# Interpretation of the Schur–Cohn test
in terms of canonical systems

Masatoshi Suzuki

###### Abstract.

We solve direct and inverse problems for two-dimensional (quasi) canonical
systems related to exponential polynomials of a specific but sufficiently
general type. The approach to the inverse problem in this paper provides an
interpretation of the matrices and their determinants in the classical Schur-
Cohn test for polynomials in terms of Hamiltonians of canonical systems.

###### Key words and phrases:

Schur–Cohn test, canonical systems, inverse problem

###### 2000 Mathematics Subject Classification:

30C15, 34A55

##  1\. Introduction

This paper generalizes the results in [10] by considering a finite-dimensional
or discretized version of the theory of quasi-canonical systems in [11, 12],
but is presented in an almost self-contained fashion.

The subject of this paper is direct and inverse problems of quasi-canonical
systems, but we begin by stating the relation with the classical Schur–Cohn
test obtained from the main results, because it may be of interest to readers
in a wider field. On this account, we review the Schur–Cohn test originate
from Schur [8, 9] and Cohn [2]. Let
$f(x)=a_{d}x^{d}+a_{d-1}x^{d-1}+\cdots+a_{1}x+a_{0}$ be a complex polynomial
of degree $d$. Using the triangular matrices

$M_{n}(f):=\begin{bmatrix}a_{d}&a_{d-1}&\cdots&a_{d-n+1}\\\
&a_{d}&\cdots&a_{d-n+2}\\\ &&\ddots&\vdots\\\ &&&a_{d}\end{bmatrix},\quad
N_{n}(f):=\begin{bmatrix}a_{0}&a_{1}&\cdots&a_{n-1}\\\
&a_{0}&\cdots&a_{n-2}\\\ &&\ddots&\vdots\\\ &&&a_{0}\end{bmatrix},$

we define the matrices

$L_{n}^{\pm}(f):=\begin{bmatrix}{}^{\rm t}M_{n}(f)&\pm{}^{\rm
t}\overline{N_{n}(f)}\\\ \pm N_{n}(f)&\overline{M_{n}(f)}\end{bmatrix},$ (1.1)

and denote their determinants as

$D_{n}(f):=\det L_{n}^{\pm}(f)$

for $1\leq n\leq d$, where the bar means taking the complex conjugate of each
entry. Also define $D_{0}(f)=1$ for convenience. (We find that $\det
L_{n}^{+}(f)=\det L_{n}^{-}(f)$ by multiplying each of the $(n+1)$th to
$(2n)$th columns of $\det L_{n}^{-}(f)$ by $-1$ and then multiplying each of
the $(n+1)$th to $(2n)$th rows by $-1$.) The Schur–Cohn test associates the
sign changes of $D_{n}(f)$ with the distribution of the roots of $f$. Suppose
that the determinants $D_{n}(f)$ are all different from zero and that the
number of sign changes in the sequence $(D_{0}(f),D_{1}(f),\dots,D_{d}(f))$ is
$q$. Then $f$ has no roots on the unit circle
$\mathbb{T}=\\{z\in\mathbb{C}~{}:~{}|z|=1\\}$ and $d-q$ roots inside
$\mathbb{T}$ counting multiplicities. In particular, all roots of $f$ are
inside $\mathbb{T}$ if and only if $D_{n}(f)>0$ for all $n$ ([7, Corollaries
11.5.14 and 11.5.15]). For the history and related results on the Schur–Cohn
test, see Rahman–Schmeisser [7, §11.5 and pp.395–396] or Marden [6, §43], for
example. English translations of [8] and [9] are found in [4, pp. 31–60] and
[4, pp. 61–88], respectively.

To explain an interpretation of $D_{n}(f)$ in terms of quasi-canonical
systems, we introduce the exponential polynomial

$E_{f}(z)=e^{irdz/2}f(e^{-irz}),$ (1.2)

where $r=1$ if $d$ is even and $r=2$ if $d$ is odd. If all roots of $f$ are
inside $\mathbb{T}$, the exponential polynomial $E_{f}$ belongs to the
Hermite–Biehler class $\mathbb{HB}$, which is the class of all entire
functions satisfying the inequality

$|E^{\sharp}(z)|<|E(z)|\quad\text{for all $z\in\mathbb{C}_{+}$}$

and having no real zeros, where
$\mathbb{C}_{+}=\\{z\in\mathbb{C}:\Im(z)>0\\}$. Then, de Branges’ inverse
theorem in the theory of canonical systems asserts that there exists a
positive semi-definite quadratic real symmetric matrix-valued function $H_{f}$
defined on a subinterval $[t_{0},t_{1})$ of the real line such that a solution
$(A(t,z),B(t,z))$ of the canonical system

$-\frac{d}{dt}\begin{bmatrix}A(t,z)\\\
B(t,z)\end{bmatrix}=z\begin{bmatrix}0&-1\\\
1&0\end{bmatrix}H(t)\begin{bmatrix}A(t,z)\\\
B(t,z)\end{bmatrix}\quad(z\in\mathbb{C})$ (1.3)

for $H=H_{f}$ satisfying the boundary condition

$\lim_{t\to
t_{1}}\frac{\overline{A(t,z)}B(t,w)-\overline{B(t,z)}A(t,w)}{\pi(w-\bar{z})}=0$

recovers the original $E_{f}$ as $E_{f}(z)=A(t_{0},z)-iB(t_{0},z)$ ([1,
Theorem 40]).

In [10], we studied a method to construct $H_{f}$ for a class of polynomials
with real coefficient, since de Branges’ inverse theorem guarantees the
existence of $H_{f}$, but does not provide information about its concrete
form. By generalizing a method in [10] according to [12], we present an
explicit way to construct $H_{f}$ for many polynomials with complex
coefficient. As a result, we find that $H_{f}$ is a locally constant function
of the form:

$H_{f}(t)=\frac{1}{D_{n-1}(f)D_{n}(f)}\,\widetilde{H}_{f,n}\quad\text{for}\quad
r(n-1)/2\leq t<rn/2,\quad 1\leq n\leq d,$

where $\widetilde{H}_{f,n}$ are some positive definite matrices. In
particular, $H_{f}$ is positive definite if all roots of $f$ are inside
$\mathbb{T}$ by the Schur–Cohn test. This is consistent with the fact that a
matrix-valued function $H$ obtained by de Branges’ inverse theorem from a
function of $\mathbb{HB}$ takes values in a set of semi-positive definite
quadratic real symmetric matrices. Furthermore, the above method of
constructing $H_{f}$ works even if $E_{f}$ does not belong to $\mathbb{HB}$ if
at least $f$ and $f^{\sharp}=x^{d}\,\overline{f(1/x)}$ have no common roots,
in which case the sign change of $H_{f}$ describes the distribution of the
roots of $f$ by the Schur–Cohn test. This interpretation of $H_{f}$ by the
classical result is what was expected in [10, §7.5]. As the converse of the
above, that is, by solving a direct problem of quasi-canonical systems, we
obtain a polynomial $f$ having a specified number of roots in $\mathbb{T}$
from an appropriately chosen locally constant matrix valued function $H$
taking values in ${\rm Sym}_{2}(\mathbb{R})\cap{\rm SL}_{2}(\mathbb{R})$.

By associating the Schur–Cohn test with the theory of quasi-canonical systems
as described above, we find a correspondence between the set of all
polynomials $f$ of degree $d$ with $n$ roots in $\mathbb{T}$ and
$D_{d}(f)\not=0$ and the set of all sequence $(H_{1},\dots,H_{d})$ of
$H_{i}\in{\rm Sym}_{2}(\mathbb{R})\cap{\rm SL}_{2}(\mathbb{R})$ in which the
number of sign changes of the determinants is $d-n$:

$f~{}\overset{\text{inverse problem}}{\underset{\text{direct
problem}}{\rightleftarrows}}~{}(H_{1},\dots,H_{d}).$

This is rigorously stated as a one-to-one correspondence by using the main
theorems (Theorems 1.1, 1.2, and 1.3) stated below and by arranging the
settings appropriately.

To state the main results precisely, we explain the notion of quasi-canonical
systems. Let $H(t)$ be a quadratic real symmetric matrix-valued function
defined on a finite interval $I=[t_{0},t_{1})$. We refer to the first-order
system of differential equations (1.3) on $I$ parametrized by $z\in\mathbb{C}$
as a quasi-canonical system (on $I$) as well as [10] (but, as a difference, we
deal with the additive variable $t$ instead of a multiplicative variable, and
do not specify the condition at the right end of the interval $I$ when using
the word). A column vector-valued function ${}^{\rm
t}[A(\cdot,z)\,\,B(\cdot,z)]:I\to\mathbb{C}^{2\times 1}$ is called a solution
if it consists of absolutely continuous functions and satisfies (1.3) almost
everywhere on $I$ for every fixed $z\in\mathbb{C}$. A quasi-canonical system
(1.3) is called a canonical system if $H(t)$ is a real positive semi-definite
symmetric matrix for almost all $t$, $H\not\equiv 0$ on any subset of $I$ with
positive Lebesgue measure, and $H$ is locally integrable on $I$ with respect
to the Lebesgue measure $dt$. The matrix-valued function $H$ is called a
Hamiltonian of a canonical system. Abusing language, if it cause no confusion,
we often call $H$ a Hamiltonian if a quasi-canonical system (1.3) is not a
canonical system.

Let $d$ be a positive integer and set

$(L,r):=\begin{cases}(d/2,1)&\text{if $d$ is even},\\\ (d,2)&\text{if $d$ is
odd}.\end{cases}$ (1.4)

Then $2L=rd$. For a sequence $\mathcal{C}$ of complex numbers of length $d+1$
indexed as

$\mathcal{C}=(C_{L},C_{L-r},C_{L-2r},\cdots,C_{-L})\in\mathbb{C}^{d+1}\quad\text{with}\quad
C_{L}C_{-L}\not=0,$ (1.5)

we consider the exponential polynomial

$E(z):=E_{\mathcal{C}}(z):=\sum_{j=0}^{d}C_{L-rj}e^{i(L-rj)z}$ (1.6)

along with associated functions

$A(z):=A_{\mathcal{C}}(z):=\frac{1}{2}(E_{\mathcal{C}}(z)+E_{\mathcal{C}}^{\sharp}(z)),\quad
B(z):=B_{\mathcal{C}}(z):=\frac{i}{2}(E_{\mathcal{C}}(z)-E_{\mathcal{C}}^{\sharp}(z)).$
(1.7)

We also consider the polynomial

$f_{\mathcal{C}}(T):=\sum_{j=0}^{d}C_{-(L-rj)}\,T^{d-j}~{}\in\mathbb{C}[T]$

and denote related matrices and its determinants as

$L_{n}^{\pm}({\mathcal{C}}):=L_{n}^{\pm}(f_{\mathcal{C}}),\quad
D_{n}({\mathcal{C}}):=D_{n}(f_{\mathcal{C}}).$ (1.8)

An exponential polynomial $E_{\mathcal{C}}$ of (1.6) belongs to $\mathbb{HB}$
if and only if it has no zeros in the closed upper half-plane
$\overline{\mathbb{C}_{+}}=\\{z\in\mathbb{C}:\Im(z)\geq 0\\}$ ([5, Chapter
VII, Theorem 6]). The latter is equivalent that $f_{\mathcal{C}}$ has no roots
in the closed unit disk $\overline{\mathbb{D}}=\\{z\in\mathbb{C}:|z|\leq
1\\}$, since $E_{\mathcal{C}}$ and $f_{\mathcal{C}}$ are related as (1.2),
$E_{\mathcal{C}}(z)=e^{iLz}f_{\mathcal{C}}(e^{-irz})$, by definition.

If $E_{\mathcal{C}}$ belongs to $\mathbb{HB}$, there exists a Hamiltonian of a
canonical system corresponding to $E_{\mathcal{C}}$ in the sense of de
Branges’ inverse theorem. In the following, we describe an explicit method for
associating a Hamiltonian of a quasi-canonical system with exponential
polynomial $E_{\mathcal{C}}$, which does not necessarily belong to
$\mathbb{HB}$.

For every $1\leq n\leq d$, using the solutions of linear equations

$L_{n}^{\pm}({\mathcal{C}})\begin{bmatrix}z_{n}^{\pm}(1)\\\ z_{n}^{\pm}(2)\\\
\vdots\\\ z_{n}^{\pm}(n)\\\ \overline{z_{n}^{\pm}(n)}\\\
\overline{z_{n}^{\pm}(n-1)}\\\ \vdots\\\
\overline{z_{n}^{\pm}(1)}\end{bmatrix}=\mp\begin{bmatrix}0\\\ \vdots\\\ 0\\\
2\overline{C_{L}}\\\ 2C_{L}\\\ 0\\\ \vdots\\\ 0\end{bmatrix}$ (1.9)

for unknowns $z_{n}^{\pm}(1)\,\dots,z_{n}^{\pm}(n)$, where $2\overline{C_{L}}$
and $2C_{L}$ are $n$th and $(n+1)$th entries, respectively, we define a
quadratic real symmetric matrix $H_{n}=H_{n}(\mathcal{C})$ by

$\displaystyle\begin{bmatrix}\Re(z_{n}^{+}(1))&\Im(z_{n}^{+}(1))\\\
-\Im(z_{n}^{-}(1))&\Re(z_{n}^{-}(1))\end{bmatrix}\cdots\begin{bmatrix}\Re(z_{1}^{+}(1))&\Im(z_{1}^{+}(1))\\\
-\Im(z_{1}^{-}(1))&\Re(z_{1}^{-}(1))\end{bmatrix}\begin{bmatrix}0&-1\\\
1&0\end{bmatrix}H_{n}$ (1.10) $\displaystyle\quad=\begin{bmatrix}0&1\\\
-1&0\end{bmatrix}\begin{bmatrix}\Re(z_{n}^{+}(1))&\Im(z_{n}^{+}(1))\\\
-\Im(z_{n}^{-}(1))&\Re(z_{n}^{-}(1))\end{bmatrix}\cdots\begin{bmatrix}\Re(z_{1}^{+}(1))&\Im(z_{1}^{+}(1))\\\
-\Im(z_{1}^{-}(1))&\Re(z_{1}^{-}(1))\end{bmatrix}.$

Then, we obtain the following results for the inverse problem of quasi-
canonical system associated with exponential polynomials of the form (1.6).

