lambda^{m},\\\ &&\\\
h_{m+3}(z)&=&h_{m+2}-\mu\lambda^{m}\,h_{2}=z-3\mu\lambda^{m},\\\ &&\\\
&\vdots&\\\ &&\\\
h_{2m+1}(z)&=&h_{2m}-\mu\lambda^{m}\,h_{m}=z-(m+1)\mu\lambda^{m},\end{array}$

hence

$x_{m+k,1}=k\,\mu\lambda^{m}>0,\quad k=1,2,\ldots,m+1.$

Observe that if $d-1=0$ $h_{m}\equiv 1$ does not vanish. This implies that the
two statements are trivially true

$x_{m+1,1}<x_{m+2,1}<\cdots<x_{2m+1,1}.$

We assume that the three statements in Lemma 5.5 when $d\geq 1$ are true.
Consider $n=d(m+1)+m$. Combining the recurrence relations in (5.11) for each
$k=1,2,\ldots,m+1$ we have that

(5.12)
$h_{n+k}(z)=zh_{n}(z)-\lambda\mu^{m}\left(h_{n-m}(z)+h_{n-m+1}(z)+\cdots+h_{n-m+k-1}(z)\right).$

From assumption we have that the polynomial

$\mu\lambda^{m}\left(h_{n-m}(z)+h_{n-m+1}(z)+\cdots+h_{n-m+k-1}(z)\right)$

does not vanish at the origin, hence the equation (5.12) yields

$h_{n+k}(0)=-\mu\lambda^{m}\left(h_{n-m}+h_{n-m+1}+\cdots+h_{n-m+k-1}\right)(0)\not=0.$

The property (1) is already proved. We continuous with the two others (2) and
(3).

Note that

$h_{n+k}(x_{n,j})=-\mu\lambda^{m}\left(h_{n-m}+h_{n-m+1}+\cdots+h_{n-m+k-1}\right)(x_{n,j}),\quad
j=1,2,\ldots,d.$

Then the polynomial $h_{n+k}$ also changes sign at a point between two
consecutive zeros of $h_{n}$. Observe that all the polynomials $h$’s are
monic. Let us take into account that

$\deg
h_{n+k}=\deg\mu\lambda^{m}\left(h_{n-m}+h_{n-m+1}+\cdots+h_{n-m+k-1}\right)+2.$

This implies that

$\lim_{z\to-\infty}h_{n+k}(z)=\lim_{z\to-\infty}\lambda\mu^{m}\left(h_{n-m}+h_{n-m+1}+\cdots+h_{n-m+k-1}\right)(z)=\pm\infty.$

Assuming that the polynomial
$\mu\lambda^{m}\left(h_{n-m}+h_{n-m+1}+\cdots+h_{n-m+k-1}\right)(z)$ does not
vanish before $x_{n,1}$, and taking into account that the polynomials
$h_{n+k}(x_{n,1})$ and
$\mu\lambda^{m}\left(h_{n-m}+h_{n-m+1}+\cdots+h_{n-m+k-1}\right)(x_{n,1})$

$h_{n+k}(x_{n,1})=-\mu\lambda^{m}\left(h_{n-m}+h_{n-m+1}+\cdots+h_{n-m+k-1}\right)(x_{n,1}),$

have different sing, hence $h_{n,k}$ changes sing ones before $x_{n,1}$. We
have proved that given $k\in\\{2,\ldots,m+1\\}$, $x_{n,k-1}<x_{n+1,k}<x_{n,k}$
and $x_{n+1,1}<x_{n,1}$.

In order to complete the proof we only need to show that given
$k\in\\{1,\ldots,m\\}$, $x_{n+k,j}<x_{n+k+1,j}$. We procede similarly as we
have done. Consider again the recurrence relations in (5.11). We have that

$h_{n+k+1}=h_{n+k}(z)-\mu\lambda^{m}h_{n-m+k}(z).$

From the assumption that the polynomial $h_{n-m}$ changes sign between two
consecutive zeros of $h_{n+k}$, then $h_{n+k+1}$ also interlace zeros with
$h_{n+k}$. Since $h_{n-m+k}$ does not vanishes after $x_{n+k,d}$ and taking
into account that both $h_{n+k}$ and $h_{n-m+k}$ are monic polynomials and
also from the fact that they have different sign at $x_{n+k,d}$, we have that
$h_{n+k+1}$ vanishes ones after $x_{n+k,d}$. This completes the proof. ∎

######  Remark 5.6.

Combining Lemma 4.7, Lemma 5.4, and Lemma 5.5 we have that given any
$n=d(m+1)+r$ its corresponding polynomial $T_{n}(z)=z^{r}h_{n}(z^{m+1})$ has a
zero of order $r$ at the origin and $d(m+1)$ simple roots more shared in a
starlike set

$\displaystyle\Sigma\subset\bigcup_{k=0}^{m}\left[-\left(1+\mu\lambda^{n}\right)^{1/m+1}\exp\left(\frac{2\pi
ik}{m+1}\right)\,,\,\left(1+\mu\lambda^{n}\right)^{1/m+1}\exp\left(\frac{2\pi
ik}{m+1}\right)\right].$

There are $d$ different real zeros

$x_{n,j}\in\left[-1-\mu\lambda^{m}\,,\,1+\mu\lambda^{m}\right],\quad
j=1,2,\ldots,d,$

and for each $k\in\\{1,\ldots,m\\}$, $T_{n}$ vanishes at

$x_{n,j}\exp\left(\frac{2\pi ik}{m+1}\right),\quad j=1,\ldots,d.$

In Remark 5.11 we explain why we can state that $\Sigma=S_{0}$.

We now study an interpolatory quadrature formula in this context of operators

######  Lemma 5.7.

Fix $n=d(m+1)+r$ with $d\in\mathbb{Z}_{+}$ and $r\in\\{0,1,\ldots,m\\}$. Set a
polynomial $P$ with $\displaystyle\deg P\leq 2(m+1)d+r-1$. Denote

$\left(x_{n,j}\right)^{1/m+1}=\zeta_{n,j}\exp\frac{2\pi ik}{m+1},\quad
k=0,1,\ldots,m,\quad j=1,2,\ldots,d,$

the roots of the composite polynomial $h_{n}(z^{m+1})$.

(5.13) $P({\bf T}){\bf T}^{r}{\bf e}_{0}\cdot{\bf
e}_{0}=\sum_{k=0}^{m}\sum_{j=1}^{d}\lambda_{n,j,k}\frac{P\left(\zeta_{n,j}\exp\frac{2\pi
ik}{m+1}\right)}{(m+1)\left(\zeta_{n,j}\exp\frac{2\pi
ik}{m+1}\right)^{m+1-r}},$

where $k=0,1,\ldots,m,$ $j=1,\ldots,d,$

(5.14) $\lambda_{n,j,k}=\frac{h_{n-1}\left(x_{n,j}\right)}{\displaystyle
h_{n}^{\prime}\left(x_{n,j}\right)}>0$

such that

(5.15) $\sum_{k=0}^{m}\sum_{j=1}^{d}\lambda_{n,j,k}=(m+1)\mu\lambda^{m}.$

###### Proof.

The Lagrange interpolatory polynomial that interpolates $P$ at the roots of
$h_{n}(z^{m+1})$ has the following form

$\mathcal{L}_{n}\left[P\right]({\bf
T})=\sum_{k=0}^{m}\sum_{j=1}^{d}\frac{P\left(\zeta_{n,j}\exp\frac{2\pi
ik}{m+1}\right)}{\displaystyle(m+1)\left(\zeta_{n,j}\exp\frac{2\pi
ik}{m+1}\right)^{m}h_{n}^{\prime}\left(x_{n,j}\right)}$ $\times\left({\bf
T}-\zeta_{n,j}\exp\frac{2\pi
ik}{m+1}\mathbb{I}\right)^{-1}h_{n}\left(\left({\bf T}\right)^{m+1}\right).$

Form Lemma 5.5 we have the denominator never becomes zero. Recall that when
$n=d(m+1)+r$, $\displaystyle T_{n}(z)=z^{r}\,h_{n}\left(z^{m+1}\right)$. Then
we obtain that

(5.16) $\mathcal{L}_{n}\left[P\right]({\bf T})\,{\bf
T}^{r}=\sum_{k=0}^{m}\sum_{j=1}^{d}\frac{P\left(\zeta_{n,j}\exp\frac{2\pi
ik}{m+1}\right)}{\displaystyle(m+1)\left(\zeta_{n,j}\exp\frac{2\pi
ik}{m+1}\right)^{m}h_{n}^{\prime}\left(x_{n,j}\right)}$ $\times\left({\bf
T}-\zeta_{n,j}\exp\frac{2\pi ik}{m+1}\mathbb{I}\right)^{-1}T_{n}\left({\bf
T}\right),$

and also there is a polynomial $\mathcal{Q}$ with degree $\deg\mathcal{Q}\leq
n-1$

$\left[\mathcal{L}_{n}\left[P\right]({\bf T})-P({\bf T})\right]{\bf
T}^{r}=\mathcal{Q}({\bf T})\,h_{n}\left({\bf T}^{m+1}\right)\,{\bf
T}^{r}=\mathcal{Q}({\bf T})\,T_{n}({\bf T}).$

Now using the identities (4.10) in Lemma 4.2 and (4.9)

$\left[\mathcal{L}_{n}\left[P\right]({\bf T})-P({\bf T})\right]{\bf T}^{r}{\bf
e}_{0}\cdot{\bf e}_{0}=\mathcal{Q}({\bf T})\,T_{n}({\bf T}){\bf
e}_{0}\cdot{\bf e}_{0}$ $=\mathcal{Q}({\bf T}){\bf e}_{0}\cdot\,T_{n}({\bf
T}^{\top}){\bf e}_{0}=\mathcal{Q}({\bf T}){\bf e}_{0}\cdot\,{\bf e}_{n}=0$