######  Theorem 1.1.

Let $\mathcal{C}$ be a sequence of complex numbers of length $d+1$ as in (1.5)
and let $(L,r)$ be as in (1.4). Let $E=E_{\mathcal{C}}$ be an exponential
polynomial defined by (1.6). Suppose that $D_{d}(\mathcal{C})\not=0$. Then,

  1. (1)

matrices $H_{n}=H_{n}(\mathcal{C})$ of (1.10) are well-defined for all $1\leq
n\leq d$;

  2. (2)

the pair of functions $(A(t,z),B(t,z))$ defined in (2.10) below satisfies a
quasi-canonical system (1.3) associated with $H(t)$ defined by

$H(t)=H_{\mathcal{C}}(t):=H_{n}\quad\text{for}\quad r(n-1)/2\leq t<rn/2$
(1.11)

on the interval $t\in[0,L)$ together with the boundary conditions

$\displaystyle\begin{bmatrix}A(0,z)\\\
B(0,z)\end{bmatrix}=\begin{bmatrix}A(z)\\\ B(z)\end{bmatrix},\quad\lim_{t\to
L}\begin{bmatrix}A(t,z)\\\ B(t,z)\end{bmatrix}=\begin{bmatrix}A(0)\\\
B(0)\end{bmatrix},$ (1.12)

where $A(z)$ and $B(z)$ are functions of (1.7);

  3. (3)

functions $A(t,z)$ and $B(t,z)$ have the forms

$\displaystyle A(t,z)$
$\displaystyle=\frac{1}{2}\sum_{j=0}^{d-n}\Bigl{[}\,a_{n}(L-rj)\,e^{i(L-rj-t)z}+\overline{a_{n}(L-rj)}\,e^{-i(L-rj-t)z}\,\Bigr{]},$
(1.13) $\displaystyle B(t,z)$
$\displaystyle=\frac{1}{2}\sum_{j=0}^{d-n}\Bigl{[}\,b_{n}(L-rj)\,e^{i(L-rj-t)z}+\overline{b_{n}(L-rj)}\,e^{-i(L-rj-t)z}\,\Bigr{]},$

if $r(n-1)/2\leq t<rn/2$ and $1\leq n\leq d$, where $a_{n}(k)$ and $b_{n}(k)$
are explicit complex numbers depending only on $\\{H_{n}\\}_{1\leq n\leq d}$.

  4. (4)

there exist positive definite quadratic real symmetric matrices
$\widetilde{H}_{n}$ such that

$H_{n}=\frac{1}{D_{n-1}({\mathcal{C}})D_{n}({\mathcal{C}})}\,\widetilde{H}_{n}$
(1.14)

holds for all $1\leq n\leq d$. In particular, the positivity of
$H_{\mathcal{C}}(t)$ is equivalent to that of $D_{n}(\mathcal{C})$ for all
$1\leq n\leq d$.

######  Remark 1.1.

If $D_{d}(\mathcal{C})\not=0$, $E_{\mathcal{C}}$ has no real zeros (Lemma
3.4), especially $E_{\mathcal{C}}(0)\not=0$. Therefore, we can normalize as
$E_{\mathcal{C}}(0)=1$ or equivalent
$(A_{\mathcal{C}}(0),B_{\mathcal{C}}(0))=(1,0)$ by multiplying it by an
appropriate constant.

We mention another way of constructing $(H(t),A(t,z),B(t,z))$ in Section 5.

As mentioned above, $E_{\mathcal{C}}$ of (1.6) belongs to $\mathbb{HB}$ if and
only if $f_{\mathcal{C}}$ has no zeros in $\overline{\mathbb{D}}$. The latter
is equivalent that $D_{n}(\mathcal{C})$ are positive for all $1\leq n\leq d$
by Schur–Cohn test. Therefore, if one of these three equivalent conditions is
satisfied, $H(t)$ in Theorem 1.1 is defined and positive definite by (1.14):

######  Corollary 1.2.

For $\mathcal{C}$ of (1.5), the following are equivalent to each other:

  1. (1)

$E_{\mathcal{C}}$ belongs to $\mathbb{HB}$;

  2. (2)

$f_{\mathcal{C}}$ has no roots in $\overline{\mathbb{D}}$;

  3. (3)

$D_{n}(\mathcal{C})>0$ for all $1\leq n\leq d$;

  4. (4)

$H_{\mathcal{C}}(t)$ is positive definite for all $0\leq t<L$.

As a result of Theorem 1.1 and Corollary 1.2, if an exponential polynomial $E$
of (1.6) belongs to $\mathbb{HB}$, it is recovered as $E(z)=A(0,z)-iB(0,z)$ by
solving the canonical system attached to $H$ defined in (1.11), and, the
condition at the right-endpoint in (1.12) guarantees that this $H$ is nothing
but the one whose existence is stated in de Branges’ inverse theorem.

The descent of the order of $E(t,z)=A(t,z)-iB(t,z)$ given by Theorem 1.1 (3)
starting from $E(z)=E(0,z)$ is reminiscent of the relation with the Schur
transformation $f\mapsto\bar{a}_{0}f-a_{d}f^{\ast}$ and Cohn’s algorithm ([7,
§11.5]), but it is not known at present whether there is a concrete relation.

The converse of Theorem 1.1 is the direct problem for quasi-canonical systems
(1.3) with the Hamiltonians of the form (1.11). It is easier than the inverse
problem, because the Hamiltonians is a locally constant function.

######  Theorem 1.2.

Let $d\in\mathbb{Z}_{>0}$ and let $(H_{1},H_{2},\dots,H_{d})$ be a sequence of
matrices $H_{n}$ in ${\rm Sym}_{2}(\mathbb{R})\cap{\rm SL}_{2}(\mathbb{R})$.
Define a locally constant matrix-valued function $H(t)$ on $[0,L)$ by

$H(t)=H_{n}\quad\text{for}\quad r(n-1)/2\leq t<rn/2\quad(1\leq n\leq d),$
(1.15)

where $(L,r)$ are numbers in (1.4). Then the quasi-canonical system (1.3)
associated with $H(t)$ on $[0,L)$ together with the boundary condition

$\lim_{t\to L}\begin{bmatrix}A(t,z)\\\ B(t,z)\end{bmatrix}=\begin{bmatrix}A\\\
B\end{bmatrix}\not=0\quad(A,B\in\mathbb{R})$

has a unique solution ${}^{\rm t}[A(t,z)~{}B(t,z)]$ whose components have the
form (1.13). Therefore, for $r(n-1)/2\leq t<rn/2$, $E(t,z):=A(t,z)-iB(t,z)$ is
the exponential polynomial

$\displaystyle E(t,z)$
$\displaystyle=\frac{1}{2}\sum_{j=0}^{d-n}\Bigl{[}\,(a_{n}(L-rj)-ib_{n}(L-rj))e^{i(L-rj-t)z}$
$\displaystyle\qquad\qquad+\overline{(a_{n}(L-rj)+ib_{n}(L-rj))}e^{-i(L-rj-t)z}\,\Bigr{]}.$

Moreover, $E(t,0)=A-iB$ and $E(t,z)$ has no real zeros for any fixed $0\leq
t\leq L$. In particular, each Hamiltonian of the form (1.15), namely $H_{n}>0$
for all $n$, yields an exponential polynomial $E(0,z)=A(0,z)-iB(0,z)$
belonging to $\mathbb{HB}$.

######  Remark 1.3.

According to the normalization in Remark 1.1, we can normalize the initial
condition as $\begin{bmatrix}A\\\ B\end{bmatrix}=\begin{bmatrix}1\\\
0\end{bmatrix}$ by transformations $\begin{bmatrix}A(t,z)\\\
B(t,z)\end{bmatrix}\mapsto M\begin{bmatrix}A(t,z)\\\ B(t,z)\end{bmatrix}$ and
$H(t)\mapsto MH(t)M^{-1}$ for some $M\in{\rm GL}_{2}(\mathbb{R})$.

The choice of intervals in (1.15) depending on the parity of $d$ is only
adopted so that $H(t)$ has the same shape as the Hamiltonians obtained by
solving the inverse problem as in Theorem 1.1, and is not essential for
solving the direct problem.

Theorem 1.2 does not guarantee that the exponential polynomial $E(0,z)$ has
the form (1.6). In fact, $H(t)$ on $[0,1)$ with $H(t)=I_{2}$ for $0\leq t<1/2$
and $H(t)=-I_{2}$ for $1/2\leq t<1$ yields the constant function
$E(0,z)=A-iB$, and $H(t)$ on $[0,1)$ with $H(t)=I_{2}$ for $0\leq t<1$ yields
the function $E(0,z)=(A-iB)e^{-iz}$. It can be discriminated as follows
whether $E(0,z)$ has the form (1.6).

######  Theorem 1.3.

With the notation of Theorem 1.2, $E(0,z)$ is an exponential polynomial of the
form (1.6) with (1.5) if and only if

$\frac{1}{2}(I-iJH_{1})\cdot\frac{1}{2}(I-iJH_{2})\cdots\frac{1}{2}(I-iJH_{d})\begin{bmatrix}A\\\
B\end{bmatrix}$

is not proportional or equal to any of three vectors

${}^{\rm t}\\!\begin{bmatrix}1&i\end{bmatrix},\quad{}^{\rm
t}\\!\begin{bmatrix}1&-i\end{bmatrix},\quad{}^{\rm
t}\\!\begin{bmatrix}0&0\end{bmatrix}.$ (1.16)

If $E(0,z)$ has the form (1.6) with (1.5), we define $f(x)$ by
$f(e^{-irz})=e^{-irdz/2}E(0,z)$. Then $f$ is a polynomial of degree $d$ and
has $d-q$ roots inside $\mathbb{T}$ counting multiplicity, where $q$ is the
number of sign changes in $(H_{1},\dots,H_{d})$.

Theorem 1.3 generalizes a sufficient condition [10, Theorem 1.5] dealing with
the case $A=1$, $B=0$, $H_{i}={\rm diag}(1/\gamma_{i},\gamma_{i})$. In fact,
in that case, we have

$\displaystyle\begin{bmatrix}1&i\gamma_{1}\\\
-i/\gamma_{1}&1\end{bmatrix}\begin{bmatrix}1&i\gamma_{2}\\\
-i/\gamma_{2}&1\end{bmatrix}\cdots\begin{bmatrix}1&i\gamma_{d}\\\
-i/\gamma_{d}&1\end{bmatrix}\begin{bmatrix}1\\\ 0\end{bmatrix}$
$\displaystyle\quad=(\gamma_{1}\gamma_{2}\cdots\gamma_{d})^{-1}\begin{bmatrix}\gamma_{1}(\gamma_{1}+\gamma_{2})(\gamma_{2}+\gamma_{3})\cdots(\gamma_{d-1}+\gamma_{d})\\\
-i(\gamma_{1}+\gamma_{2})(\gamma_{2}+\gamma_{3})\cdots(\gamma_{d-1}+\gamma_{d})\end{bmatrix}.$

This can not be proportional to any vectors in (1.16) if $\gamma_{i}>0$ and
$\gamma_{1}\not=1$.

From the proof of the above results, we find that the correspondence

$E_{\mathcal{C}}~{}\mapsto~{}(H_{1},H_{2},\dots,H_{d}),\quad H_{n}\in{\rm
Sym}_{2}(\mathbb{R})\cap{\rm SL}_{2}(\mathbb{R}),$

obtained by Theorems 1.1, 1.2, and 1.3 is one-to-one under the normalization
$E(0)=1$ and the choice of intervals as in (1.11). This is compatible with the
uniqueness of Hamiltonians obtained in de Branges’ inverse theorem for entire
functions in the Hermite–Biehler class. Hence, the exponential polynomials
(1.6) belonging to the class $\mathbb{HB}$ are characterized in terms of the
positive-deﬁniteness of Hamiltonians as well as the case of real coefficient
in [10].

According to Corollary 1.2, there is nothing newer than the Schur–Cohn test
regarding the criteria by which a given $E_{\mathcal{C}}$ belongs to
$\mathbb{HB}$, and the results [10, Corollary 1.3, Theorems 1.6 and 1.7] are
reduced to the Schur–Cohn test. However, the method of associating
$E_{\mathcal{C}}$ with Hamiltonians of quasi-canonical systems and the
relation between the Hamiltonian $H_{\mathcal{C}}$ and determinants
$D_{n}(\mathcal{C})$ are new. The former is undoubtedly important for direct
and inverse problems for quasi-canonical systems, which is the subject of this
paper. The latter shows the existence of an interesting class of quasi-
canonical systems that are not necessarily canonical, and also contributes to
the simplification of the proofs of the main results. Conversely, by proving
the main theorems without using the Schur–Cohn test, another proof of the
Schur–Cohn test may be obtained, but this will not be discussed in this paper.

To prove Theorem 1.1, we assumed that $C_{L}C_{-L}\not=0$, but considering the
relation with the Schur–Cohn test, it is expected that it can be removed. In
fact, as in the case

$H_{1}(\mathcal{C})=\frac{1}{D_{1}({\mathcal{C}})}\begin{bmatrix}|C_{-L}-C_{L}|^{2}&2\,\Im(C_{L}C_{-L})\\\
2\,\Im(C_{L}C_{-L})&|C_{-L}+C_{L}|^{2}\end{bmatrix},$

it is observed for small $n$ that $H_{n}(\mathcal{C})$ makes sense even if one
of $C_{L}$ and $C_{-L}$ is zero. However, we have no idea to prove it for
general $n$ at present.

The paper is organized as follows. We outline the proof of Theorem 1.1 in
Section 2 after preparing the settings similar to [10, §2], and complete the
proof by filling in the details of Section 2 in Section 3. The discussion in
Section 3 is a generalization of [10], but the linear equations mainly studied
are changed (by considering a theory analogous to [12]), special matrices
handled in the proof are also changed, and the argument of proof is largely
simplified. In Section 4, we prove Theorems 1.2 and 1.3. In Section 5, we
mention an inductive way of constructing a triple $(H(t),A(t,z),B(t,z))$ in
(1.3) which is different from the way of Sections 2 and 3. The discussions of
these two sections are straightforward generalizations of [10, §5-6] according
to Section 3.

Acknowledgments This work was supported by JSPS KAKENHI Grant Number
JP17K05163 and was also supported by the Research Institute for Mathematical
Sciences, an International Joint Usage/Research Center located in Kyoto
University.

##  2\. Outline of the proof of Theorem 1.1

###  2.1. Hilbert spaces and operators.

Let $L^{2}(\mathbb{R}/(2\pi\mathbb{Z}))$ be the completion of the space of
$2\pi$-periodic continuous functions on $\mathbb{R}$ with respect to the
$L^{2}$-norm $\|f\|_{L^{2}}^{2}:=\langle f,f\rangle_{L^{2}}$, where $\langle
f,g\rangle_{L^{2}}:=(2\pi)^{-1}\int_{0}^{2\pi}f(z)\overline{g(z)}\,dz$. Every
$f\in L^{2}(\mathbb{R}/(2\pi\mathbb{Z}))$ has the Fourier expansion
$f(z)=\sum_{k\in\mathbb{Z}}u(k)e^{ikz}$ with $\\{u(k)\\}_{k\in\mathbb{Z}}\in
l^{2}(\mathbb{Z})$ and $\|f\|_{L^{2}}^{2}=\sum_{k\in\mathbb{Z}}|u(k)|^{2}$,
where $l^{2}(\mathbb{Z})$ is the Hilbert space of sequences
$\\{u(k)\in\mathbb{C}\,:\,k\in\mathbb{Z}\\}$ satisfying
$\sum_{k\in\mathbb{Z}}|u(k)|^{2}<\infty$.

Fix a positive integer $d$ and set $(L,r)$ as (1.4). For
$t\in\mathbb{R}\setminus((r/2)\mathbb{Z})$, we define the vector space

$V_{t}:=\left\\{\left.\phi(z)=e^{-itz}f(z)+e^{itz}g(z)\,\right|~{}f,\,g\in
L_{d}^{2}(\mathbb{R}/(2\pi\mathbb{Z}))\right\\}$

of functions of $z\in\mathbb{R}$, where
$L_{d}^{2}(\mathbb{R}/(2\pi\mathbb{Z}))=L^{2}(\mathbb{R}/(2\pi\mathbb{Z}))$ if
$d$ is even and $L_{d}^{2}(\mathbb{R}/(2\pi\mathbb{Z}))$ is the subspace of
$L^{2}(\mathbb{R}/(2\pi\mathbb{Z}))$ consisting of all Fourier series with odd
indices $k$ if $d$ is odd. We define the inner product on $V_{t}$ by

$\langle\phi_{1},\phi_{2}\rangle=\langle f_{1},g_{1}\rangle_{L^{2}}+\langle
g_{1},g_{2}\rangle_{L^{2}}$

for $\phi_{j}(z)=e^{-itz}f_{j}(z)+e^{itz}g_{j}(z)$ ($j=1,2$). Then $V_{t}$
with this inner product is a Hilbert space and is isomorphic to the
(orthogonal) direct sum $L^{2}(\mathbb{R}/(2\pi\mathbb{Z}))\oplus
L^{2}(\mathbb{R}/(2\pi\mathbb{Z}))$ of Hilbert spaces as well as [10, §2]. The
maps $p_{1}:(e^{-itz}f(z)+e^{itz}g(z))\mapsto e^{-itz}f(z)$ and
$p_{2}:(e^{-itz}f(z)+e^{itz}g(z))\mapsto e^{itz}g(z)$ are projections from
$V_{t}$ to the first and the second components of the direct sum,
respectively. We put

$X(k):=e^{i(r(k+1)-1-t)z},\quad Y(l):=e^{-i(r(l+1)-1-t)z}$ (2.1)

for $k,l\in\mathbb{Z}$ and $t\in\mathbb{R}$. We regard $X(k)$ and $Y(l)$ as
functions of $z$, functions of $(t,z)$, or symbols, depending on the
situation. For a fixed $t\in\mathbb{R}\setminus((r/2)\mathbb{Z})$, the
countable set consisting of all $X(k)$ and $Y(l)$ is linearly independent over
$\mathbb{C}$ as a set of functions of $z$, since the linear dependence of
$\\{X(k),\,Y(l)\\}_{k,l\in\mathbb{Z}}$ implies the existence of a nontrivial
pair of functions $f,g\in L^{2}(\mathbb{R}/(2\pi\mathbb{Z}))$ satisfying
$e^{-itz}f(z)+e^{itz}g(z)=0$. Using these vectors, $V_{t}$ is written as

$\displaystyle V_{t}$
$\displaystyle=\left\\{\phi=\sum_{k\in\mathbb{Z}}u(k)X(k)+\sum_{l\in\mathbb{Z}}v(l)Y(l-r+1)\,:~{}\\{u(k)\\}_{k\in\mathbb{Z}},\,\\{v(l)\\}_{l\in\mathbb{Z}}\in
l^{2}(\mathbb{Z})\right\\},$

and

$\langle\phi_{1},\phi_{2}\rangle=\sum_{k\in\mathbb{Z}}u_{1}(k)\overline{u_{2}(k)}+\sum_{l\in\mathbb{Z}}v_{1}(l)\overline{v_{2}(l)}\quad(\phi_{1},\phi_{2}\in
V_{t}),$
$\|\phi_{1}\|^{2}=\langle\phi_{1},\phi_{1}\rangle=\sum_{k\in\mathbb{Z}}|u_{1}(k)|^{2}+\sum_{l\in\mathbb{Z}}|v_{1}(l)|^{2}\quad(\phi_{1}\in
V_{t}),$ (2.2)

if

$\phi_{i}=\sum_{k\in\mathbb{Z}}u_{i}(k)X(k)+\sum_{l\in\mathbb{Z}}v_{i}(l)Y(l-r+1)\quad(i=1,2).$

On the other hand, we have

$\|\phi\|^{2}=\frac{1}{2\pi}\int_{0}^{2\pi}p_{1}\phi(z)\overline{p_{1}\phi(z)}\,dz+\frac{1}{2\pi}\int_{0}^{2\pi}p_{2}\phi(z)\overline{p_{2}\phi(z)}\,dz$

for $\phi\in V_{t}$, since

$\frac{1}{2\pi}\int_{0}^{2\pi}(e^{\pm itz}f_{1}(z))\overline{(e^{\pm
itz}f_{2}(z))}\,dz=\sum_{k\in\mathbb{Z}}u_{1}(k)\overline{u_{2}(k)}=\langle
f_{1},f_{2}\rangle_{L^{2}}$

for $f_{j}(z)=\sum_{k\in\mathbb{Z}}u_{j}(k)e^{ikz}\in
L^{2}(\mathbb{R}/(2\pi\mathbb{Z}))$ ($j=1,2$). Note that, for $\phi\in V_{t}$,
$p_{1}\phi$ and $p_{2}\phi$ are not periodic functions of $z$, but the
integrals $\int_{I}p_{j}\phi(z)\overline{p_{j}\phi^{\prime}(z)}\,dz$ ($j=1,2$,
$\phi,\phi^{\prime}\in V_{t}$) are independent of the intervals
$I=[\alpha,\alpha+2\pi]$ ($\alpha\in\mathbb{R}$). We write $\phi\in V_{t}$ as
$\phi(z)$ (respectively, $\phi(t,z)$) to emphasize that $\phi$ is a function
of $z$ (respectively, $(t,z)$). If we regard $X(k)$ and $Y(l)$ as symbols,
$V_{t}$, endowed with the norm defined by (2.2), is an abstract Hilbert space
isomorphic to $l^{2}(\mathbb{Z})\oplus l^{2}(\mathbb{Z})$.

For each nonnegative integer $n$, we define the closed subspace $V_{t,n}$ of
$V_{t}$ by

$\displaystyle{V_{t,n}=\left\\{\phi_{n}=\sum_{k=0}^{\infty}u_{n}(k)X(k)+\sum_{l=-\infty}^{n-1}v_{n}(l)Y(l-r+1)\,:~{}\\{u_{n}(k)\\}_{k=0}^{\infty},\,\\{v_{n}(l)\\}_{l=-\infty}^{n-1}\in
l^{2}(\mathbb{Z})\right\\}.}$

Define the projection ${\mathsf{P}}_{n}:V_{t}\to V_{t,n}$ by
${\mathsf{P}}_{0}=0$ and by

${\mathsf{P}}_{n}\phi=\sum_{k=0}^{n-1}u(k)X(k)+\sum_{l=0}^{n-1}v(l)Y(l-r+1)\quad(\phi\in
V_{t}),$

for $n\in\mathbb{Z}_{>0}$ (this ${\mathsf{P}}_{n}$ corresponds to
${\mathsf{P}}_{n}{\mathsf{P}}_{n}^{\ast}$ of [10, §2]). Also define the
involution ${\mathsf{J}}:\phi(z)\mapsto\overline{\phi(\bar{z})}$. Then,

$\displaystyle{\mathsf{J}}{\mathsf{P}}_{n}\phi$
$\displaystyle=\sum_{l=0}^{n-1}\overline{v(l)}X(l-r+1)+\sum_{k=0}^{n-1}\overline{u(k)}Y(k)$
(2.3)

for $\phi\in V_{t}$ and $n\in\mathbb{Z}_{>0}$.

Let $\mathcal{C}\in\mathbb{C}^{d+1}$ as in (1.5). Using the modified function
$E_{0}(z)=e^{-i(r-1)z}E_{\mathcal{C}}(z)$, we define two multiplication
operators

${\mathsf{E}}:\phi(z)\mapsto
E_{0}(z)\phi(z),\quad{\mathsf{E}}^{\sharp}:\phi(z)\mapsto
E_{0}^{\sharp}(z)\phi(z)$ (2.4)

on $V_{t}$. These operators map $V_{t}$ into $V_{t}$, because ${\mathsf{E}}$
and ${\mathsf{E}}^{\sharp}$ are expressed as

${\mathsf{E}}=\sum_{j=0}^{d}C_{L-rj}{\mathsf{T}}_{(L-rj-r+1)/r},\qquad{\mathsf{E}}^{\sharp}=\sum_{j=0}^{d}\overline{C_{L-rj}}\,{\mathsf{T}}_{-(L-rj-r+1)/r}$

by using shift operators ${\mathsf{T}}_{m}:V_{t}\to V_{t}$ ($m\in\mathbb{Z}$)
defined by

${\mathsf{T}}_{m}v=\sum_{k=-\infty}^{\infty}u(k)X(k+m)+\sum_{l=-\infty}^{\infty}v(l)Y(l-r+1-m).$

Both ${\mathsf{E}}$ and ${\mathsf{E}}^{\sharp}$ are bounded on $V_{t}$, since
$\|{\mathsf{E}}\|_{\rm
op}\leq\sum_{j=0}^{d}|C_{L-rj}|\cdot\|\mathsf{T}_{(L-rj-r+1)/r}\|_{\rm op}\leq
dM$ and $\|{\mathsf{E}}^{\sharp}\|_{\rm
op}\leq\sum_{j=0}^{d}|C_{L-rj}|\cdot\|\mathsf{T}_{-(L-rj-r+1)/r}\|_{\rm
op}\leq dM$ for $M=\max\\{|C_{L-rj}|\,|\,0\leq j\leq d\\}$. If
$E_{\mathcal{C}}$ has no zeros on the real line, ${\mathsf{E}}$ is invertible
on $V_{t}$ (Lemma 3.1). Thus the operator

$\Theta:={\mathsf{E}}^{-1}{\mathsf{E}}^{\sharp}$ (2.5)

is well-defined on $V_{t}$, and we have
$(\Theta\phi)(z)=(E_{0}^{\sharp}(z)/E_{0}(z))\phi(z)$ for $\phi\in V_{t}$.

###  2.2. Quasi-canonical systems associated with exponential polynomials.

Under the above settings, a quasi-canonical system associated with an
exponential polynomial $E(z)$ of (1.6) is constructed starting from solutions
of linear equations

$\left\\{\begin{aligned}
~{}({\mathsf{I}}+\Theta{\mathsf{J}}{\mathsf{P}}_{n})\,\phi_{n}^{+}&=X(0)-\Theta
Y(0),\\\
~{}({\mathsf{I}}-\Theta{\mathsf{J}}{\mathsf{P}}_{n})\,\phi_{n}^{-}&=X(0)+\Theta
Y(0),\end{aligned}\right.\quad(\phi_{n}^{\pm}\in
V_{t,n}+\Theta\mathsf{J}{\mathsf{P}}_{n}V_{t,n},\,0\leq n\leq d),$ (2.6)

where ${\mathsf{I}}$ is the identity operator. Note that the constant terms on
the right-hand sides are different from that of [10, §2–§3]. Suppose that
$D_{d}(\mathcal{C})\not=0$. Then both
$\mathsf{I}\pm\Theta\mathsf{J}{\mathsf{P}}_{n}$ are invertible on
$V_{t,n}+\Theta\mathsf{J}{\mathsf{P}}_{n}V_{t,n}$ for every $0\leq n\leq d$,
that is, $(\mathsf{I}\pm\Theta\mathsf{J}{\mathsf{P}}_{n})^{-1}$ exist as
bounded operators on $V_{t,n}+\Theta\mathsf{J}{\mathsf{P}}_{n}V_{t,n}$ (Lemma
3.2). Using unique solutions of (2.6), we define

$\displaystyle A_{n}^{\ast}(t,z):$
$\displaystyle=\frac{1}{2}(({\mathsf{I}}+{\mathsf{J}}){\mathsf{E}}\,(\phi_{n}^{+}+X(0)))(t,z),$
(2.7) $\displaystyle B_{n}^{\ast}(t,z):$
$\displaystyle=\frac{i}{2}(({\mathsf{I}}-{\mathsf{J}}){\mathsf{E}}\,(\phi_{n}^{-}+X(0)))(t,z).$

The functions $A_{n}^{\ast}(t,z)$ and $B_{n}^{\ast}(t,z)$ are entire functions
of $z$ and extend to functions of $t$ on $\mathbb{R}$ (by formula (3.5)). In
particular, for $n=0$,

$\displaystyle A_{0}^{\ast}(t,z)$
$\displaystyle=\frac{1}{2}\left(\mathsf{E}X(0)+\mathsf{E}^{\sharp}Y(0)\right)(t,z)=\frac{1}{2}\left(E_{0}(z)e^{i(r-t-1)}+E_{0}^{\sharp}(z)e^{-i(r-t-1)}\right),$
$\displaystyle B_{0}^{\ast}(t,z)$
$\displaystyle=\frac{i}{2}\left(\mathsf{E}X(0)-\mathsf{E}^{\sharp}Y(0)\right)(t,z)=\frac{i}{2}\left(E_{0}(z)e^{i(r-t-1)}-E_{0}^{\sharp}(z)e^{-i(r-t-1)}\right),$

since $\mathsf{P}_{0}=0$ by definition, and thus $A_{0}^{\ast}(0,z)=A(z)$ and
$B_{0}^{\ast}(0,z)=B(z)$.

In general, the equality $A_{n}^{\ast}(rn/2,z)=A_{n+1}^{\ast}(rn/2,z)$ may not
be hold and the same is true about $B_{n}^{\ast}(t,z)$. However, we will see
that the connection formula

$\begin{bmatrix}A_{n+1}^{\ast}(rn/2,z)\\\
B_{n+1}^{\ast}(rn/2,z)\end{bmatrix}=P_{n+1}^{\ast}\begin{bmatrix}A_{n}^{\ast}(rn/2,z)\\\
B_{n}^{\ast}(rn/2,z)\end{bmatrix}$ (2.8)

holds for some real matrix $P_{n+1}^{\ast}$, which is independent of $z$ for
every $1\leq n\leq d$ (Proposition 3.10). Therefore, we obtain functions
$A(t,z)$ and $B(t,z)$ of $(t,z)\in[0,L)\times\mathbb{C}$ which are continuous
for $t$ and entire for $z$ by defining

$\begin{bmatrix}A_{n}(t,z)\\\
B_{n}(t,z)\end{bmatrix}:=P_{n}\begin{bmatrix}A_{n}^{\ast}(t,z)\\\
B_{n}^{\ast}(t,z)\end{bmatrix}$ (2.9)

for $1\leq n\leq d$, where

$P_{n}:=(P_{1}^{\ast})^{-1}\cdots(P_{n}^{\ast})^{-1},$

and

$A(t,z):=A_{n}(t,z),\quad B(t,z):=B_{n}(t,z)$ (2.10)

for $r(n-1)/2\leq t<rn/2$. We find that $A(t,z)$ and $B(t,z)$ have the form
(1.13) ((2.9) and Lemma 3.6). Moreover, $(A(t,z),B(t,z))$ satisfies a quasi-
canonical system (1.3) for the locally constant quadratic real symmetric
matrix-valued function $H(t)$ defined by

$H(t):=H_{\mathcal{C}}(t):=H_{n}=\begin{bmatrix}\alpha_{n}&\beta_{n}\\\
\beta_{n}&\gamma_{n}\end{bmatrix}\quad\text{if}\quad r(n-1)/2\leq a<rn/2,$

where $H_{n}$ is defined by

$\displaystyle H_{n}$ $\displaystyle=\begin{bmatrix}0&1\\\
-1&0\end{bmatrix}P_{n}\begin{bmatrix}0&1\\\ -1&0\end{bmatrix}P_{n}^{-1},$
(2.11)

together with the boundary conditions (1.12) (Proposition 3.9 with (2.9) and
(2.10)). These $H_{n}$ are equal to the matrices defined in (1.10)
(Propositions 3.10 and 3.12). Equality (1.14) is obtained by studying the
solutions of equations in (1.9) (Proposition 3.12). As a summary of the above
argument, we obtain Theorem 1.1. See Section 3 for details.