In the last equality we took into account that $m\deg\mathcal{Q}\leq n-1$ and
(4.9). Let us study the equality

$P({\bf T})\left({\bf T}\right)^{r}{\bf e}_{0}\cdot{\bf
e}_{0}=\mathcal{L}_{n}\left[P\right]({\bf T}){\bf T}^{r}{\bf e}_{0}\cdot{\bf
e}_{0}.$

Form (5.16) we have that

$P({\bf T}){\bf T}^{r}{\bf e}_{0}\cdot{\bf
e}_{0}=\sum_{k=0}^{m}\sum_{j=1}^{d}\frac{P\left(\zeta_{n,j}\exp\frac{2\pi
ik}{m+1}\right)}{\displaystyle(m+1)\left(\zeta_{n,j}\exp\frac{2\pi
ik}{m+1}\right)^{m}h_{n}^{\prime}\left(x_{n,j}\right)}$ $\times\left({\bf
T}-x_{n,j}\exp\frac{2\pi ik}{m+1}\mathbb{I}\right)^{-1}T_{n}\left({\bf
T}\right){\bf e}_{0}\cdot{\bf e}_{0},$

which combined with (5.1) yields

$P({\bf T})\left({\bf T}\right)^{r}{\bf e}_{0}\cdot{\bf e}_{0}$
$=\sum_{k=0}^{m}\sum_{j=1}^{d}\frac{P\left(\zeta_{n,j}\exp\frac{2\pi
ik}{m+1}\right)}{\displaystyle(m+1)\left(\zeta_{n,j}\exp\frac{2\pi
ik}{m+1}\right)^{m}h_{n}^{\prime}\left(x_{n,j}\right)}T_{n,1}\left(\zeta_{n,j}\exp\frac{2\pi
ik}{m+1}\right)$
$=\sum_{k=0}^{m}\sum_{j=1}^{d}\frac{\left(\zeta_{n,j}\exp\frac{2\pi
ik}{m+1}\right)^{r-1}P\left(\zeta_{n,j}\exp\frac{2\pi
ik}{m+1}\right)}{\displaystyle(m+1)\left(\zeta_{n,j}\exp\frac{2\pi
ik}{m+1}\right)^{m}h_{n}^{\prime}\left(x_{n,j}\right)}h_{n-1}\left(x_{n,j}\right).$

which proves (5.13) and (5.14). Taking $\displaystyle P\left({\bf
T}\right)=\left({\bf T}\right)^{m+1-r}$ we obtain that

$\sum_{k=0}^{m}\sum_{j=1}^{d}\frac{h_{n-1}\left(x_{n,j}\right)}{\displaystyle(m+1)h_{n}^{\prime}\left(x_{n,j}\right)}={\bf
T}^{m+1-r}{\bf T}^{r}{\bf e}_{0}\cdot{\bf e}_{0}={\bf e}_{0}\cdot\left({\bf
T}^{\top}\right)^{m+1}{\bf e}_{0}=\mu\lambda^{m}.$

This proves (5.15). The relations in (5.14) are deduced from Lemma 5.5. Since
$\displaystyle h_{n}$ has simple zeros its derivatives changes sign between
two consecutive roots of such polynomial. In addition we observe that
$h_{n-1}$ and $h_{n}^{\prime}$ have same sign at $x_{n,1}$, hence they also
have same sign at every $x_{n,j}$, $j=1,\ldots,d$, and their quotient is
always positive. ∎

######  Proposition 5.8.

Let $\displaystyle\left\\{T_{n}\right\\}_{n\in\mathbb{Z}_{+}}$ be the sequence
of polynomials satisfying the recurrence relation (4.7) with initial
conditions (4.8). There exists a subset $\Sigma$ of the starlike set

(5.17)
$\bigcup_{k=0}^{m}\left[-\left(1+\lambda\mu^{m}\right)^{1/m+1}\exp\left(\frac{2\pi
ik}{m+1}\right)\,,\,\left(1+\lambda\mu^{m}\right)^{1/m+1}\exp\left(\frac{2\pi
ik}{m+1}\right)\right].$

where given an arbitrary compact set
$\mathcal{K}\subset\mathbb{C}\setminus\Sigma$ the following equality holds
uniformly

(5.18) $\lim_{n\to\infty}\frac{T_{n-1}}{T_{n}}=\frac{1}{\omega_{0}}$

where $\omega_{0}$ is the branch solution of the algebraic equation (2.1) with
$c=\lambda\mu^{m}$, as in Proposition 2.1. This is

(5.19) $0=\omega^{m+1}-z\omega^{m}+\mu\lambda^{m}.$

###### Proof.

From Remark 5.6 we have that all zeros of $T_{n}$,
$n=d(m+1)+r\in\mathbb{Z}_{+}$ belong to a subset $\Sigma$ to the starlike set
in (5.17). The quotient

(5.20)
$\frac{T_{n-1}(z)}{T_{n}(z)}=z^{\kappa}\frac{h_{n-1}(z^{m+1})}{h_{n}(z^{m+1})}\in\mathcal{H}\left(\overline{\mathbb{C}}\setminus\Sigma\right),\quad\kappa\in\\{-1,m\\}.$

Take a compact domain $\mathcal{K}=\mathbb{C}\setminus\Sigma$. Let us
introduce the following magnitudes:

$\mbox{dist}(\mathcal{K},\Sigma)=\min_{(z,\zeta)\in\mathcal{K}\times\Sigma}\left|z-\zeta\right|,$

and

$M(\mathcal{K})=\left\\{\begin{array}[]{l c
l}\displaystyle\max_{z\in\mathcal{K}}\left|z^{m}\right|&\mbox{if}&\kappa=m\\\
&&\\\
\displaystyle\frac{1}{\mbox{dist}(\mathcal{K},\Sigma)}&\mbox{if}&\kappa=-1\end{array}\right.$

Considering Lemma 5.7 we decompose the above quotient in simple fractions

$\left|\frac{T_{n-1}(z)}{T_{n}(z)}\right|=M(\mathcal{K})\sum_{k=0}^{m}\sum_{j=1}^{d}\left|\frac{\lambda_{n,j,k}}{z-\zeta_{n,j}\exp\frac{2\pi
i}{m+1}}\right|$
$\leq\frac{M(\mathcal{K})}{\mbox{dist}(\mathcal{K},\Sigma)}\sum_{k=0}^{m}\sum_{j=1}^{d}\left|\lambda_{n,j,k}\right|=\frac{M(\mathcal{K})}{\mbox{dist}(\mathcal{K},\Sigma)}\sum_{k=0}^{m}\sum_{j=1}^{d}\lambda_{n,j,k}$
$=\frac{M(\mathcal{K})}{\mbox{dist}(\mathcal{K},\Sigma)}(m+1)\mu\lambda^{m}.$

This implies that the sequence of quotients $T_{n-1}(z)/T_{n}$,
$n\in\mathbb{Z}_{+}$ is a normal family of holomorphic functions in
$\mathcal{K}$, hence each subsequence of it contains a convergent subsequence.
We then prove that all convergent subsequence has the limit stated in (5.18).

Consider a convergent subsequence $T_{n^{\prime}-1}(z)/T_{n^{\prime}}$,
$n^{\prime}\in\Lambda\subset\mathbb{Z}_{+}$ with limit

$\lim_{n^{\prime}\to\infty}\frac{T_{n^{\prime}-1}(z)}{T_{n^{\prime}}(z)}=\ell(z)$

From recurrence relation (4.7) we obtain that

$0=1-z\ell(z)+\mu\lambda^{m}\ell^{m+1}\quad\mbox{or equivelently}\quad
0=\left(\frac{1}{\ell}\right)^{m+1}-z\left(\frac{1}{\ell}\right)^{m}+\mu\lambda^{m}.$

The function $1/\ell$ is a solution of (5.19). The branch that admites
analytic extension to $\overline{\mathbb{C}}\setminus\Sigma$ is
$1/\omega_{0}$. We have proved (5.20) for any compact set in
$\overline{\mathbb{C}}\setminus\Sigma$. This completes the proof. ∎

In Remark 5.11, we explain that the equality in (5.18) holds in
$\overline{\mathbb{C}}\setminus S_{0}$, where $S_{0}$ is defined in (2.2),
with $c=\mu\lambda^{m}$

$S_{0}=\bigcup_{k=0}^{m}\left[0,\,a\exp{\frac{2\pi ik}{m+1}}\right],\quad
a=\frac{m+1}{m}\left(\frac{\mu\lambda^{m}}{m}\right)^{m/m+1}.$

Let us recall the second kind polynomials in (5.1). The vector function

(5.21) ${\bf
R}_{n}(z)=\left(\frac{T_{n,1}}{T_{n}},\ldots,\frac{T_{n,m}}{T_{n}}\right),\quad
n\in\mathbb{Z}_{+}$

is often called Hermite-Padé approximant associated to the system of functions
$\displaystyle{\bf f}=\left(f_{1},\ldots,f_{m}\right)$ and the multi-index
$\displaystyle{\bf
n}=\left(\kappa+1,\ldots,\kappa+1,\kappa,\ldots,\kappa\right)\in\mathbb{Z}_{+}^{m}$
with $\ell^{1}$-norm $|{\bf n}|=n=m\,\kappa+s$. In [7] we find that O. Perron
developed a formulation of Hermite-Padé approximation in terms of generalized
continued fractions. Later on V. I Parusnikov [26] considered a called Jacobi-
Perron’s algorithm based on this formulation. Finally V. Kalyaguin, in several
publications such as [16, 17] generalized such algorithm to extend Favar’s
theorem to the field of multi-orthogonality of polynomials. Since our problems
deals with an easier particular case, we arrive at explicit expressions of the
resolvent functions via corresponding sequences of Hermite-Padé approximants.

Let us recall (5.7) in Lemma 5.3. Each component of $\displaystyle{\bf
R}_{n}=\left(\frac{T_{n-1}}{T_{n}},\ldots,\frac{T_{n-m}}{T_{n}}\right)$ can be
written as

$\frac{T_{n,j}}{T_{n}}=\frac{T_{n-1}}{T_{n}}\frac{T_{n-2}}{T_{n-1}}\cdots\frac{T_{n-j}}{T_{n_{j}+1}},\quad
j\in\\{1,\ldots,m\\}.$

From Lemma 5.8 we obtain the following limits uniformly

(5.22)
$\lim_{n\to\infty}\frac{T_{n,j}}{T_{n}}(z)=\frac{1}{\omega_{0}^{j}(z)},\quad
j\in\\{1,\ldots,m\\},\quad\mathcal{K}\subset\overline{\mathbb{C}}\setminus\Sigma,$

with $\mathcal{K}$ being compact. As we mentioned before the equality can be
extended to compact sets in $\mathcal{K}\subset\overline{\mathbb{C}}\setminus
S_{0}$.

######  Proposition 5.9.

Set the resolvent functions $f_{j}(z),$ $j=1,\ldots,m$ as in (4.14)
corresponding to the operator ${\bf T}$. Then

(5.23) $f_{j}(z)=\frac{1}{\omega_{0}^{j}(z)},\quad j\in\\{1,\ldots,m\\},$

in every compact $\mathcal{K}\subset\overline{\mathbb{C}}\setminus\Sigma.$

###### Proof.

We already have (5.22), then we only need to prove that the components of
Hermite-Padé approximants $T_{n,j}/T_{n}$ converge to their corresponding
resolvent functions $f_{j}$. Let us recall Lemma 5.1

(5.24)
$f_{j}(z)-\frac{T_{n,j}(z)}{T_{n}(z)}=\mathcal{O}\left(\frac{1}{z^{n+\tau(n,j)}}\right)\quad\mbox{as}\quad
z\to\infty,$

with

$\tau(n,j)=\left\\{\begin{array}[]{l l
l}\kappa+m+j-s+1&\mbox{if}&j\in\\{1,\ldots,m\\}\\\ &&\\\
\kappa+j-s+1&\mbox{if}&j\in\\{s+1,\ldots,m\\}.\end{array}\right.$

Using the results enumerated in Remark 5.6 we have that all zeros of $T_{n}$
belong to a subset of the the starlike set $\Sigma$ and also $\mbox{supp}({\bf
T})\subset\Sigma$. This yields

(5.25)
$f_{j}(z)-\frac{T_{n,j}(z)}{T_{n}(z)}\in\mathcal{H}\left(\overline{\mathbb{C}}\setminus\Sigma)\right),$

Combining (5.24) and (5.25) we obtain

$\left(f_{j}-\frac{T_{n,j}}{T_{n}}\right)\omega_{0}^{n+\tau(n,j)}\in\mathcal{H}\left(\overline{\mathbb{C}}\setminus\Sigma\right).$

Let us consider $\alpha>1$ such that the level curve

$\gamma_{\alpha}=\left\\{z\in\mathbb{C}\setminus
S_{0}:\left|\frac{1}{\omega_{0}(z)}\right|=\alpha\right\\}\subset\mathbb{C}\setminus\Sigma.$

Note that $\gamma_{\alpha}$ surrounds $\Sigma$. We take a compact set
$\mathcal{K}\subset\overline{\mathbb{C}}\setminus\Sigma$ lying in the
unbounded domain (connected and open region) whose boundary is
$\gamma_{\alpha}$. Using maximum principle for $1/\omega_{0}$, the fact that
$f_{j}$ is analytic on $\mathcal{K}$, which yields there exists $C>0$ such
that $|f_{j}(z)|\leq C$, $z\in\mathcal{K}$, and Lemma 5.13, we obtain that

$\left|\left|\left(f_{j}(z)-\frac{T_{n,j}(z)}{T_{n}(z)}\right)\right|\right|_{\mathcal{K}}\leq\frac{C+(m+1)\mu\lambda^{m}}{\mbox{dist}(\Sigma,\gamma_{\alpha})}\left(\frac{1}{\displaystyle\alpha\min_{z\in\mathcal{K}}\left|\omega_{0}(z)\right|}\right)^{n+\tau(n,j)}.$

Since $\displaystyle\alpha\min_{z\in\mathcal{K}}\left|\omega_{0}(z)\right|>1$,
we obtain that

$\limsup_{n\to\infty}\left|\left|\left(f_{j}(z)-\frac{T_{n,j}(z)}{T_{n}(z)}\right)\right|\right|_{\mathcal{K}}^{1/(n+\kappa)}=0.$

This proves the equality (5.23) in $\overline{\mathbb{C}}\setminus\Sigma$. ∎

Combining Proposition 2.3 and Proposition 5.9 we deduce immediately the
following result.

######  Corollary 5.10.

(5.26) $\displaystyle
f_{j}(z)=\frac{1}{\omega_{0}^{j}(z)}=\frac{1}{z^{j-1}}\int
x^{j-1}\,\widetilde{\rho}_{j-1}(x)\,\widetilde{\rho}(x)\,\frac{d\,x}{z-x},\quad
j=1,\ldots,m,$

in every compact $\mathcal{K}\subset\overline{\mathbb{C}}\setminus\Sigma$ and
$\widetilde{\rho},\widetilde{\rho}_{0},\widetilde{\rho}_{1},\ldots,\widetilde{\rho}_{m-1}$
being the functions introduced in Proposition 2.3.

######  Remark 5.11.

Despite our purpose does not need a description of $\Sigma$ beyond it is
bounded we state that $\Sigma=S_{0}$. This implies that Proposition 5.8,
Proposition 2.3, and Corollary 5.10 are extended to the set
$\overline{\mathbb{C}}\setminus S_{0}.$ According to [3] the functions
$\widetilde{\rho}_{1},\ldots,\widetilde{\rho}_{m}$ conforms a Nikishin system.
If we now consider the results given in publications [21, 22, 23] about
Nikishin systems on starlike sets, we conclude that every polynomial in the
sequence $\displaystyle\left\\{T_{n}\right\\}_{n\in\mathbb{Z}_{+}}$ vanishes
in the interior of $S_{0}$, being the set of those zeros dense in the whole
$S_{0}$. Hence $\mbox{supp}\left({\bf T}\right)=S_{0}$. Certainly these papers
require of such Nikishin systems to be generated by measures with bounded
supports, however the results on $T_{n}$’s zeros location are easily extended
for the case of unbounded supports, as we procede in [12] with respect to the
previous [11] in the real line.

######  Corollary 5.12.

For each $j\in\\{1,2,\ldots,m\\}$ the moments corresponding to ${\bf T}$ as in
(4.16) have integral forms

$c_{\nu,j}=\left({\bf T}\right)^{\nu}{\bf e}_{j-1}\cdot{\bf e}_{0}=\int
x^{\nu}\widetilde{\rho}_{j}(x)\,\widetilde{\rho}(x)\,dx,\quad\nu\in\mathbb{Z}_{+}.$

When $\nu.\in\\{0,\ldots,j-1\\}$ $c_{\nu,j}=0$.

###### Proof.

Fix $j\in\\{1,\ldots,m\\}$. The first $j-1$th cases are deduced by combining
Proposition 2.4 and Lemma 4.9. Let us recall the definition of the resolvent
function in (4.14) corresponding to the operator ${\bf T}$. Explicitly

$f_{j}(z)=\left(z\mathbb{I}-{\bf T}\right)^{-1}{\bf e}_{j-1}\cdot{\bf
e}_{0},\quad j=1,\ldots,m.$

Since $\displaystyle\mbox{supp}({\bf T})$ is bounded Neumann’s series for
operator inversion to $\displaystyle\left(z\mathbb{I}-{\bf T}\right)^{-1}$
satisfies the equality

$f_{j}(z)=\sum_{\nu=0}^{\infty}\frac{({\bf T})^{\nu}{\bf e}_{j-1}\cdot{\bf
e}_{0}}{z^{\nu+1}}=\sum_{\nu=0}^{\infty}\frac{c_{\nu,j}}{z^{\nu+1}}=\frac{1}{z^{j-1}}\sum_{\nu=0}^{\infty}\frac{c_{\nu+j-1,j}}{z^{\nu+1}}$

uniformly in any compact set contained in the widest open disk of
$\overline{\mathbb{C}}$ centered at $z=\infty$ leaving outside
$\mbox{supp}\left({\bf T}\right)$. Now using the equality (5.26) in Corollary
5.10 we obtain that uniformly in an open disk centered at $z=\infty$ we obtain
that

$f_{j}(z)=\frac{1}{z^{j-1}}\sum_{\nu=0}^{\infty}\frac{c_{\nu+j-1,j}}{z^{\nu+1}}=\frac{1}{z^{j-1}}\sum_{\nu=0}^{\infty}\frac{1}{z^{\nu+1}}\int
x^{\nu+j-1}\widetilde{\rho}_{j-1}(x)\,\widetilde{\rho}(x)\,dx,$

uniformly. This completes the proof. ∎

##  6\. Proof of Theorem 1.3

We deal with the resolvent functions in (4.14) corresponding to the operators
${\bf A}$, ${\bf T}$ and ${\bf L}$. In order to distinguish them for each
$j\in\\{1,\ldots,m\\}$ we relabel $f_{j}(\cdot,{\bf A})$, $f_{j}(\cdot,{\bf
T})$ and $f_{j}(\cdot,{\bf L})$. Let us recall the starlike $\Sigma_{0}$ in
Theorem 1.3