On the other hand, Theorems 1.2 and 1.3 follow from the standard properties of
quasi-canonical systems as described in Section 4 .

##  3\. Proof of Theorem 1.1.

We complete the proof of Theorem 1.1 in this section by filling in the details
of the outline described in the previous section. We fix $d\in\mathbb{Z}_{>0}$
and a sequence $\mathcal{C}\in\mathbb{C}^{d+1}$ as in (1.5) throughout this
section.

######  Lemma 3.1.

Let ${\mathsf{E}}$ be the multiplication operator defined by (2.4) for
$E=E_{\mathcal{C}}$. Suppose that $E$ has no real zeros. Then ${\mathsf{E}}$
is invertible on $V_{t}$, and thus $\Theta$ of (2.5) is well-defined as a
bounded operator on $V_{t}$. Moreover $\|\Theta\|_{\rm op}=1$.

###### Proof.

It is sufficient to prove that ${\mathsf{E}}$ is invertible on
$L_{d}^{2}(\mathbb{R}/(2\pi\mathbb{Z}))$, since $V_{t}$ is a direct sum of
$e^{\pm itz}L_{d}^{2}(\mathbb{R}/(2\pi\mathbb{Z}))$ and $e^{\pm
2itz}(1/E_{0}(z))f(z)=g(z)$ is impossible for any $0\not=f,g\in
L_{d}^{2}(\mathbb{R}/(2\pi\mathbb{Z}))$. We have $1/E_{0}(z)\in
L^{\infty}(\mathbb{R}/(2\pi\mathbb{Z}))$ by assumption. Therefore,
multiplication by $1/E_{0}(z)$ defines a bounded operator ${\mathsf{E}}^{-1}$
on $L^{2}(\mathbb{R}/(2\pi\mathbb{Z}))$ with the norm
$\|{\mathsf{E}}^{-1}\|_{\rm op}=\|1/E_{0}\|_{L^{\infty}}$. Moreover
$\|\Theta\|_{\rm op}=\|E_{0}^{\sharp}/E_{0}\|_{L^{\infty}}=1$. Hence the case
of even $d$ is proved. For odd $d$, we find that $(1/E_{0}(z))f(z)\in
L_{d}^{\infty}(\mathbb{R}/(2\pi\mathbb{Z}))$ for $f\in
L_{d}^{\infty}(\mathbb{R}/(2\pi\mathbb{Z}))$, since $f(z)=E_{0}(z)g(z)$ is
impossible for a Fourier series $g\in L^{2}(\mathbb{R}/(2\pi\mathbb{Z}))$
containing $e^{ikz}$ of an even index. Hence the claim holds as well. ∎

######  Lemma 3.2.

Let $t\not\in(r/2)\mathbb{Z}$. Suppose that $E=E_{\mathcal{C}}$ has no real
zeros. Then, $\Theta\mathsf{J}\mathsf{P}_{n}$ defines a compact anti-linear
(conjugate linear) operator on
$V_{t,n}+\Theta\mathsf{J}{\mathsf{P}}_{n}V_{t,n}$ for each $0\leq n\leq d$.
Additionally suppose that $D_{d}({\mathcal{C}})\not=0$. Then,
$\mathsf{I}\pm\Theta\mathsf{J}{\mathsf{P}}_{n}$ are invertible on
$V_{t,n}+\Theta\mathsf{J}{\mathsf{P}}_{n}V_{t,n}$, and (2.6) have unique
solutions in $V_{t,n}+\Theta\mathsf{J}{\mathsf{P}}_{n}V_{t,n}$ for each $0\leq
n\leq d$.

######  Remark 3.3.

If $r\in r\mathbb{Z}/2$, $\mathsf{I}\pm\Theta\mathsf{J}\mathsf{P}_{n}$ may not
be invertible.

###### Proof.

The assertion is trivial for $n=0$, since ${\mathsf{P}}_{0}=0$ as an operator.
Let $n\geq 1$ and write
$W_{n}=V_{t,n}+\Theta\mathsf{J}{\mathsf{P}}_{n}V_{t,n}$. By definition,
$\mathsf{P}_{n}$ is a projection from $V_{t}$ into $V_{t,n}$, so
$\Theta\mathsf{J}{\mathsf{P}}_{n}$ is an operator on $W_{n}$. The image of
$W_{n}$ by $\mathsf{E}^{\sharp}\mathsf{J}\mathsf{P}_{n}$ is finite dimensional
by definition of $\mathsf{E}^{\sharp}$ and (2.3), thus
$\Theta\mathsf{J}\mathsf{P}_{n}=\mathsf{E}^{-1}(\mathsf{E}^{\sharp}\mathsf{J}\mathsf{P}_{n})$
is a finite rank operator which is compact. On the other hand,
$\|\Theta\mathsf{J}{\mathsf{P}}_{n}\|_{\rm op}\leq\|\Theta\|_{\rm
op}\cdot\|\mathsf{J}{\mathsf{P}}_{n}\|_{\rm op}\leq 1$. Therefore, if
$\Theta\mathsf{J}{\mathsf{P}}_{n}|_{W_{n}}$ has no eigenvalues of modulus one,
$\|\Theta\mathsf{J}{\mathsf{P}}_{n}|_{W_{n}}\|_{\rm op}<1$ and thus
$\mathsf{I}\pm\Theta\mathsf{J}{\mathsf{P}}_{n}$ are invertible on $W_{n}$ by
the convergence of Neumann series.

Assume that $\Theta{\mathsf{J}}{\mathsf{P}_{n}}\phi=\lambda\phi$ and
$|\lambda|=1$ for $\phi\in W_{n}$. Because $\Theta$ is an isometry on $V_{t}$
by Lemma 3.1, we have $\|\Theta{\mathsf{J}}{\mathsf{P}_{n}}\phi\|=\|\phi\|$,
and
$\|\Theta{\mathsf{J}}{\mathsf{P}_{n}}\phi\|^{2}=\|{\mathsf{J}}{\mathsf{P}_{n}}\phi\|^{2}=\sum_{k=0}^{n-1}|u(k)|^{2}+\sum_{l=0}^{n-1}|v(l)|^{2}$
by (2.3), while
$\|\phi\|^{2}=\sum_{k\in\mathbb{Z}}|u(k)|^{2}+\sum_{l\in\mathbb{Z}}|v(l)|^{2}$.
Thus,

$\phi=\sum_{k=0}^{n-1}u_{n}(k)X(k)+\sum_{l=0}^{n-1}v_{n}(l)Y(l-r+1).$

For such $\phi$,

$\displaystyle\mathsf{E}^{\sharp}{\mathsf{J}}{\mathsf{P}_{n}}\phi$
$\displaystyle=\sum_{j=0}^{d}\sum_{l=0}^{n-1}\overline{C_{-(L-rj)}}\overline{v(l)}X(l+(d-r+1)/2-j)$
$\displaystyle\quad+\sum_{j=0}^{d}\sum_{k=0}^{n-1}\overline{C_{-(L-rj)}}\overline{u(k)}Y(k-(d+r-1)/2+j)$

and

$\displaystyle\mathsf{E}\phi$
$\displaystyle=\sum_{j=0}^{d}\sum_{k=0}^{n-1}C_{L-rj}u(k)X(k+(d-r+1)/2-j)$
$\displaystyle\quad+\sum_{j=0}^{d}\sum_{l=0}^{n-1}C_{L-rj}v(l)Y(l-(d+r-1)/2+j).$

Comparing $2n$ coefficient of $X(k)$ with indices $-(d+r-1)/2\leq
k\leq-(d+r-1)/2+n-1$ and $(d-r+1)/2\leq k\leq(d-r+1)/2+n-1$ in the equality
$\lambda\mathsf{E}\phi-\mathsf{E}^{\sharp}\mathsf{J}\mathsf{P}_{n}\phi=0$, we
obtain the linear equation

$M_{\lambda}\cdot{}^{t}\\!\begin{bmatrix}u(0)&\cdots&u(n-1)&\overline{v(0)}&\cdots&\overline{v(n-1)}\end{bmatrix}=0,$
(3.1)

where $\displaystyle{M_{\lambda}=\begin{bmatrix}\lambda\cdot{}^{\rm
t}\\!M_{n}({\mathcal{C}})&\overline{{}^{\rm t}\\!N_{n}({\mathcal{C}})}\\\
\lambda\cdot
N_{n}({\mathcal{C}})&\overline{M_{n}({\mathcal{C}})}\end{bmatrix}.}$ Here,
$\det M_{\lambda}\not=0$ by assumption for $D_{d}({\mathcal{C}})$. Therefore
(3.1) has no nontrivial solutions, which implies $\phi=0$. Consequently, none
of $\lambda\in\mathbb{C}$ with modulus $1$ is an eigenvalue of
$\Theta\mathsf{J}{\mathsf{P}}_{n}|_{W_{n}}$, and hence complete the proof. ∎

######  Lemma 3.4.

Let $E=E_{\mathcal{C}}$.

  1. (1)

Suppose that $D_{d}(\mathcal{C})\not=0$. Then $E$ and $E^{\sharp}$ have no
common zeros. In particular, $E$ has no real zeros.

  2. (2)

Suppose that $E$ belongs to the Hermite–Biehler class $\mathbb{HB}$. Then
$D_{d}(\mathcal{C})\not=0$.

###### Proof.

The determinant $D_{d}(\mathcal{C})$ is zero if and only if
$f_{\mathcal{C}}(T)$ and
$f_{\mathcal{C}}^{\sharp}(T):=T^{d}\overline{f_{\mathcal{C}}(T^{-1})}$ have a
common root ([7, Lemmas 11.5.11 and 11.5.12]). The latter is equivalent that
$E$ and $E^{\sharp}$ have a common zero, since
$E(z)=e^{iLz}f_{\mathcal{C}}(e^{-irz})$ and
$E^{\sharp}(z)=e^{iLz}f_{\mathcal{C}}^{\sharp}(e^{-irz})$. In general, if an
entire function $F(z)$ has a real zero, it is also a zero of $F^{\sharp}(z)$.
Hence (1) holds. If $E$ belongs to $\mathbb{HB}$, it has no real zeros and
$|E(\bar{z})|<|E(z)|$ in $\mathbb{C}_{+}$ by definition of $\mathbb{HB}$.
Therefore $E$ and $E^{\sharp}$ have no common zeros. Hence (2) holds. ∎

In the remaining part of this section, we assume that $\mathcal{C}$ is taken
as in (1.5) and satisfies

$D_{d}(\mathcal{C})\not=0$

so that both $\mathsf{I}\pm\Theta\mathsf{J}\mathsf{P}_{n}$ are invertible on
$V_{t,n}+\Theta\mathsf{J}{\mathsf{P}}_{n}V_{t,n}$ for every $0\leq n\leq d$ by
Lemmas 3.1 and 3.2. Note that this assumption is satisfied if
$E_{\mathcal{C}}$ belongs to $\mathbb{HB}$ by Lemma 3.4.

Under the above assumption, we consider the equations

$\left\\{\begin{aligned}
~{}({\mathsf{E}}+{\mathsf{E}}^{\sharp}{\mathsf{J}}{\mathsf{P}}_{n})\,\phi_{n}^{+}&={\mathsf{E}}X(0)-{\mathsf{E}}^{\sharp}Y(0),\\\
~{}({\mathsf{E}}-{\mathsf{E}}^{\sharp}{\mathsf{J}}{\mathsf{P}}_{n})\,\phi_{n}^{-}&={\mathsf{E}}X(0)+{\mathsf{E}}^{\sharp}Y(0),\end{aligned}\right.\quad(\phi_{n}^{\pm}\in
V_{t,n}+\Theta\mathsf{J}{\mathsf{P}}_{n}V_{t,n},\,0\leq n\leq d),$ (3.2)

which is equivalent to (2.6), since ${\mathsf{E}}$ is invertible. Firstly, we
note that each $\phi\in V_{t,n}+\Theta\mathsf{J}{\mathsf{P}}_{n}V_{t,n}$ has
the absolutely convergent expansion

$\phi=\sum_{k=0}^{\infty}u(k)X(k)+\sum_{l=-\infty}^{n-1}v(l)Y(l-r+1)$

as a function of $z$ if $\Im(z)>0$ is large enough. This is trivial for
$\phi\in V_{t,n}$ and follows for
$\phi\in\Theta\mathsf{J}{\mathsf{P}}_{n}V_{t,n}$ from (2.3) and the expansion

$\frac{E_{0}^{\sharp}(z)}{E_{0}(z)}=\frac{\overline{C_{L}}+\sum_{j=1}^{d}\overline{C_{L-rj}}e^{irjz}}{C_{-L}+\sum_{j=1}^{d}C_{-(L-rj)}e^{irjz}}\cdot
e^{2i(r-1)z}=e^{2i(r-1)z}\sum_{m=0}^{\infty}\widetilde{C}_{m}e^{irmz}$

that holds if $\Im(z)>0$ is large enough. Secondly, we introduce several
special matrices to study (3.2). We define the square matrix
$\mathfrak{E}_{0}$ of size $8d$ by

$\mathfrak{E}_{0}=\mathfrak{E}_{0}(\mathcal{C}):=\left[\begin{array}[]{cc|cc|cc|cc}&&&&&&&\\\
\lx@intercol\hfil\raisebox{6.45831pt}[0.0pt]{$\mathfrak{e}_{0}(\mathcal{C})$}\hfil\lx@intercol\vrule\lx@intercol&&&&&&\\\
\hline\cr&&&&&&&\\\
&&\lx@intercol\hfil\raisebox{6.45831pt}[0.0pt]{$\mathfrak{e}_{0}(\mathcal{C})$}\hfil\lx@intercol\vrule\lx@intercol&&&&\\\
\hline\cr&&&&&&&\\\<EMAIL_ADDRESS>t}\overline{\mathfrak{e}_{0}(\mathcal{C})}$}\hfil\lx@intercol\vrule\lx@intercol&&\\\
\hline\cr&&&&&&&\\\
<EMAIL_ADDRESS>t}\overline{\mathfrak{e}_{0}(\mathcal{C})}$}\hfil\lx@intercol\end{array}\right],$

where $\mathfrak{e}_{0}(\mathcal{C})$ is the lower triangular matrix of size
$2d$ defined by

$\mathfrak{e}_{0}=\mathfrak{e}_{0}(\mathcal{C}):=\left[\begin{array}[]{llllllll}C_{-L}&&&&&&&\\\
C_{-L+r}&\ddots&&&&&&\\\ \vdots&\ddots&C_{-L}&&&&&\\\
C_{L-r}&\ddots&C_{-L+r}&C_{-L}&&&&\\\
C_{L}&\ddots&\vdots&C_{-L+r}&C_{-L}&&&\\\
0&\ddots&C_{L-r}&\vdots&\ddots&\ddots&&\\\
\vdots&\ddots&C_{L}&C_{L-r}&\ddots&C_{-L+r}&C_{-L}&\\\
0&\cdots&0&C_{L}&C_{L-r}&\cdots&C_{-L+r}&C_{-L}\\\ \end{array}\right],$

and define the square matrix $\mathfrak{E}_{n}^{\sharp}$ of size $8d$ by

$\mathfrak{E}_{n}^{\sharp}:=\mathfrak{E}_{n}^{\sharp}(\mathcal{C}):=\left[\begin{array}[]{cc|cc|cc|cc}&&&&&&&\\\
&&&&\lx@intercol\hfil\raisebox{6.45831pt}[0.0pt]{$\overline{\mathfrak{e}_{2,n}(\mathcal{C})}$}\hfil\lx@intercol\vrule\lx@intercol&&\\\
\hline\cr&&&&&&&\\\
&&&&&&\lx@intercol\hfil\raisebox{5.16663pt}[0.0pt]{$\overline{\mathfrak{e}_{1,n}(\mathcal{C})}$}\hfil\lx@intercol\\\
\hline\cr&&&&&&&\\\
\lx@intercol\hfil\raisebox{5.16663pt}[0.0pt]{$J_{2d}\cdot\mathfrak{e}_{1,n}(\mathcal{C})\cdot
J_{2d}$}\hfil\lx@intercol\vrule\lx@intercol&&&&&&\\\ \hline\cr&&&&&&&\\\
&&\lx@intercol\hfil\raisebox{4.30554pt}[0.0pt]{$J_{2d}\cdot\mathfrak{e}_{2,n}(\mathcal{C})\cdot
J_{2d}$}\hfil\lx@intercol\vrule\lx@intercol&&&&\end{array}\right]$

with

$\displaystyle\mathfrak{e}_{1,n}$
$\displaystyle=\mathfrak{e}_{1,n}(\mathcal{C}):=\left[\begin{array}[]{cc|lll|ccc}&&C_{L}&&&&\\\
&&C_{L-r}&\ddots&&&\\\ &&\vdots&\ddots&C_{L}&&\\\
<EMAIL_ADDRESS>0}}\hfil\lx@intercol\vrule\lx<EMAIL_ADDRESS>0}}\hfil\lx@intercol\\\ &&C_{-L}&\ddots&\vdots&&\\\ &&&\ddots&C_{-L+r}&&\\\
&&&&C_{-L}&&\\\ \hline\cr&&&&&&\\\
<EMAIL_ADDRESS>0}}\hfil\lx@intercol\vrule\lx<EMAIL_ADDRESS>0}}\hfil\lx@intercol\vrule\lx<EMAIL_ADDRESS>0}}\hfil\lx@intercol\\\ \end{array}\right]=\begin{array}[]{c|c|c|c}d&n&d-n&\\\
\hline\cr&&&d+n\\\ \hline\cr&&&d-n\end{array},$
$\displaystyle\mathfrak{e}_{2,n}$
$\displaystyle=\mathfrak{e}_{2,n}(\mathcal{C}):=\left[\begin{array}[]{cc|lll}&&&&\\\
<EMAIL_ADDRESS>0}}\hfil\lx@intercol\vrule\lx<EMAIL_ADDRESS>0}}\hfil\lx@intercol\\\ \hline\cr&&C_{L}&&\\\ &&C_{L-r}&\ddots&\\\
&&\vdots&\ddots&C_{L}\\\
<EMAIL_ADDRESS>0}}\hfil\lx@intercol&&C_{-L+r}&\ddots&C_{L-r}\\\ &&C_{-L}&\ddots&\vdots\\\
&&&\ddots&C_{-L+r}\\\
&&&&C_{-L}\end{array}\right]=\begin{array}[]{c|c|c}2d-n&n&\\\
\hline\cr&&d-n\\\ \hline\cr&&d+n\end{array},$

where the right-hand sides mean the size of each block of matrices in middle
terms and $J_{n}$ is the anti-diagonal matrix of size $n$:

$J_{n}=\begin{bmatrix}&&1\\\ &\iddots&\\\ 1&&\end{bmatrix}.$

We also define the column vector $\chi$ of length $8d$ by

$\chi=\chi_{8d}={}^{\rm t}\\!\begin{bmatrix}1&0&\cdots&0&1\,\,\end{bmatrix}$

and

$\mathfrak{J}=\mathfrak{J}^{(8d)}=\begin{bmatrix}&I_{4d}\\\
I_{4d}&\end{bmatrix}.$

where $I_{4d}$ is the identity matrix of size $4d$.

Let
$\phi_{n}^{\pm}=\sum_{k=0}^{\infty}u_{n}^{\pm}(k)X(k)+\sum_{l=-\infty}^{n-1}v_{n}^{\pm}(l)Y(l-r+1)$
be absolutely convergent expansions of the solutions of (3.2) for $0\leq n\leq
d$, where it is assumed that $\Im(z)>0$ is large enough. Using these
coefficient of $\phi_{n}^{\pm}$ and putting

$v_{n}^{\pm}(n)=v_{n}^{\pm}(n+1)\cdots=v_{n}^{\pm}(d-1)=0$

if $0\leq n\leq d-1$, we define the column vectors $\Phi_{n}^{\pm}$ of length
$8d$ by

$\Phi_{n}^{\pm}=\begin{bmatrix}\Phi_{n,1}^{\pm}\\\ \Phi_{n,2}^{\pm}\\\
J_{2d}\cdot\overline{\Phi_{n,2}^{\pm}}\\\
J_{2d}\cdot\overline{\Phi_{n,1}^{\pm}}\end{bmatrix},\quad\left\\{\begin{aligned}
\Phi_{n,1}^{\pm}&={}^{\rm
t}\\!\begin{bmatrix}u_{n}^{\pm}(0)&u_{n}^{\pm}(1)&\cdots&u_{n}^{\pm}(2d-1)\end{bmatrix},\\\
\Phi_{n,2}^{\pm}&={}^{\rm
t}\\!\begin{bmatrix}v_{n}^{\pm}(d-1)&v_{n}^{\pm}(d-2)&\cdots&v_{n}^{\pm}(-d)\end{bmatrix}.\end{aligned}\right.$
(3.3)

Substituting the above expansion of $\phi_{n}^{\pm}$ into (3.2), we obtain
linear equations

$(\mathfrak{E}_{0}\pm\mathfrak{E}_{n}^{\sharp})\cdot\Phi_{n}^{\pm}=\mathfrak{E}_{0}\cdot\chi\mp\mathfrak{J}\cdot\mathfrak{E}_{0}\cdot\chi\\\
\quad(0\leq n\leq d)$ (3.4)

by comparing coefficient of $X(k)$ and $Y(l)$ for $-(d+r-1)/2\leq
k,l\leq(3d-r-1)/2$, and

$\sum_{j=0}^{d}C_{-(L-rj)}u_{n}^{\pm}(J_{+}-j)=0,\quad\sum_{j=0}^{d}C_{-(L-rj)}v_{n}^{\pm}(J_{-}+j)=0$

for every $J_{+}\geq 2d$ and $J_{-}\leq-d-1$ by comparing other coefficient.

######  Lemma 3.5.

Let $0\leq n\leq d$. Then
$\det(\mathfrak{E}_{0}\pm\mathfrak{E}_{n}^{\sharp})\not=0$ if
$\mathsf{I}\pm\Theta\mathsf{J}\mathsf{P}_{n}$ is invertible on
$V_{t,n}+\Theta\mathsf{J}\mathsf{P}_{n}V_{t,n}$ or equivalently
$D_{n}(\mathcal{C})\not=0$.

###### Proof.

Let

$\displaystyle\mathbf{k}$
$\displaystyle=(n+1,n+2,\dots,2d;2d+n+1,2d+n+2,\dots,4d;$
$\displaystyle\qquad\qquad\qquad\qquad 4d+1,4d+2,\dots
6d-n;6d+1,6d+2,\dots,8d-n)$

be a list of indices of columns of
$\mathfrak{E}_{0}\pm\mathfrak{E}_{n}^{\sharp}$ and let

$\displaystyle\mathbf{k}_{1}=(2d;4d;4d+1;6d+1),~{}\mathbf{k}_{2}=(2d-1;4d-1;4d+2;6d+2),~{}\dots{}$

be sublists of $\mathbf{k}$. Expanding
$\det(\mathfrak{E}_{0}\pm\mathfrak{E}_{n}^{\sharp})$ with respect to columns
with indices $\mathbf{k}_{1}$, $\mathbf{k}_{2}$, $\dots$ in this order, we
have

$\det(\mathfrak{E}_{0}\pm\mathfrak{E}_{n}^{\sharp})=|C_{-L}|^{4(2d-n)}D_{n}(\mathcal{C})^{2}$

Therefore, we obtain the conclusion by Lemma 3.2, since $C_{-L}\not=0$ by
assumption. ∎

On the other hand, by (3.2), we have

$\displaystyle{\mathsf{E}}\,\phi_{n}^{\pm}$
$\displaystyle={\mathsf{E}}X(0)\mp{\mathsf{E}}^{\sharp}Y(0)\mp{\mathsf{E}}^{\sharp}{\mathsf{J}}{\mathsf{P}}_{n}\phi_{n}^{\pm}$
$\displaystyle=\sum_{j=0}^{d}C_{L-rj}X((d-r+1)/2-j)\mp\sum_{j=0}^{d}\overline{C_{L-rj}}Y((d-r+1)/2-j)$
$\displaystyle\quad\mp\sum_{j=0}^{d}\sum_{l=0}^{n-1}\overline{C_{L-rj}}\overline{v_{n}^{\pm}(l)}X(l-(d+r-1)/2+j)$
$\displaystyle\quad\mp\sum_{j=0}^{d}\sum_{k=0}^{n-1}\overline{C_{L-rj}}\overline{u_{n}^{\pm}(k)}Y(k+(d-r+1)/2-j).$

Therefore, we can write

${\mathsf{E}}(\phi_{n}^{\pm}+X(0))=\sum_{k=-(d+r-1)/2}^{(d-r+1)/2+n-1}\Bigl{(}p_{n}^{\pm}(k)X(k)+q_{n}^{\pm}(k)Y(k)\Bigr{)}\quad$
(3.5)

for some complex numbers $p_{n}^{\pm}(k)$ and $q_{n}^{\pm}(k)$. Hence
$({\mathsf{E}}(\phi_{n}^{\pm}+X(0)))(t,z)$ extend to smooth functions of $t$
on $\mathbb{R}$ by the right-hand side of (3.5). We use the same notation for
such extended functions.

We put $p_{n}^{\pm}(k)=q_{n}^{\pm}(k)=0$ for every $d+n\leq k\leq 2d-1$ if
$0\leq n\leq d-1$ and define the column vectors $\Psi_{n}^{\pm}$ of length
$8d$ by

$\Psi_{n}^{\pm}=\begin{bmatrix}\Psi_{n,1}^{\pm}\\\ \Psi_{n,2}^{\pm}\\\
J_{2d}\cdot\overline{\Psi_{n,2}^{\pm}}\\\
J_{2d}\cdot\overline{\Psi_{n,1}^{\pm}}\end{bmatrix},\quad\left\\{\begin{aligned}
\Psi_{n,1}^{\pm}&={}^{\rm
t}\\!\begin{bmatrix}p_{n}^{\pm}(-\tfrac{d+r-1}{2})&p_{n}^{\pm}(-\tfrac{d+r-3}{2})&\cdots&p_{n}^{\pm}(\tfrac{3d-r-1}{2})\end{bmatrix},\\\
\Psi_{n,2}^{\pm}&={}^{\rm
t}\\!\begin{bmatrix}q_{n}^{\pm}(\tfrac{3d-r-1}{2})&q_{n}^{\pm}(\tfrac{3d-r-3}{2})&\cdots&q_{n}^{\pm}(-\tfrac{d+r-1}{2})\end{bmatrix}.\end{aligned}\right.$

Then we have

$\Psi_{n}^{\pm}=\mathfrak{E}_{0}(\Phi_{n}^{\pm}+\chi)$ (3.6)

by comparing the right-hand sides of (3.5) with

$\displaystyle{\mathsf{E}}(\phi_{n}^{\pm}+X(0))$
$\displaystyle=\sum_{j=0}^{d}\sum_{k=0}^{\infty}C_{L-rj}u_{n}^{\pm}(k)X(k+(d-r+1)/2-j)$
$\displaystyle\quad+\sum_{j=0}^{d}\sum_{l=-\infty}^{n-1}C_{L-rj}v_{n}^{\pm}(l)Y(l-(d+r-1)/2+j)$
$\displaystyle\quad+\sum_{j=0}^{d}C_{L-rj}X((d-r+1)/2-j),$

which is obtained by a direct calculation of the left-hand side with the
expansion
$\phi_{n}^{\pm}=\sum_{k=0}^{\infty}u_{n}^{\pm}(k)X(k)+\sum_{l=-\infty}^{n-1}v_{n}^{\pm}(l)Y(l-r+1)$.