$\Sigma_{0}=\bigcup_{k=0}^{m}\left[-\lambda-\mu,-\lambda-\mu+\frac{m+1}{m}\left(\frac{\mu\lambda^{m}}{m}\right)^{1/(m+1)}\exp\frac{2\pi
ik}{m+1}\right].$

We also consider the starlike set

$V=$
$=\bigcup_{k=0}^{m}\left[\left(-\lambda-\mu-\sqrt[m+1]{1+\lambda\mu}\right)\exp\frac{2\pi
ik}{m+1},\left(-\lambda-\mu+\sqrt[m+1]{1+\lambda\mu}\right)\exp\frac{2\pi
ik}{m+1}\right].$

######  Lemma 6.1.

For each $j\in\\{1,\ldots,m\\}$ there is a set $U\subset V$ where

(6.1) $.f_{j}(z,{\bf A})=\frac{1}{\lambda^{j-1}}f_{j}(z+\lambda+\mu,{\bf
L}),\quad z\in\overline{\mathbb{C}}\setminus U.$

###### Proof.

Fix $j\in\\{1,\ldots,m\\}$. We use the identity (4.11) of Remark 4.3 in (4.14)
to obtain

$f_{j}(z,{\bf A})=\left[z\mathbb{I}-{\bf A}\right]^{-1}{\bf e}_{j-1}\cdot{\bf
e}_{0}=\left[z\mathbb{I}-{\bf\Lambda}^{-1}\left({\bf
L}-(\lambda+\mu)\right){\bf\Lambda}\right]^{-1}{\bf e}_{j-1}\cdot{\bf e}_{0}$
$={\bf\Lambda}\left[\left(z+\lambda+\mu\right)\mathbb{I}-{\bf
L}\right]^{-1}{\bf\Lambda}^{-1}{\bf e}_{j-1}\cdot{\bf
e}_{0}=\left[\left(z+\lambda+\mu\right)\mathbb{I}-{\bf
L}\right]^{-1}{\bf\Lambda}^{-1}{\bf e}_{j-1}\cdot{\bf\Lambda}^{\top}{\bf
e}_{0}$
$=\frac{1}{\lambda^{j-1}}\left[\left(z+\lambda+\mu\right)\mathbb{I}-{\bf
L}\right]^{-1}{\bf e}_{j-1}\cdot{\bf e}_{0}.$

This completes the proof. ∎

######  Lemma 6.2.

For each $j\in\\{1,\ldots,m\\}$ there is an open dick $\mathcal{V}$ in the
Riemann sphere $\overline{\mathbb{C}}$ centred at $z=\infty$ such that
$\mathcal{V}\cap\left(\Sigma\cup\\{\mu\\}\right)=\emptyset$ where on any
compact set the following equalities hold

(6.2) $\displaystyle f_{j}(z,{\bf
L})=\frac{1}{z^{j-1}}\int\frac{x^{j-1}\,\widetilde{\rho}_{j-1}(x)\,\widetilde{\rho}(x)\,d\,x}{z-x}+\int\frac{d(\delta_{\mu}-\delta_{0})(x)}{(z-x)^{j-1}},$

with $\widetilde{\rho},\widetilde{\rho}_{0},\ldots,\widetilde{\rho}_{m-1}$
being the functions introduced in Proposition 2.3. Equivalently:

(6.3) $\displaystyle f_{j}(z,{\bf
L})=\int\frac{\left(\widetilde{\rho}_{j-1}(x)\,\widetilde{\rho}(x)+\frac{(-1)^{j}}{(j-2)!}(\delta_{\mu}-\delta_{0})^{(j-2)}(x)\right)dx}{z-x}.$

###### Proof.

Note ${\bf L}={\bf T}-\mu\mathbb{I}_{m}$ where the infinite diagonal matrix
$\displaystyle\mathbb{I}_{m}=[e_{i,j}]_{(i,j)\in\mathbb{Z}_{+}^{2}}$ has form

$\mathbb{I}_{m}=[{\bf e}_{0}\,{\bf e}_{1}\,\cdots{\bf e}_{m-1}\,{\bf
0}]=\left\\{\begin{array}[]{l l
l}e_{j,j}=1&\mbox{if}&j\in\\{0,1,\ldots,m-1\\},\\\ &&\\\
e_{i,j}=0,&\mbox{otherwise}.&\end{array}\right.$

Then for each $j\in\\{1,2,\ldots,m\\}$

$f_{j}(z,{\bf L})={\bf e}_{j-1}\cdot\left[z\mathbb{I}-{\bf
L}^{\top}\right]^{-1}{\bf e}_{0}={\bf e}_{j}\cdot\left[z\mathbb{I}-{\bf
T}^{\top}+\mu\mathbb{I}_{m}\right]^{-1}{\bf e}_{0}$ $={\bf
e}_{j-1}\cdot\left[\left(\mathbb{I}+\mu\mathbb{I}_{m}\left(z\mathbb{I}-{\bf
T}^{\top}\right)^{-1}\right)\left(z\mathbb{I}-{\bf
T}^{\top}\right)\right]^{-1}{\bf e}_{0}.$

Hence

(6.4) $f_{j}(z,{\bf L})={\bf e}_{j-1}\cdot\left[z\mathbb{I}-{\bf
T}^{\top}\right]^{-1}\left[\mathbb{I}+\mu\mathbb{I}_{m}\left(z\mathbb{I}-{\bf
T}^{\top}\right)^{-1}\right]^{-1}{\bf e}_{0}$

The operator

$\mu\mathbb{I}_{m}\left(z\mathbb{I}-{\bf T}^{\top}\right)^{-1}\to
0\quad\mbox{as}\quad z\to\infty,$

uniformly. Taking into account its support is also bounded we have that there
is a neighborhood of $z=\infty$ where the following equalities hold uniformly

$\displaystyle\left[\mathbb{I}+\mu\mathbb{I}_{m}\left(z\mathbb{I}-{\bf
T}^{\top}\right)^{-1}\right]^{-1}=\mathbb{I}+\sum_{k=1}^{\infty}\mu^{k}\mathbb{I}_{m}\left[\left(z\mathbb{I}-{\bf
T}^{\top}\right)^{-1}\right]^{k}$
$=\mathbb{I}+\sum_{k=1}^{\infty}\frac{\mu^{k}}{z^{k}}\mathbb{I}_{m}\left[\sum_{l=0}^{\infty}\frac{{\bf
T}^{\top
l}}{z^{l}}\right]^{k}=\mathbb{I}+\sum_{k=1}^{\infty}\frac{\mu^{k}}{z^{k}}\mathbb{I}_{m}\sum_{l=0}^{\infty}{l+k-1\choose
k-1}\frac{{\bf T}^{\top l}}{z^{l}}.$

Let us consider the product

$\left[z\mathbb{I}-{\bf
T}^{\top}\right]^{-1}\left[\mathbb{I}+\mu\mathbb{I}_{m}\left(z\mathbb{I}-{\bf
T}^{\top}\right)^{-1}\right]^{-1}$ $=\left[z\mathbb{I}-{\bf
T}^{\top}\right]^{-1}\left(\mathbb{I}+\sum_{k=1}^{\infty}\frac{\mu^{k}}{z^{k}}\mathbb{I}_{m}\sum_{l=0}^{\infty}{l+k-1\choose
k-1}\frac{{\bf T}^{\top l}}{z^{l}}\right)$ $=\left[z\mathbb{I}-{\bf
T}^{\top}\right]^{-1}+\sum_{\tau=0}^{\infty}\sum_{k=1}^{\infty}\frac{\mu^{k}}{z^{k}}\sum_{l=0}^{\infty}{l+k-1\choose
k-1}\frac{{\bf T}^{\top\tau}{\mathbb{I}_{m}\bf T}^{\top l}}{z^{l+\tau}}.$

Fix $j\in\\{1,2,\ldots,m\\}$, and using (6.4) we obtain that

$f_{j}(z,{\bf L})=f_{j}(z,{\bf T})+{\bf
e}_{j-1}\cdot\sum_{k=1}^{\infty}\sum_{\ell=0}^{\infty}\sum_{\tau=0}^{\infty}\frac{\mu^{k}}{z^{k}}{\bf
T}^{\top\tau}\mathbb{I}_{m}{\ell+k-1\choose k-1}\frac{{\bf
T}^{\top\ell}}{z^{\ell+\tau}}{\bf e}_{0}.$

Then we have that

(6.5) $f_{j}(z,{\bf L})=f_{j}(z,{\bf T})+{\bf
e}_{j-1}\cdot\sum_{k=1}^{\infty}\sum_{\ell=0}^{m-1}\sum_{\tau=0}^{\infty}\frac{\mu^{k}}{z^{k}}{\ell+k-1\choose
k-1}\frac{{\bf T}^{\top\tau}{\bf e}_{\ell}}{z^{\ell+\tau}}.$

Observe that

${\bf e}_{j-1}\cdot{\bf T}^{\tau}{\bf e}_{\ell}={\bf e}_{j-1}\cdot{\bf
e}_{\ell+\tau}=\delta_{j-1,\ell+\tau}.$

Hence the expression (6.5) is then reducible to

$f_{j}(z,{\bf L})=f_{j}(z,{\bf
T})+\frac{1}{z^{j-1}}\sum_{\tau=0}^{j-1}\sum_{k=1}^{\infty}{j-1-\tau+k-1\choose
k-1}\frac{\mu^{k}}{z^{k}}$ $=f_{j}(z,{\bf
T})+\frac{1}{z^{j-1}}\sum_{k=1}^{\infty}\frac{\mu^{k}}{z^{k}}\sum_{\tau=0}^{j-1}{j-1-\tau+k-1\choose
k-1}$ $=f_{j}(z,{\bf
T})+\frac{1}{z^{j-1}}\sum_{k=1}^{\infty}\frac{\mu^{k}}{z^{k}}\sum_{\tau=0}^{j-1}{\tau+k-1\choose
k-1}=f_{j}(z,{\bf
T})+\frac{1}{z^{j-1}}\sum_{k=1}^{\infty}\frac{\mu^{k}}{z^{k}}\sum_{\tau=0}^{j-1}{\tau+k-1\choose\tau}$
$=f_{j}(z,{\bf T})+\frac{1}{z^{j-1}}\,\sum_{k=1}^{\infty}{j-1+k\choose
k}\frac{\mu^{k}}{z^{k}}=f_{j}(z,{\bf
T})+\left(\frac{1}{z-\mu}\right)^{j-1}-\frac{1}{z^{j-1}}.$

The proof of identity in (6.2) are derived from Corollary 5.10.