######  Lemma 3.6.

Let $0\leq n\leq d$. Then $p_{n}^{\pm}(k)\pm\overline{q_{n}^{\pm}(k)}=0$ if
$(d-r+1)/2+1\leq k\leq(d-r+1)/2+n-1$ or $-(d+r-1)/2\leq k\leq-(d+r-1)/2+n-1$,
and $p_{n}^{\pm}((d-r+1)/2)\pm\overline{q_{n}^{\pm}((d-r+1)/2)}=2C_{L}$.
Therefore,

$\displaystyle(\mathsf{I}$
$\displaystyle\pm\mathsf{J}){\mathsf{E}}(\phi_{n}^{\pm}+X(0))$ (3.7)
$\displaystyle\quad=\sum_{k=-(d+r-1)/2+n}^{(d-r+1)/2}\Bigl{(}(p_{n}^{\pm}(k)\pm\overline{q_{n}^{\pm}(k)})X(k)\pm(\overline{p_{n}^{\pm}(k)}\pm
q_{n}^{\pm}(k))Y(k)\Bigr{)}.$

###### Proof.

Linear equations (3.4) are equivalent to
$(\mathfrak{E}_{0}\pm\mathfrak{E}_{n}^{\sharp})(\Phi_{n}^{\pm}+\chi)=2\mathfrak{E}_{0}\cdot\chi$.
They are written in matrix forms as

$\begin{aligned}
\,&\begin{bmatrix}\mathfrak{e}_{0}&&\pm\overline{\mathfrak{e}_{2,n}}&\\\
&\mathfrak{e}_{0}&&\pm\overline{\mathfrak{e}_{1,n}}\\\ \pm
J_{2d}\cdot\mathfrak{e}_{1,n}\cdot J_{2d}&&{}^{\rm
t}\overline{\mathfrak{e}_{0}}&\\\ &\pm J_{2d}\cdot\mathfrak{e}_{2,n}\cdot
J_{2d}&&{}^{\rm
t}\overline{\mathfrak{e}_{0}}\end{bmatrix}(\Phi_{n}^{\pm}+\chi)=2\begin{bmatrix}\mathfrak{e}_{0}&&&\\\
&\mathfrak{e}_{0}&&\\\ &&{}^{\rm t}\overline{\mathfrak{e}_{0}}&\\\ &&&{}^{\rm
t}\overline{\mathfrak{e}_{0}}\end{bmatrix}\\!\chi\end{aligned}$

(3.8)

by definition of matrices. On the other hand, we have

$\begin{bmatrix}\mathfrak{e}_{0}&&\pm{}^{\rm t}\overline{\mathfrak{e}_{0}}&\\\
&\mathfrak{e}_{0}&&\pm{}^{\rm t}\overline{\mathfrak{e}_{0}}\\\
\pm\mathfrak{e}_{0}&&{}^{\rm t}\overline{\mathfrak{e}_{0}}&\\\
&\pm\mathfrak{e}_{0}&&{}^{\rm
t}\overline{\mathfrak{e}_{0}}\end{bmatrix}(\Phi_{n}^{\pm}+\chi)=\begin{bmatrix}R_{n}^{\pm}\\\
\pm J_{2d}\cdot\overline{R_{n}^{\pm}}\\\ \pm R_{n}^{\pm}\\\
J_{2d}\cdot\overline{R_{n}^{\pm}}\end{bmatrix},\quad
R_{n}^{\pm}=\begin{bmatrix}p_{n}^{\pm}(-\tfrac{d+r-1}{2})\pm\overline{q_{n}^{\pm}(-\tfrac{d+r-1}{2})}\\\
p_{n}^{\pm}(-\tfrac{d+r-3}{2})\pm\overline{q_{n}^{\pm}(-\tfrac{d+r-3}{2})}\\\
\vdots\\\
p_{n}^{\pm}(\tfrac{3d-r-1}{2})\pm\overline{q_{n}^{\pm}(\tfrac{3d-r-1}{2})}\\\
\end{bmatrix},$

since
$\displaystyle{(I_{8d}\pm\mathfrak{J})\cdot\mathfrak{E}_{0}=\begin{bmatrix}\mathfrak{e}_{0}&&\pm{}^{\rm
t}\overline{\mathfrak{e}_{0}}&\\\ &\mathfrak{e}_{0}&&\pm{}^{\rm
t}\overline{\mathfrak{e}_{0}}\\\ \pm\mathfrak{e}_{0}&&{}^{\rm
t}\overline{\mathfrak{e}_{0}}&\\\ &\pm\mathfrak{e}_{0}&&{}^{\rm
t}\overline{\mathfrak{e}_{0}}\end{bmatrix}}$ by definition of
$\mathfrak{E}_{0}$ and

$(I_{8d}\pm\mathfrak{J})\mathfrak{E}_{0}(\Phi_{n}^{\pm}+\chi)=(I_{8d}\pm\mathfrak{J})\Psi_{n}^{\pm}=\begin{bmatrix}R_{n}^{\pm}\\\
\pm J_{2d}\cdot\overline{R_{n}^{\pm}}\\\ \pm R_{n}^{\pm}\\\
J_{2d}\cdot\overline{R_{n}^{\pm}}\end{bmatrix}$

by (3.6). In addition,

$\begin{bmatrix}\mathfrak{e}_{0}&&\pm{}^{\rm t}\overline{\mathfrak{e}_{0}}&\\\
&\mathfrak{e}_{0}&&\pm{}^{\rm t}\overline{\mathfrak{e}_{0}}\\\
\pm\mathfrak{e}_{0}&&{}^{\rm t}\overline{\mathfrak{e}_{0}}&\\\
&\pm\mathfrak{e}_{0}&&{}^{\rm
t}\overline{\mathfrak{e}_{0}}\end{bmatrix}(\Phi_{n}^{\pm}+\chi)=\begin{bmatrix}\mathfrak{e}_{0}&&\pm
J_{2d}\cdot{}^{\rm t}\overline{\mathfrak{e}_{0,n}}\cdot J_{2d}&\\\
&\mathfrak{e}_{0,n}&&\pm{}^{\rm t}\overline{\mathfrak{e}_{0}}\\\
\pm\mathfrak{e}_{0}&&J_{2d}\cdot{}^{\rm t}\overline{\mathfrak{e}_{0,n}}\cdot
J_{2d}&\\\ &\pm\mathfrak{e}_{0,n}&&{}^{\rm
t}\overline{\mathfrak{e}_{0}}\end{bmatrix}(\Phi_{n}^{\pm}+\chi),$

(3.9)

where $\mathfrak{e}_{0,n}$ on the right-hand side is obtained by replacing $n$
columns from the left of $\mathfrak{e}_{0}$ with zero columns, since
$v_{n}^{\pm}(n)=v_{n}^{\pm}(n+1)\cdots=v_{n}^{\pm}(d-1)=0$ for $1\leq n\leq
d-1$ by definition of $\Phi_{n}^{\pm}$. Therefore,

$M_{\pm}(\Phi_{n}^{\pm}+\chi_{8d})=\begin{bmatrix}R_{n}^{\pm}\\\ \pm
J_{2d}\cdot\overline{R_{n}^{\pm}}\\\ \pm R_{n}^{\pm}\\\
J_{2d}\cdot\overline{R_{n}^{\pm}}\end{bmatrix}~{}\text{with}~{}\,\,M_{\pm}^{\prime}=\begin{bmatrix}\mathfrak{e}_{0}&&\mp
J_{2d}\cdot{}^{\rm t}\overline{\mathfrak{e}_{0,n}}\cdot J_{2d}&\\\
&\mathfrak{e}_{0,n}&&\mp{}^{\rm t}\overline{\mathfrak{e}_{0}}\\\
\mp\mathfrak{e}_{0}&&J_{2d}\cdot{}^{\rm t}\overline{\mathfrak{e}_{0,n}}\cdot
J_{2d}&\\\ &\mp\mathfrak{e}_{0,n}&&{}^{\rm
t}\overline{\mathfrak{e}_{0}}\end{bmatrix}.$

(3.10)

Here we find that $d$ rows of both
$\mathfrak{E}_{0}\pm\mathfrak{E}_{n}^{\sharp}$ and $M_{\pm}$ with indices
$(d+1,d+2,\cdots,2d)$ and $n$ rows of both
$\mathfrak{E}_{0}\pm\mathfrak{E}_{n}^{\sharp}$ and $M_{\pm}$ with indices
$(4d-n+1,4d-n+2,\cdots,4d)$ have the same entries. Therefore, by comparing $d$
rows of (3.8) and (3.10) with indices $(d+1,d+2,\cdots,2d)$, we obtain
$p_{n}^{\pm}((d-r+1)/2)\pm\overline{q_{n}^{\pm}((d-r+1)/2)}=2C_{L}$ and
$p_{n}^{\pm}(k)\pm\overline{q_{n}^{\pm}(k)}=0$ for $(d-r+1)/2+1\leq
k\leq(d-r+1)/2+d-1=(3d-r-1)/2$. Similarly, by comparing $n$ rows of (3.8) and
(3.10) with indices $(4d-n+1,4d-n+2,\cdots,4d)$, we obtain
$p_{n}^{\pm}(k)\pm\overline{q_{n}^{\pm}(k)}=0$ for $-(d+r-1)/2\leq
k\leq-(d+r-1)/2+n-1$. ∎

######  Lemma 3.7.

We have

$\displaystyle u_{n}^{+}(k)=u_{n}^{-}(k)\quad(0\leq k\leq 2d-1),\quad
v_{n}^{+}(k)=-v_{n}^{-}(k)\quad(-d\leq k\leq d-1)$ (3.11)

for every $0\leq n\leq d$.

###### Proof.

We have $\Phi_{n}^{\pm}+\chi=2M_{\pm}^{-1}\cdot\mathfrak{E}_{0}\cdot\chi$ with

$M_{\pm}=\mathfrak{E}_{0}\pm\mathfrak{E}_{n}^{\sharp}=\begin{bmatrix}\mathfrak{e}_{0}&&\pm\overline{\mathfrak{e}_{2,n}}&\\\
&\mathfrak{e}_{0}&&\pm\overline{\mathfrak{e}_{1,n}}\\\ \pm
J_{2d}\cdot\mathfrak{e}_{1,n}\cdot J_{2d}&&{}^{\rm
t}\overline{\mathfrak{e}_{0}}&\\\ &\pm J_{2d}\cdot\mathfrak{e}_{2,n}\cdot
J_{2d}&&{}^{\rm t}\overline{\mathfrak{e}_{0}}\end{bmatrix}$

by (3.8). Put

$A=\begin{bmatrix}\mathfrak{e}_{0}&\\\ &\mathfrak{e}_{0}\end{bmatrix},\quad
B=\begin{bmatrix}\pm\overline{\mathfrak{e}_{2,n}}&\\\
&\pm\overline{\mathfrak{e}_{2,n}}\end{bmatrix},$ $C=\begin{bmatrix}\pm
J_{2d}\cdot\mathfrak{e}_{1,n}\cdot J_{2d}&\\\ &\pm
J_{2d}\cdot\mathfrak{e}_{2,n}\cdot J_{2d}\end{bmatrix},\quad
D=\begin{bmatrix}{}^{\rm t}\overline{\mathfrak{e}_{0}}&\\\ &{}^{\rm
t}\overline{\mathfrak{e}_{0}}\end{bmatrix}.$

Then $\det A=C_{-L}^{2d}\not=0$ (resp. $\det D=\overline{C_{-L}}^{2d}\not=0$)
by assumption. Therefore the identity for the Schur complement $\det
M^{\pm}=\det A\det(D-CA^{-1}B)$ (resp. $\det M^{\pm}=\det D\det(A-BD^{-1}C)$)
shows that $\det(D-CA^{-1}B)\not=0$ (resp. $\det(A-BD^{-1}C)\not=0$), since
$M^{\pm}$ are invertible. Also, $A-BD^{-1}C$ and $D-CA^{-1}B$ are block-
diagonal matrices, and thus, their inverse matrices are also block-diagonal.
Therefore, applying the inversion formula for block matrices ([10, Lemma 3.2])
to $M^{\pm}$, we obtain

$\displaystyle\Phi_{n}^{\pm}+\chi$
$\displaystyle=\begin{bmatrix}(A-BD^{-1}C)^{-1}&\mp
A^{-1}B(D-CA^{-1}B)^{-1}\\\ \mp
D^{-1}C(A-BD^{-1}C)^{-1}&(D-CA^{-1}B)^{-1}\end{bmatrix}\cdot\mathfrak{E}_{0}\cdot\chi$
$\displaystyle=\begin{bmatrix}A_{11}&O&\pm A_{13}&O\\\ O&A_{22}&O&\pm
A_{24}\\\ \pm A_{31}&O&A_{33}&O\\\ O&\pm A_{42}&O&A_{44}\\\
\end{bmatrix}\cdot\mathfrak{E}_{0}\cdot\chi,$

where $A_{ij}$ are some square matrices of size $2d$. Recalling definition
(3.3) of $\Phi_{n}^{\pm}$, this establishes Lemma 3.7, since all $4d$ entries
of the column vector $\mathfrak{E}_{0}\cdot\chi$ with indices
$(2d+1,2d+2,\cdots,6d)$ are zero. ∎

######  Lemma 3.8.

We have

$p_{n}^{+}(k)+\overline{q_{n}^{+}(k)}=p_{n}^{-}(k)-\overline{q_{n}^{-}(k)}\quad\left(-\frac{d+r-1}{2}\leq
k\leq\frac{d-r+1}{2}+n-1\right).$

for every $0\leq n\leq d$, where
$p_{n}^{\pm}(k)\pm\overline{q_{n}^{\pm}(k)}=0$ if $-(d+r-1)/2\leq
k\leq-(d+r-1)/2+n-1$ or $(d-r+1)/2+1\leq k\leq(d-r+1)/2+n-1$ by Lemma 3.6.

###### Proof.

According to Lemma 3.7, we write $u_{n}(k)=u_{n}^{\pm}(k)$ and $v_{n}(k)=\pm
v_{n}^{\pm}(k)$. By (3.2),

$\displaystyle{\mathsf{E}}\,\phi_{n}^{\pm}$
$\displaystyle={\mathsf{E}}X(0)\mp{\mathsf{E}}^{\sharp}Y(0)\mp{\mathsf{E}}^{\sharp}{\mathsf{J}}{\mathsf{P}}_{n}\phi_{n}^{\pm}$
$\displaystyle=\sum_{j=0}^{d}C_{L-rj}X((d-r+1)/2-j)\mp\sum_{j=0}^{d}\overline{C_{L-rj}}\,Y((d-r+1)/2-j)$
$\displaystyle\quad-\sum_{j=0}^{d}\sum_{l=0}^{n-1}\overline{C_{L-rj}}\overline{v_{n}(l)}X(l-(d+r-1)/2+j)$
$\displaystyle\quad\mp\sum_{j=0}^{d}\sum_{k=0}^{n-1}\overline{C_{L-rj}}\overline{u_{n}(k)}Y(k+(d-r+1)/2-j),$

where we understand that the double sums on the right-hand side are zero when
$n$ is zero. Therefore,

$\displaystyle\mathsf{J}\mathsf{E}\,\phi_{n}^{\pm}$
$\displaystyle=\mp\sum_{j=0}^{d}C_{L-rj}X((d-r+1)/2-j)+\sum_{j=0}^{d}\overline{C_{L-rj}}Y((d-r+1)/2-j)$
$\displaystyle\quad\mp\sum_{j=0}^{d}\sum_{k=0}^{n-1}C_{L-rj}u_{n}(k)X(k+(d-r+1)/2-j)$
$\displaystyle\quad-\sum_{j=0}^{d}\sum_{l=0}^{n-1}C_{L-rj}v_{n}(l)Y(l-(d+r-1)/2+j).$

Combining the above,

$\displaystyle({\mathsf{I}}\pm{\mathsf{J}}){\mathsf{E}}\,\phi_{n}^{\pm}$
$\displaystyle+({\mathsf{E}}X(0)\pm{\mathsf{E}}^{\sharp}Y(0))$ (3.12)
$\displaystyle=\sum_{j=0}^{d}C_{L-rj}X((d-r+1)/2-j)\pm\sum_{j=0}^{d}\overline{C_{L-rj}}\,Y((d-r+1)/2-j)$
$\displaystyle\quad-\sum_{j=0}^{d}\sum_{k=0}^{n-1}(C_{L-rj}u_{n}(k)+\overline{C_{-(L-rj)}}\overline{v_{n}(k)})X(k+(d-r+1)/2-j)$
$\displaystyle\quad\mp\sum_{j=0}^{d}\sum_{k=0}^{n-1}(\overline{C_{L-rj}}\overline{u_{n}(k)}+C_{-(L-rj)}v_{n}(k))Y(k+(d-r+1)/2-j),$

where we understand that the double sums on the right-hand side are zero when
$n$ is zero. Comparing the right-hand sides of the above formulas of
$({\mathsf{I}}\pm{\mathsf{J}}){\mathsf{E}}(\phi_{n}^{\pm}+X(0))$ with (3.7),
we obtain Lemma 3.8. ∎

We define the column vectors $A_{n}^{\ast}$ and $B_{n}^{\ast}$ of length $8d$
by

$\displaystyle A_{n}^{\ast}=A_{n}^{\ast}(\mathcal{C})$
$\displaystyle:=(I+\mathfrak{J})\Psi_{n}^{+}=(I+\mathfrak{J})\mathfrak{E}_{0}(\Phi_{n}^{+}+\chi),$
(3.13) $\displaystyle B_{n}^{\ast}=B_{n}^{\ast}(\mathcal{C})$
$\displaystyle:=(I-\mathfrak{J})\Psi_{n}^{-}=(I-\mathfrak{J})\mathfrak{E}_{0}(\Phi_{n}^{-}+\chi),$

where $I=I_{8d}$ is the identify matrix of size $8d$. We define the row
vectors $F^{\pm}(t,z)$ of length $2d$ by

$\displaystyle F^{+}(t,z):=$
$\displaystyle\begin{bmatrix}X(-\tfrac{d+r-1}{2})~{}X(-\tfrac{d+r-1}{2}+1)~{}\cdots~{}X(\tfrac{d-r+1}{2})\quad
0~{}\cdots~{}0\end{bmatrix},$ $\displaystyle F^{-}(t,z):=$
$\displaystyle\begin{bmatrix}0~{}\cdots~{}0\quad
Y(\tfrac{d-r+1}{2})~{}Y(\tfrac{d-r+1}{2}-1)~{}\cdots~{}Y(-\tfrac{d+r-1}{2})\end{bmatrix},$

and the row vector $F(t,z)$ of length $4d$ by

$F(t,z):=\begin{bmatrix}F^{+}(t,z)&F^{-}(t,z)\end{bmatrix}.$

Then, we obtain

$\displaystyle
A_{n}^{\ast}(t,z)=\frac{1}{4}\begin{bmatrix}F(t,z)&F(t,z)\end{bmatrix}\cdot
A_{n}^{\ast},\quad
B_{n}^{\ast}(t,z)=\frac{i}{4}\begin{bmatrix}F(t,z)&-F(t,z)\end{bmatrix}\cdot
B_{n}^{\ast}$ (3.14)

by (2.7), (3.6), and (3.7).