In order to prove the equivalence between (6.2) and (6.3) we observe that

$\frac{1}{(j-2)!}\frac{\partial^{j-2}}{\partial
x^{j-2}}\left(\frac{1}{z-x}\right)=\frac{1}{(z-x)^{j-1}}.$

From the definition of the distribution derivative corresponding to
$\delta_{\zeta}$ in (1.15) we obtain that

$\int\frac{(\delta_{\mu}-\delta_{0})(x)\,dx}{(z-x)^{j-1}}=\int\frac{1}{(j-2)!}\frac{\partial^{j-2}}{\partial
x^{j-2}}\left(\frac{1}{z-x}\right)\,(\delta_{\mu}-\delta_{0})\,dx$
$=\frac{(-1)^{j-2}}{(j-2)!}\int\frac{(\delta_{\mu}-\delta_{0})^{(j-2)}(x)\,dx}{z-x}.$

This chain of equalities completes the proof. ∎

For each $j=1,\ldots,m$, the integral form of the resolvent function
$f_{j}(z,{\bf L})$ in (6.2) contains a term which is

$f_{j}(z,{\bf
T})=\int\frac{\widetilde{\rho}_{j-1}(x)\,\widetilde{\rho}(x)\,dx}{z-x},$

as in (5.26) of Corollary 5.10 which is often called the Markov function
corresponding to the absolutely continuous measure with respect Lebesgue’s
whose weight is $x^{j-1}\,\widetilde{\rho}_{j}(x)\,\widetilde{\rho}(|x|)$. The
other term is a rational function with poles at a point $z=\mu$. There is no
publication on the behavior of its sequence of Hermite-Padé approximants and
their denominators $L_{n}$, $n\in\mathbb{Z}_{+}$ for these meromorphic
variation of Markov functions corresponding to measures supported on starlike
sets. Nevertheless since in our case the coefficients of the recurrence
relation are constant, we have obtained explicit expression in terms of the
solution of the algebraic equation in (5.19). In the case of measures
supported on sets of the real line we have several the papers on this topics
such as [8, 20] where the authors obtain convergence conditions of the
Hermite-Padé approximants in the sense of capacity.

From the disk $\mathcal{V}$ in Lemma 6.2 we construct
$\mathcal{U}=\\{z\in\overline{\mathbb{C}}:z+\lambda+\mu\in\mathcal{V}\\}$.

######  Proposition 6.3.

There exists a system of functions $(\rho,\rho_{1},\ldots,\rho_{m})$ such that
for each $j=1,\ldots,m$, $x^{j-1}\rho_{j}(x)\rho(|x|)$ conforms a weight on
$\Sigma_{0}$, and such that

(6.6) $\displaystyle f_{j}(z,{\bf
A})=\int\frac{\left(\rho_{j-1}(x)\,\rho(x)+\frac{(-1)^{j}}{\lambda^{j-1}(j-2)!}(\delta_{-\lambda}-\delta_{-\mu-\lambda})^{(j-2)}(x)\right)dx}{z-x}.$

with $z\in\mathcal{U}$, $j=1,\ldots,m$.

###### Proof.

We combine Lemma 6.1 and Lemma 6.2. Considering (6.1) let us plug
$\zeta+\lambda+\mu=z$ in (6.3) to obtain

$f_{j}(\zeta,{\bf
A})=\frac{1}{\lambda^{j-1}}\int\frac{\left(\widetilde{\rho}_{j-1}(x)\,\widetilde{\rho}(x)+\frac{(-1)^{j}}{(j-2)!}(\delta_{\mu}-\delta_{0})^{(j-2)}(x)\right)dx}{\zeta+\mu+\lambda-x}.$

We complete the proof taking the change of variable in the integral
$u=x-\lambda-\mu$, and denoting

$\rho(u)=\widetilde{\rho}(u+\lambda+\mu)\quad\mbox{and}\quad\rho_{j}(u)=\frac{1}{\lambda^{j-1}}\widetilde{\rho}_{j}(u+\lambda+\mu),\quad
j=1,\ldots,m.$

In order to avoid to use too many different letters in our notation in the
statement we recover $x$ and $z$ instead of $u$ and $\zeta$. ∎

######  Remark 6.4.

Since $\rho,\rho_{1},\ldots,\rho_{m}$ and
$\widetilde{\rho},\widetilde{\rho}_{1},\ldots,\widetilde{\rho}_{m}$ are
related by the change of variables $x=u+\lambda+\mu$, the functions
$\rho,\rho_{1},\ldots,\rho_{m}$ can be written in terms of the solutions of
the algebraic equation given in (1.9) analogously
$\widetilde{\rho},\widetilde{\rho}_{1},\ldots,\widetilde{\rho}_{m}$ are
expressed in terms of the ones corresponding to the equation in (2.1), Section
2.

Let us denote as $c_{\nu,j}({\bf T})$ and $c_{\nu,j}({\bf A})$,
$\nu\in\mathbb{Z}_{+}$, $j\in\\{1,\ldots,m\\}$ the momenta corresponding to
the operators ${\bf T}$ and ${\bf A}$, respectively, as in (4.16)

(6.7) $c_{\nu,j}({\bf T})={\bf T}^{\nu}{\bf e}_{j-1}\cdot{\bf
e}_{0}\quad\mbox{and}\quad c_{\nu,j}({\bf A})={\bf A}^{\nu}{\bf
e}_{j-1}\cdot{\bf e}_{0},\quad j=1,\ldots,m.$

######  Proposition 6.5.

Set the sequences of momenta $\displaystyle\left\\{c_{\nu,j}({\bf
A})\right\\}_{\nu\in\mathbb{Z}_{+}}$, $j=1,\ldots,m$. Then

(6.8) $c_{\nu,j}({\bf A})=\int
x^{\nu}\left[\rho(x)\,\rho_{j-1}(x)+\frac{(-1)^{j}}{\lambda^{j-1}(j-2)!}\left(\delta_{-\lambda}-\delta_{-\lambda-\mu}\right)^{(j-2)}(x)\right]\,dx.$

When $\nu.\in\\{0,\ldots,j-1\\}$ $c_{\nu,j}({\bf A})=0$.

###### Proof.

Fix $j\in\\{1,\ldots,m\\}$. Combining (6.6) and Neumann’s series for matrix
inversion to the operator $\displaystyle\left(z\mathbb{I}-{\bf
A}^{\top}\right)^{-1}$ we have that

$f_{j}(z,{\bf A})=\sum_{\nu=0}^{\infty}\frac{c_{\nu,j}({\bf A})}{z^{\nu+1}}$
$=\sum_{\nu=0}^{\infty}\frac{1}{z^{\nu+1}}\int
x^{\nu}\left[\rho(x)\,\rho_{j-1}(x)+\frac{(-1)^{j}}{\lambda^{j-1}(j-2)!}\left(\delta_{-\lambda}-\delta_{-\lambda-\mu}\right)^{(j-2)}(x)\right]\,dx.$

∎

The equality (1.16) can be deduced combining the second equality in (6.7), the
identity (4.12) in Proposition 4.5, and (6.8) in Proposition 6.5. This
completes the proof of Theorem 1.3.

##  7\. Proof of Theorem 1.4

Let us start by proving that the functions $P_{n,r}$,
$(n,r)\in\mathbb{Z}_{+}^{2}$ in (1.18) are entries of a solution of (1.3). The
integral expression in (1.18) is reduced to (1.16) when $t=0$, hence the
initial condition of the problem (1.3) holds. When we take the derivative
${\bf P}^{\prime}=[P^{\prime}_{n,r}]$ we obtain:

$P_{n,r}^{\prime}(t)=\int
e^{xt}x\,Q_{n}(x)\,\sum_{j=0}^{m-1}q_{j,r}(x)\,d\sigma_{j}(x).$

Using the recurrence relations that define $Q_{n}$ (1.5) with initial
conditions (1.7) in Section 1, we arrive at

$P_{i,j}^{\prime}(t)=\left\\{\begin{array}[]{l}-\lambda P_{i,j}(t)+\lambda
P_{i+1,j}(t),\quad(i,j)\in\left\\{0,1,\ldots,m-1\right\\}\times\mathbb{Z}_{+}=\mathbb{F},\\\
\\\ -(\lambda+\mu)P_{i,j}(t)+\lambda P_{i+1,j}(t)+\mu
P_{i-m,j}(t),\quad(i,j)\in\mathbb{Z}_{+}^{2}\setminus\mathbb{F},\end{array}\right.,$

which are the relations in (1.2). This proves that $P_{n,r}$,
$(n,r)\in\mathbb{Z}_{+}^{2}$ as in (1.18) are entries of a matrix ${\bf P}$
which satisfies (1.3).

We observe that $P_{n,r}$, $(n,r)\in\mathbb{Z}_{+}^{2}$ in (1.18) have
analytic continuations to the whole complex plane $\mathbb{C}$. The rest of
this Section concerns with showing that the initial value problem (1.3) has a
unique solution with analytic functions as entries, and they must be real on
$\mathbb{R}_{+}$. We also show that any other solution does not contain
transition probability functions in the entries. Finally we show that
$P_{n,r}$, $(n,r)\in\mathbb{Z}_{+}^{2}$ satisfy the four properties of a
transition probability functions sated in Remark 1.1. We divide this process
in two Subsections. In Subsection 7.1 we prove that there is a unique analytic
solution whose entries are uniformly bounded. In Subsection 7.2 we show that
the four properties corresponding to transition probability functions hold.

###  7.1. Solution of an infinite matrix initial value problem

We use techniques that appear in [30]. Consider initial value problem (1.3).
Observe that for each row of ${\bf P}$ its entries satisfy the same system of
equations, hence in order to simplify the notation for this Subsection, we
omit the dependence in $i\in\mathbb{Z}_{+}$. Set the infinite column vector
function

${\bf
X}(x)=\left(x_{0}(x),\,x_{1}(x),\ldots\right)^{\top}=\left(P_{i,0}(x),P_{i,1}(x),\ldots\right)^{\top}.$

Then ${\bf X}$ satisfies an initial value problem:

(7.1) $\left\\{\begin{array}[]{l l}{\bf X}^{\prime}(t)={\bf A}{\bf
X}(t),&t\geq 0,\\\ &\\\ {\bf X}(0)={\bf e}_{i},\end{array}\right.$

where ${\bf e}_{i}$ is a column vector of the identity matrix.

######  Proposition 7.1.

There is a unique solution $\displaystyle{\bf
X}(t)=\left(x_{0}(t),\,x_{1}(t),\ldots\right)^{\top}$ of the initial value
problem (7.1) that satisfies the two following equivalent conditions

  1.   2. (1)

The set of entries functions
$\displaystyle\left\\{x_{0}(t),\,x_{1}(t),\ldots\right\\}$ is uniformly
bounded in any compact disk centered at the origin with radius
$r\in\mathbb{R}_{+}$, $\displaystyle D_{r}=\left\\{z\in\mathbb{C}:|z|\leq
r\right\\}$.

  3.   4. (2)

The functions in $\displaystyle\left\\{x_{0}(t),\,x_{1}(t),\ldots\right\\}$
are analytic on $\mathbb{C}$ ($x_{j}\in\mathcal{H}(\mathbb{C})$,
$j\in\mathbb{Z}_{+}$).

  5.

This solution $\displaystyle{\bf X}$ has real values function components on
$\mathbb{R}_{+}$ ($x_{j}:\mathbb{R}_{+}\rightarrow\mathbb{R}$,
$j\in\mathbb{Z}_{+}$).

######  Remark 7.2.

Since the functions in (1.18) are analytic on $\mathbb{C}$, the above result
yields it is the only analytic solution stated in Theorem 1.4.

The proof of Proposition 7.1 could be extend to more general cases. We
consider the a linear system of infinitely many differential equations whose
unknown are now functions $y_{j}:D\subset\mathbb{C}\rightarrow\mathbb{C}$,
$j\in\mathbb{Z}_{+}$ as follows

(7.2) $\left\\{\begin{array}[]{r c
l}y_{0}^{\prime}&=&a_{0,u_{0}}\,y_{u_{0}}+a_{0,u_{0}+1}\,y_{u_{0}+1}+\cdots+a_{0,v_{0}}\,y_{v_{0}}\\\
&&\\\
y_{1}^{\prime}&=&a_{1,u_{1}}\,y_{u_{1}}+a_{1,u_{1}+1}\,y_{u_{1}+1}+\cdots+a_{1,v_{1}}\,y_{v_{1}}\\\
&\vdots&\end{array}\right.\quad t\geq 0,$

with the initial values $y_{j}(0)=b_{j},\quad j\in\mathbb{Z}_{+}$, and
coefficients

$a_{i,j}\in\mathbb{R},\quad j=u_{i},\ldots,v_{i},\quad\mbox{and}\quad
b_{i}\in\mathbb{R},\quad i\in\mathbb{Z}_{+},$

with $(u_{i},v_{i})\in\mathbb{Z}^{2}$, $i\in\mathbb{Z}_{+}$ such that there
exists $\ell\in\mathbb{Z}_{+}$

$v_{i}-u_{i}\leq\ell,\qquad i\in\mathbb{Z}_{+}.$

We also require that there exists an ordered pair $(M,B)\in\mathbb{R}_{+}^{2}$

(7.3) $\left|a_{i,j}\right|\leq
M,\quad(i,j)\in\mathbb{Z}_{+}^{2},\quad\mbox{and}\quad\left|b_{j}\right|\leq
B,\quad j\in\mathbb{Z}_{+}.$

######  Lemma 7.3.

There is a unique system of functions $y_{j}$ $j\in\mathbb{Z}_{+}$, which is
uniformly bounded in any compact interval of $\mathbb{R}_{+}$, that conforms a
solution of the initial value problem (7.2). In addition each $y_{j}$
$j\in\mathbb{Z}_{+}$ has an analytic continuation to the whole complex plane
$\mathbb{C}$ (Denote $y_{j}\in\mathcal{H}\left(\mathbb{C}\right)$, such that:

(7.4) $\left|y_{j}(z)\right|<\ell Be^{\ell Mr},\qquad
j\in\mathbb{Z}_{+},\quad|z|\leq r,\quad\forall r\in\mathbb{R}_{+}.$

###### Proof.

We use the method of successive approximations, as it is also applied to
systems of finitely many equations. Accordingly, for each $j\in\mathbb{Z}_{+}$
we define the iterated functions, which are all entier functions:

(7.5) $\left\\{\begin{array}[]{r c l}y_{j,0}&=&b_{j}\\\ &&\\\
\displaystyle\frac{y_{j,k}}{dt}&=&a_{j,u_{j}}\,y_{u_{j},k-1}+a_{j,u_{j}+1}\,y_{u_{j}+1,k-1}+\cdots+a_{j,v_{j}}\,y_{v_{j},-1}\\\
&&\\\ y_{j,k}(0)&=&b_{j}\end{array}\right.$

further

(7.6) $\left\\{\begin{array}[]{r c l}g_{j,0}&=&y_{j,0}=b_{j}\\\ &&\\\
g_{j,k}&=&y_{j,k}-y_{j,k-1}\end{array}\right..$

We will prove that

$\lim_{k\to\infty}y_{j,k}=\sum_{k=0}^{\infty}g_{j,k}=y_{j}$

exists and has the claimed properties. The limit is considered uniformly in a
compact disk of $\mathbb{C}$ centered at the origin with radius $r$, which
would proves that $y_{j}\in\mathcal{H}\left(\mathbb{C}\right)$,
$j\in\mathbb{Z}_{+}$.

First we estimate $g_{j,k}$ and prove the inequality

(7.7) $\left|g_{j,k}(t)\right|<\frac{(\ell Mt)^{k}}{k!}B\leq\frac{({\ell
Mr)}^{k}}{k!}B.$

This is evident when $k=0$. Observe the relations (7.3) and (7.6). Applying
complete induction, we assume that it has already been proven for $k-1$. Then,
combining (7.5) and (7.6) we obtain that

$\frac{dg_{j,k}}{dt}=\frac{dy_{j,k}}{dt}-\frac{dy_{j,k-1}}{dt}=\sum_{\tau=u_{j}}^{v_{j}}a_{j,\tau}\left(y_{\tau,k-1}-y_{\tau,k-2}\right)=\sum_{\tau=u_{j}}^{v_{j}}a_{j,\tau}g_{\tau,k-1}.$

So, since $g_{j,k}(0)=0$ we have that

$g_{j,k}(t)=\int_{0}^{t}\sum_{\tau=u_{j}}^{v_{j}}a_{j,\tau}g_{j,k-1}(x)\,dx.$

Taking into account the induction hypothesis and (7.3) we arrive at

(7.8)
$\left|g_{j,k}(t)\right|\leq\int_{0}^{t}\sum_{\tau=u_{j}}^{v_{j}}\left|a_{j,\tau}\right|\left|g_{j,k-1}(x)\right|\,dx$
$\leq\ell M\frac{\left(\ell
M\right)^{k-1}}{(k-1)!}B\int_{0}^{t}x^{k-1}dx=\frac{(\ell Mt)^{k}}{k!}B,$

which completes the proof of the inequality (7.7). Since the series

(7.9) $\sum_{k=0}^{\infty}\frac{(\ell Mr)^{k}}{k!}B=Be^{\ell Mr}$

converges, we have that the sequence $\displaystyle\sum_{k=0}^{n}g_{j,k}$,
$n\in\mathbb{Z}_{+}$ is uniformly convergent in any compact set of the complex
plane $\mathbb{C}$, for all $j\in\mathbb{Z}_{+}$. So denote

(7.10) $\sum_{k=0}^{n}g_{j,k}(z)=\widetilde{y}_{j}(z),\quad|z|\leq
r,\quad\forall r\in\mathbb{R}_{+}.$

Combining (7.8), (7.9) and (7.10) we also obtain that the sequence of
functions
$\displaystyle\left\\{\widetilde{y}_{j}\right\\}_{j\in\mathbb{Z}_{+}}$ is
uniformly bounded in any disk $D_{r}=\\{z\in\mathbb{C}:|z|\leq r\\}$

$\widetilde{y}_{j}(z)\leq Be^{\ell Mr},\quad|z|\leq r,\quad\forall
r\in\mathbb{R}_{+},\quad j\in\mathbb{Z}_{+}.$

Given $\varepsilon>0$ there always is a $k_{0}\in\mathbb{Z}_{+}$ such that for
every point of the disk $D_{r}=\\{z\in\mathbb{C}:|z|\leq r\\}$ the following
inequalities hold