######  Proposition 3.9.

We have

$-\frac{d}{dt}A_{n}^{\ast}(t,z)=zB_{n}^{\ast}(t,z),\quad-\frac{d}{dt}B_{n}^{\ast}(t,z)=-zA_{n}^{\ast}(t,z)$

for every $0\leq n\leq d$.

###### Proof.

According to Lemma 3.8, we write

$r_{n}(k)=p_{n}^{+}(k)+\overline{q_{n}^{+}(k)}=p_{n}^{-}(k)-\overline{q_{n}^{-}(k)}.$

Then, by (3.7) and definition of $X(k)$ and $Y(l)$,

$\displaystyle((\mathsf{I}+\mathsf{J})$
$\displaystyle{\mathsf{E}}(\phi_{n}^{+}+X(0)))(t,z)$
$\displaystyle=\sum_{k=-(d+r-1)/2+n}^{(d-r+1)/2}\Bigl{(}r_{n}(k)e^{i(r(k+1)-1-t)z}+\overline{r_{n}(k)}e^{-i(r(k+1)-1-t)z}\Bigr{)},$
$\displaystyle((\mathsf{I}-\mathsf{J})$
$\displaystyle{\mathsf{E}}(\phi_{n}^{-}+X(0)))(t,z)$
$\displaystyle=\sum_{k=-(d+r-1)/2+n}^{(d-r+1)/2}\Bigl{(}r_{n}(k)e^{i(r(k+1)-1-t)z}-\overline{r_{n}(k)}e^{-i(r(k+1)-1-t)z}\Bigr{)}.$

Therefore, the differentiability of $A_{n}^{\ast}(t,z)$ and
$B_{n}^{\ast}(t,z)$ with respect to $t$ is trivial, and

$\displaystyle-\frac{d}{dt}$
$\displaystyle(({\mathsf{I}}+{\mathsf{J}}){\mathsf{E}}(\phi_{n}^{+}+X(0)))(t,z)$
$\displaystyle=iz\sum_{k=-(d+r-1)/2+n}^{(d-r+1)/2}\Bigl{(}r_{n}(k)e^{i(r(k+1)-1-t)z}-\overline{r_{n}(k)}e^{-i(r(k+1)-1-t)z}\Bigr{)},$
$\displaystyle-\frac{d}{dt}$ $\displaystyle
i(({\mathsf{I}}-{\mathsf{J}}){\mathsf{E}}(\phi_{n}^{-}+X(0)))(t,z)$
$\displaystyle=-z\sum_{k=-(d+r-1)/2+n}^{(d-r+1)/2}\Bigl{(}r_{n}(k)e^{i(r(k+1)-1-t)z}+\overline{r_{n}(k)}e^{-i(r(k+1)-1-t)z}\Bigr{)}.$

Hence we obtain Proposition 3.9 by definition (2.7). ∎

As mentioned in Section 2, the next task is to show the connection formula
(2.8) for $A_{n-1}^{\ast}(t,z)$ and $A_{n}^{\ast}(t,z)$.

######  Proposition 3.10.

The connection formula (2.8) holds for some real matrix $P_{n}^{\ast}$
depending only on $\mathcal{C}$ for all $1\leq n\leq d$. In addition $\det
P_{n}^{\ast}\not=0$ for all $1\leq n\leq d$, which implies that $H_{n}$ of
(2.11) is well-defined and $\det H_{n}=1$.

###### Proof.

As in the proof of Lemma 3.8 we write $u_{n}(k)=u_{n}^{\pm}(k)$ and
$v_{n}(k)=\pm v_{n}^{\pm}(k)$. Taking the limit $t\to rn/2$ in (2.1), we have
$X(k):=e^{i(r(k-n/2)+r-1)z}$ and $Y(l)=e^{-i(r(l-n/2)+r-1)z}$. Therefore,
$X(k)=Y(l)$ as a function of $z$ if and only if $n=k+l+r-1$.

First, we prove (2.8) for $n\geq 1$. Evaluating (3.12) at $t=rn/2$, we get

$\displaystyle({\mathsf{I}}\pm{\mathsf{J}}){\mathsf{E}}$
$\displaystyle(\phi_{n}^{\pm}+X(0))(rn/2,z)$ (3.15)
$\displaystyle=-(u_{n}(0)-1)\sum_{j=0}^{d}C_{L-rj}X(0+(d-r+1)/2-j)$
$\displaystyle\quad-\sum_{k=1}^{n-1}(u_{n}(k)\pm
v_{n}(n-k))\sum_{j=0}^{d}C_{L-rj}X(k+(d-r+1)/2-j)$
$\displaystyle\quad-\overline{v_{n}(0)}\sum_{j=0}^{d}\overline{C_{L-rj}}\,X(0-(d+r-1)/2+j)\pm\left[\cdots\right]$

and

$\displaystyle({\mathsf{I}}\pm{\mathsf{J}}){\mathsf{E}}$
$\displaystyle(\phi_{n+1}^{\pm}+X(0))(rn/2,z)$ $\displaystyle=-(u_{n+1}(0)\pm
v_{n+1}(n)-1)\sum_{j=0}^{d}C_{L-rj}X(0+(d-r+1)/2-j)$
$\displaystyle\quad-\sum_{k=1}^{n-1}(u_{n+1}^{+}(k)\pm
v_{n+1}^{+}(n-k))\sum_{j=0}^{d}C_{L-rj}X(k+(d-r+1)/2-j)$
$\displaystyle\quad\mp(\overline{u_{n+1}(n)\pm
v_{n+1}(0)})\sum_{j=0}^{d}\overline{C_{L-rj}}X(0-(d+r-1)/2+j)\pm\left[\cdots\right],$

where the bracket parts on the right-hand sides are the conjugates of the
first half of the right-hand sides. Therefore, if we prove that the linear
relations

$\begin{aligned} \alpha_{n+1}^{\ast}&\begin{bmatrix}u_{n}(0)-1\\\
u_{n}(1)+v_{n}(n-1)\\\ u_{n}(2)+v_{n}(n-2)\\\ \vdots\\\ u_{n}(n-1)+v_{n}(1)\\\
v_{n}(0)\end{bmatrix}+i\beta_{n+1}^{\ast}\begin{bmatrix}u_{n}(0)-1\\\
u_{n}(1)-v_{n}(n-1)\\\ u_{n}(2)-v_{n}(n-2)\\\ \vdots\\\ u_{n}(n-1)-v_{n}(1)\\\
-v_{n}(0)\end{bmatrix}=\begin{bmatrix}u_{n+1}(0)+v_{n+1}(n)-1\\\
u_{n+1}(1)+v_{n+1}(n-1)\\\ u_{n+1}(2)+v_{n+1}(n-2)\\\ \vdots\\\
u_{n+1}(n)+v_{n+1}(0)\end{bmatrix},\\\
\gamma_{n+1}^{\ast}&\begin{bmatrix}u_{n}(0)-1\\\ u_{n}(1)+v_{n}(n-1)\\\
u_{n}(2)+v_{n}(n-2)\\\ \vdots\\\ u_{n}(n-1)+v_{n}(1)\\\
v_{n}(0)\end{bmatrix}+i\delta_{n+1}^{\ast}\begin{bmatrix}u_{n}(0)-1\\\
u_{n}(1)-v_{n}(n-1)\\\ u_{n}(2)-v_{n}(n-2)\\\ \vdots\\\ u_{n}(n-1)-v_{n}(1)\\\
-v_{n}(0)\\\ \end{bmatrix}=i\begin{bmatrix}u_{n+1}(0)-v_{n+1}(n)-1\\\
u_{n+1}(1)-v_{n+1}(n-1)\\\ u_{n+1}(2)-v_{n+1}(n-2)\\\ \vdots\\\
u_{n+1}(n)-v_{n+1}(0)\end{bmatrix}\end{aligned}$

(3.16)

hold, then they imply (2.8) for
$\displaystyle{P_{n+1}^{\ast}=\begin{bmatrix}\alpha_{n+1}^{\ast}&\beta_{n+1}^{\ast}\\\
\gamma_{n+1}^{\ast}&\delta_{n+1}^{\ast}\end{bmatrix}}$ and

$\displaystyle\alpha_{n+1}^{\ast}+i\beta_{n+1}^{\ast}$
$\displaystyle=\frac{u_{n+1}(0)+v_{n+1}(n)-1}{u_{n}(0)-1}=\frac{\overline{u_{n+1}(n)+v_{n+1}(0)}}{\overline{v_{n}(0)}},$
(3.17) $\displaystyle\delta_{n+1}^{\ast}-i\gamma_{n+1}^{\ast}$
$\displaystyle=\frac{u_{n+1}(0)-v_{n+1}(n)-1}{u_{n}(0)-1}=-\frac{\overline{u_{n+1}(n)-v_{n+1}(0)}}{\overline{v_{n}(0)}}.$

Hence the proof is completed if (3.16) is shown.

Subtracting $\mathsf{E}X(0)\pm\mathsf{E}^{\sharp}Y(0)$ from both sides of
(3.2) for $n$ and $n+1$, and then taking the limit $t\to rn/2$ on the left-
hand sides, we obtain

$\displaystyle(({\mathsf{E}}$
$\displaystyle\pm{\mathsf{E}}^{\sharp}{\mathsf{J}}{\mathsf{P}_{n}})(\phi_{n}-X(0)))(rn/2,z)$
$\displaystyle=(u_{n}(0)-1)\sum_{j=0}^{d}C_{L-rj}\cdot X(0+(d-r+1)/2-j)$
$\displaystyle\quad+\sum_{k=1}^{n-1}(u_{n}(k)\pm
v_{n}(n-k))\sum_{j=0}^{d}C_{L-rj}\cdot X(k+(d-r+1)/r-j)$
$\displaystyle\quad+\overline{v_{n}(0)}\sum_{j=0}^{d}\overline{C_{L-rj}}X(0-(d+r-1)/2+j)$
$\displaystyle\quad\pm\overline{(u_{n}(0)-1)}\sum_{j=0}^{d}\overline{C_{L-rj}}\cdot
X(n-(d+r-1)/2+j)$
$\displaystyle\quad\pm\sum_{k=1}^{n-1}(\overline{u_{n}(k)}\pm\overline{v_{n}(n-k)})\sum_{j=0}^{d}\overline{C_{L-rj}}\cdot
X(n-k-(d+r-1)/2+j)$ $\displaystyle\quad\pm v_{n}(0)\sum_{j=0}^{d}C_{L-rj}\cdot
X(n+(d-r+1)/2-j)$ $\displaystyle\quad+u_{n}(n)\sum_{j=0}^{d}C_{L-rj}\cdot
X(n+(d-r+1)/2-j)$ $\displaystyle\quad+\sum_{k=n+1}^{\infty}(u_{n}(k)\pm
v_{n}(n-k))\sum_{j=0}^{d}C_{L-rj}\cdot X(k+(d-r+1)/2-j)$

and

$\displaystyle(({\mathsf{E}}$
$\displaystyle\pm{\mathsf{E}}^{\sharp}{\mathsf{J}}{\mathsf{P}_{n+1}})(\phi_{n+1}-X(0)))(rn/2,z)$
$\displaystyle=(u_{n+1}(0)\pm v_{n+1}(n)-1)\sum_{j=0}^{d}C_{L-rj}\cdot
X(0+(d-r+1)/2-j)$ $\displaystyle\quad\pm(\overline{u_{n+1}(0)\pm
v_{n+1}(n)-1})\sum_{j=0}^{d}\overline{C_{L-rj}}\cdot X(n-(d+r-1)/2+j)$
$\displaystyle\quad+\sum_{k=1}^{\infty}(u_{n+1}(k)\pm
v_{n+1}(n-k))\sum_{j=0}^{d}C_{L-rj}\cdot X(k+(d-r+1)/2-j)$
$\displaystyle\quad\pm\sum_{k=1}^{n}(\overline{u_{n+1}(k)\pm
v_{n+1}(n-k)})\sum_{j=0}^{d}\overline{C_{L-rj}}\cdot X(n-k-(d+r-1)/2+j).$

In both cases of $n$ and $n+1$, the right-hand sides are

$\displaystyle\mp 2{\mathsf{E}}^{\sharp}Y(0)$ $\displaystyle=\mp
2\,\sum_{j=0}^{d}\overline{C_{L-rj}}X(n-(d+r-1)/2+j)$ $\displaystyle=\mp
2\,\overline{C_{L}}X(n-(d+r-1)/2)\mp
2\,\sum_{j=1}^{d}\overline{C_{L-rj}}X(n-(d+r-1)/2+j).$

Therefore, by comparing $(n+1)$ coefficient of $X(k-(d+r-1)/2)$ for $0\leq
k\leq n$ in equations
$(({\mathsf{E}}\pm{\mathsf{E}}^{\sharp}{\mathsf{J}}{\mathsf{P}_{n}})(\phi_{n}-X(0)))(rn/2,z)=\mp
2{\mathsf{E}}^{\sharp}Y(0)$ and
$(({\mathsf{E}}\pm{\mathsf{E}}^{\sharp}{\mathsf{J}}{\mathsf{P}_{n+1}})(\phi_{n+1}-X(0)))(rn/2,z)=\mp
2{\mathsf{E}}^{\sharp}Y(0)$, we obtain linear equations