(7.11) $\left|\widetilde{y}_{j}-y_{j,k}\right|<\varepsilon,\quad k\geq
k_{0},\quad\forall j\in\mathbb{Z}_{+}.$

Recall the identity (7.5). We have that

$y_{j,k}(t)=b_{j}+\int_{0}^{t}\sum_{\tau=u_{j}}^{v_{j}}a_{j,\tau}y_{\tau,k-1}(x)dx.$

Working with above equality we obtain that

$\displaystyle\widetilde{y}_{j}(t)=b_{j}+\int_{0}^{t}\sum_{\tau=u_{j}}^{v_{j}}a_{j,\tau}\widetilde{y}_{\tau}(x)dx+\widetilde{y}_{j}-y_{j,k}+\int_{0}^{t}\sum_{\tau=u_{j}}^{v_{j}}a_{j,\tau}\left(y_{\tau,k-1}-\widetilde{y}_{\tau}\right)(x)dx.$

We recall the condition (7.11), and we obtain that

$\displaystyle\left|\widetilde{y}_{j}(t)-b_{j}-\int_{0}^{t}\sum_{\tau=u_{j}}^{v_{j}}a_{j,\tau}\widetilde{y}_{\tau}(x)dx\right|$
$=\left|\widetilde{y}_{j}-y_{j,k}+\int_{0}^{t}\sum_{\tau=u_{j}}^{v_{j}}a_{j,\tau}\left(y_{\tau,k-1}-\widetilde{y}_{\tau}\right)(x)dx\right|$
$\leq\left|\widetilde{y}_{j}-y_{j,k}\right|+\int_{0}^{t}\sum_{\tau=u_{j}}^{v_{j}}\left|a_{j,\tau}\right|\left|y_{\tau,k-1}-\widetilde{y}_{\tau}\right|(x)dx$
$<\varepsilon\left(1+\ell M\int_{0}^{t}dx\right)\leq\varepsilon\left(1+\ell
Mr\right).$

Then all the convergent subsequence have limits satisfying that

$\widetilde{y}_{j}(t)=b_{j}+\int_{0}^{t}\sum_{\tau=u_{j}}^{v_{j}}a_{j,\tau}\widetilde{y}_{\tau}(x)dx,\quad
t\in D_{r},\quad\forall j\in\mathbb{Z}_{+}.$

We observe that

$\widetilde{y}_{j}(0)=b_{j}\quad\mbox{and}\quad\frac{\widetilde{y}_{j}}{dt}=\sum_{\tau=u_{j}}^{v_{j}}a_{j,\tau}\widetilde{y}_{\tau}(t),$

which implies that the functions $\widetilde{y}_{j}$, $j\in\mathbb{Z}_{+}$
conform a system solution of the initial value problem (7.2), and is uniformly
bounded on any disk $D_{r}$, $r\in\mathbb{R}_{+}$.

In order to prove the uniqueness, we consider another uniformly bounded
sequence $\displaystyle\left\\{z_{j}\right\\}_{j\in\mathbb{Z}_{+}}$ in any
disk $D_{r}$, $r\in\mathbb{R}_{+}$ whose elements conform a solution of the
initial value problem (7.2). Then we have that

$(\widetilde{y}_{j}-z_{j})(0)=0$

and

(7.12)
$\left(\widetilde{y}_{j}-z_{j}\right)(t)=\int_{0}^{t}\sum_{\tau=u_{j}}^{v_{j}}a_{j,\tau}\left(\widetilde{y}_{\tau}-z_{\tau}\right)(x)\,dx,\quad
t\in[0,r].$

Since both sequences
$\displaystyle\left\\{\widetilde{y}_{j}\right\\}_{j\in\mathbb{Z}_{+}}$ and
$\displaystyle\left\\{z_{j}\right\\}_{j\in\mathbb{Z}_{+}}$ are uniformly
bounded in $D_{r}$, the set of differences $|z_{j}-\widetilde{y}_{j}|$,
$j\in\mathbb{Z}_{+}$ is also uniformly bounded in $D_{r}$. Fix $b\leq r$ and
consider the sequence of

$H_{j}=\max_{t\in[0,b]}\left|\widetilde{y}_{j}(t)-z_{j}(t)\right|,\qquad
j\in\mathbb{Z}_{+}.$

Then there exists a no negative value $\displaystyle
H=\max_{j\in\mathbb{Z}_{+}}M_{j}$. Then given $\varepsilon>0$ we can find
$k\in\mathbb{Z}_{+}$ such that

$H-H_{k}<\epsilon.$

We now consider $\displaystyle\varepsilon=\frac{\ell-1}{\ell}M$. Then

$H-H_{k}<\frac{\ell-1}{\ell}H\implies\frac{1}{\ell}H<H_{k}.$

Taking into account (7.12) we obtain that

$\left|\widetilde{y}_{j}-z_{j}\right|\leq\int_{0}^{t}\sum_{\tau=u_{j}}^{v_{j}}\left|a_{j,\tau}\right|\left|\widetilde{y}_{\tau}-z_{\tau}\right|(x)\,dx\implies\frac{1}{\ell}H<HM\ell
b\implies\frac{1}{\ell^{2}M}<b.$

Due $b$ is an arbitrary nonnegative value, we arrive at a contradiction that
proves the uniqueness stated in Lemma 7.3. The proof is completed taking into
account that by construction the functions
$g_{j,k}\in\mathcal{H}(\mathbb{C})$, $(j,k)\in\mathbb{Z}_{+}^{2}$, then
$\widetilde{y}_{j}\in\mathcal{H}(\mathbb{C})$, $j\in\mathbb{Z}_{+}$. ∎

We have proved that there is a unique solution of (7.2) whose set of functions
is uniformly bounded in every disk $D_{r}$, $r\in\mathbb{R}_{+}$. These
functions must also be analytic on $D_{r}$, $r\in\mathbb{R}_{+}$. We now show
that any solution with analytic functions is uniformly bounded, hence it is
the mentioned unique solution.

######  Lemma 7.4.

Let $y_{j}$, $j\in\mathbb{Z}_{+}$ be a system of analytic functions which is
solution of the initial value (7.2). Such system is uniformly bounded in any
disck $\displaystyle D_{r}=\left\\{z\in\mathbb{C}:|z|\leq r\right\\}$,
$r\in\mathbb{R}_{+}$.

###### Proof.

Fix $r\in\mathbb{R}_{+}$, and assume that $y_{j}$, $j\in\mathbb{Z}_{+}$ are
analytic functions in $\mathbb{C}$ that solve the initial value problem in
(7.2). Take the system of functions $\widetilde{y}_{j}$, $j\in\mathbb{Z}_{+}$
constructed in the proof of Lemma 7.3. We have that
$h_{j}=y_{j}-\widetilde{y}_{j}\in\mathcal{H}(\mathbb{C})$,
$j\in\mathbb{Z}_{+}$. By linearity we have that:

$\left\\{\begin{array}[]{r c
l}h_{0}^{\prime}(t)&=&a_{0,u_{0}}\,h_{u_{0}}(t)+a_{0,u_{0}+1}\,h_{u_{0}+1}(t)+\cdots+a_{0,v_{0}}\,h_{v_{0}}(t)\\\
&&\\\
h_{1}^{\prime}(t)&=&a_{1,u_{1}}\,h_{u_{1}}(t)+a_{1,u_{1}+1}\,h_{u_{1}+1}(t)+\cdots+a_{1,v_{1}}\,h_{v_{1}}(t)\\\
&\vdots&\end{array}\right.\quad t\geq 0,$

with the initial values $h_{j}(0)=0\quad j\in\mathbb{Z}_{+}$. Differentiating
each equation $k$th times ($k\in\mathbb{Z}_{+}$) and evaluating at $t=0$ we
obtain inductively that

$\left\\{\begin{array}[]{r c
l}h_{0}^{(k+1)}(0)&=&a_{0,u_{0}}\,h_{u_{0}}^{(k)}(0)+a_{0,u_{0}+1}\,h_{u_{0}+1}^{(k)}(0)+\cdots+a_{0,v_{0}}\,h_{v_{0}}^{(k)}(0)=0\\\
&&\\\
h_{1}^{(k+1)}(0)&=&a_{1,u_{1}}\,h_{u_{1}}^{(k)}(0)+a_{1,u_{1}+1}\,h_{u_{1}+1}^{(k)}(0)+\cdots+a_{1,v_{1}}\,h_{v_{1}}^{(k)}(0)=0\\\
&\vdots&\end{array}\right..$

Then $h_{j}\equiv 0\implies y_{j}\equiv\widetilde{y}_{j},\quad
j\in\mathbb{Z}_{+}.$ Recalling (7.4) in Lemma 7.3 the proof is completed. ∎

We are now ready to prove Proposition 7.1

###### Proof.

( of Proposition 7.1)  Combining Lemmas 7.3 and 7.4 the equivalence between
conditions $(1)$ and $(2)$ in Proposition 7.1 is immediately deduced. We now
observe that if $b_{j}=b_{j,0}$, $j\in\mathbb{Z}_{+}$ in the initial
conditions $y_{j}(0)=b_{j}$ of Problem (7.2) are real, then we prove
inductively that all the derivatives $b_{k,j}=y^{(k)}_{j}(0)$,
$k\in\mathbb{Z}_{+}$ evaluated at $t=0$ must also be real

$\left\\{\begin{array}[]{r c
l}b_{0,k+1}=y_{0}^{(k+1)}(0)&=&a_{0,u_{0}}\,y_{u_{0}}^{(k)}(0)+a_{0,u_{0}+1}\,y_{u_{0}+1}^{(k)}(0)+\cdots+a_{0,v_{0}}\,y_{v_{0}}^{(k)}(0)\\\
&&\\\
b_{1,k+1}=y_{1}^{(k+1)}(0)&=&a_{1,u_{1}}\,y_{u_{1}}^{(k)}(0)+a_{1,u_{1}+1}\,h_{u_{1}+1}^{(k)}(0)+\cdots+a_{1,v_{1}}\,h_{v_{1}}^{(k)}(0)\\\
&\vdots&\end{array}\right..$

This completes the proof. ∎

###  7.2. Proofs of the four properties

Given an ordered pair $(n,r)\in\mathbb{Z}_{+}^{2}$, we now prove that the
function $P_{n,r}$ expressed with the integral form (1.18) in Theorem 1.4
satisfies the four properties enumerated in Remark 1.1. Previously we
introduce the infinite matrices $\displaystyle{\bf
F}=[f_{i,j}]_{(i,j)\in\mathbb{Z}_{+}^{2}}$ and $\displaystyle{\bf
G}=[g_{i,j}]_{(i,j)\in\mathbb{Z}_{+}^{2}}$ with forms

$f_{i,j}=\left\\{\begin{array}[]{l l}1,&\mbox{if}\quad j=i+1,\\\ &\\\
0,&\mbox{otherwise},\end{array}\right.\quad\mbox{and}\quad
g_{i,j}=\left\\{\begin{array}[]{l l}\displaystyle 1,&\mbox{if}\quad i=j+m,\\\
&\\\ 0,&\mbox{otherwise},\end{array}\right.\quad(i,j)\in\mathbb{Z}_{+}^{2}.$

Fixed $(a,b)\in\mathbb{Z}_{+}^{2}$ let us denote the matrix ${\bf F}^{a}{\bf
G}^{b}=\mathbb{E}_{a,b}=[e_{i,j,a,b}]_{(i,j)\in\mathbb{Z}_{+}^{2}}$, which has
form

$e_{i,j,a,b}=\left\\{\begin{array}[]{l l}1,&\mbox{if}\quad i=j+am-b,\\\ &\\\
0,&\mbox{otherwise},\end{array}\right.\quad(i,j)\in\mathbb{Z}_{+}^{2}.$

We accept that ${\bf E}_{0,0}=\mathbb{I}$. Consider the matrix ${\bf
T}=[t_{i,j}]_{(i,j)\in\mathbb{Z}_{+}^{2}}$ whose components are defined by

$t_{i,j}=\left\\{\begin{array}[]{l l}\lambda,&\mbox{if}\quad j=i+1,\\\ &\\\
\displaystyle\mu,&\mbox{if}\quad i=j+m,\\\ &\\\
0,&\mbox{otherwise}.\end{array}\right.\quad(i,j)\in\mathbb{Z}_{+}^{2}.$

Observe that ${\bf T}=\lambda{\bf F}+\mu{\bf G}$ and

${\bf
T}^{\nu}=\sum_{\ell=0}^{\nu}{\nu\choose\ell}\lambda^{\ell}\mu^{\nu-\ell}{\bf
F}^{\ell}{\bf G}^{\nu-\ell}.$

We also analyze the infinite matrix $\displaystyle{\bf
K}=[k_{i,j}]_{(i,j)\in\mathbb{Z}_{+}^{2}}=\mu\mathbb{I}_{m}+{\bf T}$, whose
entries explicitly are

$k_{i,j}=\left\\{\begin{array}[]{l l}\mu,&\mbox{if}\quad i=j,\quad 0\leq i\leq
m-1\\\ &\\\ \lambda,&\mbox{if}\quad j=i+1,\\\ &\\\ \mu,&\mbox{if}\quad
i=j+m,\\\ &\\\
0,&\mbox{otherwise},\end{array}\right.\quad(i,j)\in\mathbb{Z}_{+}^{2},$

with form

${\bf K}^{\top}=\left(\begin{array}[]{c c c c c c c c c c c c
c}\mu&0&0&\cdots&0&\mu&0&0&\cdots&0&0&\cdots\\\
\lambda&\mu&0&\cdots&0&0&\mu&0&\cdots&0&0&\\\
0&\lambda&\mu&\cdots&0&0&0&\mu&\ddots&0&0&\\\
0&0&\lambda&\cdots&0&0&0&0&&0&0\\\
\vdots&&\ddots&&\vdots&&\vdots&&\ddots&\ddots&\\\
0&0&\cdots&\lambda&\mu&0&0&0&&0&\mu&\\\
0&0&\cdots&0&\lambda&0&0&0&\cdots&0&0&\ddots\\\
0&0&\cdots&0&0&\lambda&0&0&\cdots&0&0&\ddots\\\
\vdots&&&\vdots&\ddots&\ddots&\ddots&\ddots&&\vdots&\vdots&\\\
0&0&\cdots&0&\cdots&0&0&\lambda&0&0&0&\\\
0&0&\cdots&0&\cdots&0&0&0&\lambda&0&0&\\\
\vdots&&&\vdots&&&\ddots&&\ddots&\ddots&&\ddots\end{array}\right).$

We rewrite the equalities (1.5), (1.7) in a formal matrix way as follows

${\bf K}\,{\bf Q}(x)=(x+\lambda+\mu){\bf Q}(x),\quad\mbox{where}\quad{\bf
Q}=\left(\begin{array}[]{c}Q_{0}\\\ Q_{1}\\\ \vdots\end{array}\right).$

For each $k\in\mathbb{Z}_{+}$, we have that $\displaystyle{\bf K}^{k}\,{\bf
Q}(x)=(x+\lambda+\mu)^{k}{\bf Q}(x)$, which is

$(x+\lambda+\mu)^{k}{\bf Q}(x)=\left(\mu\mathbb{I}_{m}+{\bf T}\right)^{k}{\bf
Q}(x)=\left({\bf T}^{k}+\sum_{j=1}^{k}{k\choose j}\mu^{j}\mathbb{I}_{m}{\bf
T}^{k-j}\right){\bf Q}(x)$
$=\left(\sum_{\ell=0}^{k}{k\choose\ell}\lambda^{\ell}\mu^{k-\ell}{\bf
F}^{\ell}{\bf G}^{k-\ell}+\sum_{j=1}^{k}\sum_{\ell=0}^{k-j}{k\choose
j}{k-j\choose\ell}\lambda^{\ell}\mu^{k-\ell}\mathbb{I}_{m}{\bf F}^{\ell}{\bf
G}^{k-j-\ell}\right){\bf Q}(x).$

Let $\Phi_{m}:\mathbb{R}\rightarrow\\{0,1\\}$ denote the Heaviside function

$\Phi_{m}(x)=\left\\{\begin{array}[]{l l}0&\mbox{if}\quad x<m,\\\ &\\\
1&\mbox{if}\quad x\geq m.\end{array}\right.$

We also denote the ceiling function by $\displaystyle\lceil
x\rceil=\min\left\\{c\in\mathbb{Z}:c\geq x\right\\}$.

######  Lemma 7.5.

Fix $(n,k)\in\mathbb{Z}_{+}^{2}$. Then

(7.13) $(x+\mu+\lambda)^{k}Q_{n}(x)=\sum_{j=0}^{k}{k\choose
j}\lambda^{j}\mu^{k-j}Q_{g(n,k,j)}(x)$
$=\sum_{\ell=\max\\{0,\lceil\frac{km-n}{m+1}\rceil\\}}^{k}{k\choose\ell}\lambda^{\ell}\mu^{k-\ell}Q_{n+\ell-m(k-\ell)}(x)$
$+\Phi_{m}(n)\sum_{j=1}^{k}\sum_{\ell=\max\\{0,\lceil\frac{(k-j)m-n}{m+1}\rceil\\}}^{k-j}{k\choose
j}{k-j\choose\ell}\lambda^{\ell}\mu^{k-\ell}Q_{n+\ell-m(k-j-\ell)}$

###### Proof.

Since $\displaystyle{\bf e}_{n}^{\top}{\bf Q}(x)=Q_{n}(x)$,
$n\in\mathbb{Z}_{+}$ we have that

$(x+\lambda+\mu)^{k}Q_{n}(x)$ $={\bf
e}_{n}^{\top}\left(\sum_{\ell=0}^{k}{k\choose\ell}\lambda^{\ell}\mu^{k-\ell}{\bf
F}^{\ell}{\bf G}^{k-\ell}+\sum_{j=1}^{k}\sum_{\ell=0}^{k-j}{k\choose
j}{k-j\choose\ell}\lambda^{\ell}\mu^{k-\ell}\mathbb{I}_{m}{\bf F}^{\ell}{\bf
G}^{k-j-\ell}\right){\bf Q}(x)$
$=\sum_{\ell=0}^{k}{k\choose\ell}\lambda^{\ell}\mu^{k-\ell}{\bf
e}_{n}^{\top}{\bf F}^{\ell}{\bf G}^{k-\ell}{\bf Q}(x)$
$+\sum_{j=1}^{k}\sum_{\ell=0}^{k-j}{k\choose
j}{k-j\choose\ell}\lambda^{\ell}\mu^{k-\ell}{\bf
e}_{n}^{\top}\mathbb{I}_{m}{\bf F}^{\ell}{\bf G}^{k-j-\ell}{\bf Q}(x)$
$=\sum_{\ell=\max\\{0,\lceil\frac{km-n}{m+1}\rceil\\}}^{k}{k\choose\ell}\lambda^{\ell}\mu^{k-\ell}{\bf
e}_{n+\ell-m(k-\ell)}^{\top}{\bf Q}(x)$
$+\sum_{j=1}^{k}\sum_{\ell=\max\\{0,\lceil\frac{(k-j)m-n}{m+1}\rceil\\}}^{k-j}{k\choose
j}{k-j\choose\ell}\lambda^{\ell}\mu^{k-\ell}\Phi_{m}(n){\bf
e}_{n+\ell-m(k-j-\ell)}^{\top}{\bf Q}(x)$