$L_{n+1}^{\pm}({\mathcal{C}})\begin{bmatrix}u_{n}(0)-1\\\ u_{n}(1)\pm
v_{n}(n-1)\\\ u_{n}(2)\pm v_{n}(n-2)\\\ \vdots\\\ u_{n}(n-1)\pm v_{n}(1)\\\
\pm v_{n}(0)\\\ \pm\overline{v_{n}(0)}\\\ \overline{u_{n}(n-1)\pm v_{n}(1)}\\\
\vdots\\\ \overline{u_{n}(1)\pm v_{n}(n-1)}\\\ \overline{u_{n}(0)-1}\\\
\end{bmatrix}=\mp\begin{bmatrix}0\\\ \vdots\\\ 0\\\ 2\overline{C_{L}}\\\
2C_{L}\\\ 0\\\ \vdots\\\ 0\end{bmatrix}-\begin{bmatrix}0\\\ \vdots\\\ 0\\\
C_{-L}u_{n}(n)\\\ \overline{C_{-L}}\,\overline{u_{n}(n)}\\\ 0\\\ \vdots\\\
0\end{bmatrix}$ (3.18)

and

$L_{n+1}^{\pm}({\mathcal{C}})\begin{bmatrix}u_{n+1}(0)\pm v_{n+1}(n)-1\\\
u_{n+1}(1)\pm v_{n+1}(n-1)\\\ u_{n+1}(2)\pm v_{n+1}(n-2)\\\ \vdots\\\
u_{n+1}(n)\pm v_{n+1}(0)\\\ \overline{u_{n+1}(n)\pm v_{n+1}(0)}\\\
\overline{u_{n+1}(n-1)\pm v_{n+1}(1)}\\\ \vdots\\\ \overline{u_{n+1}(0)\pm
v_{n+1}(n)-1}\end{bmatrix}=\mp\begin{bmatrix}0\\\ \vdots\\\ 0\\\
2\overline{C_{L}}\\\ 2C_{L}\\\ 0\\\ \vdots\\\ 0\end{bmatrix},$ (3.19)

where $L_{n}^{\pm}(\mathcal{C})$ are defined in (1.1) and (1.8), and non-zero
components of the column vectors on the right-hand side are the $(n+1)$th and
$(n+2)$th entries. Suppose that

$-2\overline{C_{L}}-C_{L}u_{n}(n)=K_{n}\cdot
i(2\overline{C_{L}}-C_{L}u_{n}(n))$ (3.20)

holds for some $1\leq n\leq d$ and $K_{n}\in\mathbb{R}\setminus\\{0\\}$. Then
$A_{n}^{\ast}(t,z)=K_{n}B_{n}^{\ast}(t,z)$ by (3.15) and (3.18). But, in this
case, it must be $K_{n}=\pm i$ by Proposition 3.9. This is a contradiction.
Therefore, (3.20) does not hold for any $K_{n}\in\mathbb{R}\setminus\\{0\\}$.
Hence, there exist real numbers $\alpha_{n+1}^{\ast}$, $\beta_{n+1}^{\ast}$,
$\gamma_{n+1}^{\ast}$, $\delta_{n+1}^{\ast}$ such that

$\displaystyle\alpha_{n+1}^{\ast}(-2\overline{C_{L}}-C_{L}u_{n}(n))+i\beta_{n+1}^{\ast}(2\overline{C_{L}}-C_{L}u_{n}(n))$
$\displaystyle=-2\overline{C_{L}},$ (3.21)
$\displaystyle\gamma_{n+1}^{\ast}(-2\overline{C_{L}}-C_{L}u_{n}(n))+i\delta_{n+1}^{\ast}(2\overline{C_{L}}-C_{L}u_{n}(n))$
$\displaystyle=2i\overline{C_{L}}$

holds. This implies relation (3.16).

We show that $\displaystyle{\det
P_{n+1}^{\ast}=\begin{bmatrix}\alpha_{n+1}^{\ast}&\beta_{n+1}^{\ast}\\\
\gamma_{n+1}^{\ast}&\delta_{n+1}^{\ast}\end{bmatrix}\not=0.}$ If $\det
P_{n+1}^{\ast}=0$, its row vectors are proportional:
$[\,\alpha_{n+1}^{\ast}\,\,\,\,\beta_{n+1}^{\ast}\,]=K_{n}^{\prime}[\,\gamma_{n+1}^{\ast}\,\,\,\,\delta_{n+1}^{\ast}\,]$,
say. Then (3.21) implies $K_{n}^{\prime}=-i$, but it is impossible for real
vectors $[\,\alpha_{n+1}^{\ast}\,\,\,\,\beta_{n+1}^{\ast}\,]$ and
$[\,\gamma_{n+1}^{\ast}\,\,\,\,\delta_{n+1}^{\ast}\,]$.

Finally we prove (2.8) for $n=0$. We have
$({\mathsf{I}}\pm{\mathsf{J}}){\mathsf{E}}(\phi_{0}^{\pm}+X(0))={\mathsf{E}}X(0)\pm{\mathsf{E}^{\sharp}}Y(0)$
by $\mathsf{E}\phi_{0}^{\pm}={\mathsf{E}}X(0)\mp{\mathsf{E}^{\sharp}}Y(0)$,
since $\mathsf{P}_{0}=0$. Evaluating
$({\mathsf{I}}\pm{\mathsf{J}}){\mathsf{E}}(\phi_{0}^{\pm}+X(0))$ and
$({\mathsf{I}}\pm{\mathsf{J}}){\mathsf{E}}(\phi_{1}^{\pm}+X(0))$ at $t=0$ by
using (3.12) for $n=0$ and $n=1$, we get

$\displaystyle({\mathsf{I}}\pm{\mathsf{J}})$
$\displaystyle{\mathsf{E}}(\phi_{0}^{\pm}+X(0))(0,z)$
$\displaystyle=\sum_{j=0}^{d}C_{L-rj}X(0+(d-r+1)/2-j)\pm\sum_{j=0}^{d}\overline{C_{L-rj}}\,X(0-(d+r-1)/2+j),$
$\displaystyle({\mathsf{I}}\pm{\mathsf{J}})$
$\displaystyle{\mathsf{E}}(\phi_{1}^{\pm}+X(0))(0,z)$
$\displaystyle=(1-u_{1}(0)\mp v_{1}(0))\sum_{j=0}^{d}C_{L-rj}X(0+(d-r+1)/2-j)$
$\displaystyle\quad\pm(\overline{1-u_{1}(0)\mp
v_{1}(0)})\sum_{j=0}^{d}\overline{C_{L-rj}}\,X(0-(d+r-1)/2+j).$

Therefore, (2.8) holds for
$\displaystyle{P_{1}^{\ast}=\begin{bmatrix}\alpha_{1}^{\ast}&\beta_{1}^{\ast}\\\
\gamma_{1}^{\ast}&\delta_{1}^{\ast}\end{bmatrix}}$ with
$\alpha_{1}^{\ast}+i\beta_{1}^{\ast}=1-u_{1}(0)-v_{1}(0)$ and
$\gamma_{1}^{\ast}+i\delta_{1}^{\ast}=1-u_{1}(0)+v_{1}(0)$. ∎

######  Lemma 3.11.

Let $\mathfrak{c}$ be the column vector of length $2n$ defined by

$\mathfrak{c}={}^{\rm
t}\begin{bmatrix}0&\cdots&0&\overline{C_{L}}&C_{L}&0&\cdots&0\end{bmatrix},$

where $\overline{C_{L}}$ and $C_{L}$ are $n$th and $(n+1)$th entries,
respectively. Then,

$\displaystyle\frac{1}{2}\Bigl{(}\det
L_{n}^{+}({\mathcal{C}};-2\mathfrak{c};1)\det
L_{n}^{-}({\mathcal{C}};2\mathfrak{c};2n)+\det
L_{n}^{+}({\mathcal{C}};-2\mathfrak{c};2n)\det
L_{n}^{-}({\mathcal{C}};2\mathfrak{c};1)\Bigr{)}$ (3.22)
$\displaystyle\quad=\begin{cases}4|C_{L}|^{4}\cdot
D_{n-2}({\mathcal{C}})D_{n}({\mathcal{C}})&n\geq 2,\\\ -4|C_{L}|^{2}\cdot
D_{1}(\mathcal{C})&n=1,\end{cases}$

where $L_{n}^{\pm}({\mathcal{C}};\mp\mathfrak{c};k)$ is a matrix obtained by
replacing the $k$th column of $L_{n}^{\pm}({\mathcal{C}})$ with
$\mp\mathfrak{c}$. Recall that $D_{0}(\mathcal{C})=1$ by convention.

###### Proof.

In the case of $n=1$, we have

$\displaystyle\frac{1}{2}$
$\displaystyle\left(\det\begin{bmatrix}-2\overline{C_{L}}&\overline{C_{L}}\\\
-2C_{L}&\overline{C_{-L}}\end{bmatrix}\det\begin{bmatrix}C_{-L}&2\overline{C_{L}}\\\
-C_{L}&2C_{L}\end{bmatrix}+\det\begin{bmatrix}C_{-L}&-2\overline{C_{L}}\\\
C_{L}&-2C_{L}\end{bmatrix}\det\begin{bmatrix}2\overline{C_{L}}&-\overline{C_{L}}\\\
2C_{L}&\overline{C_{-L}}\end{bmatrix}\right)$
$\displaystyle=-4|C_{L}|^{2}(|C_{-L}|^{2}-|C_{L}|^{2})=-4|C_{L}|^{2}D_{1}(\mathcal{C}).$

Let $n\geq 2$. Multiplying each of the $(n+1)$th to $2n$th columns of $\det
L_{n}^{-}({\mathcal{C}};2\mathfrak{c};1)$ and $\det
L_{n}^{-}({\mathcal{C}};2\mathfrak{c};2n)$ by $-1$, and then, multiplying each
of the $(n+1)$th to $2n$th rows of them by $-1$,

$\displaystyle\frac{1}{2}\Bigl{(}\det
L_{n}^{+}({\mathcal{C}};-2\mathfrak{c};1)\det
L_{n}^{-}({\mathcal{C}};2\mathfrak{c};2n)+\det
L_{n}^{+}({\mathcal{C}};-2\mathfrak{c};2n)\det
L_{n}^{-}({\mathcal{C}};2\mathfrak{c};1)\Bigr{)}$
$\displaystyle=2\,\Bigl{(}\det L_{n}^{+}({\mathcal{C}};\mathfrak{c};1)\det
L_{n}^{+}({\mathcal{C}};\mathfrak{c}^{\prime};2n)+\det
L_{n}^{+}({\mathcal{C}};\mathfrak{c};2n)\det
L_{n}^{+}({\mathcal{C}};-\mathfrak{c}^{\prime};1)\Bigr{)},$

where $\displaystyle{\mathfrak{c}^{\prime}={}^{\rm
t}\begin{bmatrix}0&\cdots&0&\overline{C_{L}}&-C_{L}&0&\cdots&0\end{bmatrix},}$
$\overline{C_{L}}$ and $-C_{L}$ are $n$th and $(n+1)$th entries, respectively.
The right-hand side is equal to

$4|C_{L}|^{2}\Bigl{(}\det L_{n}^{+}({\mathcal{C}};e_{n+1};1)\det
L_{n}^{+}({\mathcal{C}};e_{n};2n)-\det L_{n}^{+}({\mathcal{C}};e_{n};1)\det
L_{n}^{+}({\mathcal{C}};e_{n+1};2n)\Bigr{)}$

by expanding $\det L_{n}^{+}({\mathcal{C}};\mathfrak{c};1)$ and $\det
L_{n}^{+}({\mathcal{C}};-\mathfrak{c}^{\prime};1)$ along the first columns,
and by expanding $\det L_{n}^{+}({\mathcal{C}};\mathfrak{c}^{\prime};2n)$ and
$\det L_{n}^{+}({\mathcal{C}};\mathfrak{c};2n)$ along the $2n$th columns,
Therefore, what should be shown is the equality

$\displaystyle\det$ $\displaystyle L_{n}^{+}({\mathcal{C}};e_{n+1};1)\det
L_{n}^{+}({\mathcal{C}};e_{n};2n)-\det L_{n}^{+}({\mathcal{C}};e_{n};1)\det
L_{n}^{+}({\mathcal{C}};e_{n+1};2n)$ (3.23)
$\displaystyle=|C_{L}|^{2}\,D_{n-2}({\mathcal{C}})D_{n}({\mathcal{C}}).$

For a matrix $M$, we denote $M^{(a,b;c,d)}$ the matrix removing $a$-th and
$b$-th rows and $c$-th and $d$-th columns from $M$, and set

$\Delta_{n-1}({\mathcal{C}}):=\det\left(L_{n}^{+}(\mathcal{C})^{(1,n;1,n+1)}\right).$

Expanding $\det L_{n}^{+}({\mathcal{C}};e_{n+1};1)$ and $\det
L_{n}^{+}({\mathcal{C}};e_{n};1)$ along the $1$st row,

$\det
L_{n}^{+}({\mathcal{C}};e_{n+1};1)=\overline{C_{L}}D_{n-1}(\mathcal{C}),\quad\det
L_{n}^{+}({\mathcal{C}};e_{n};1)=\overline{C_{L}}\Delta_{n-1}(\mathcal{C}),$

because the only non-zero component in the $1$st row is $\overline{C_{L}}$ in
the $(n+1)$-th column. Expanding $\det L_{n}^{+}({\mathcal{C}};e_{n};2n)$ and
$\det L_{n}^{+}({\mathcal{C}};e_{n+1};2n)$ along the $2n$-th row,

$\det L_{n}^{+}({\mathcal{C}};e_{n};2n)=C_{L}D_{n-1}(\mathcal{C}),\quad\det
L_{n}^{+}({\mathcal{C}};e_{n+1};2n)=C_{L}\overline{\Delta_{n-1}(\mathcal{C})},$

because the only non-zero component in the $2n$-th row is $C_{L}$ in the
$n$-th column.

From the above, the right-hand side of (3.23) is equal to

$|C_{L}|^{2}\left(D_{n-1}(\mathcal{C})^{2}-|\Delta_{n-1}(\mathcal{C})|^{2}\right),$

but it is equal to $|C_{L}|^{2}D_{n-2}(\mathcal{C})D_{n}(\mathcal{C})$ by [4,
p.41, (12)]. Hence we complete the proof. ∎

######  Proposition 3.12.

The matrices $H_{n}=H_{n}(\mathcal{C})$ defined by (1.10) are represented by
the Schur–Cohn determinants as in (1.14) for all $1\leq n\leq d$.

###### Proof.

Fix $n$ and write
$P_{n}=(P_{1}^{\ast})^{-1}\cdots(P_{n}^{\ast})^{-1}=[\begin{smallmatrix}a&b\\\
c&d\end{smallmatrix}]$ $(a,b,c,d\in\mathbb{R})$. Then, by (2.11),

$H_{n}=-\frac{1}{\det P_{n}}\begin{bmatrix}c^{2}+d^{2}&-(ac+bd)\\\
-(ac+bd)&a^{2}+b^{2}\end{bmatrix}=-\frac{1}{\det P_{n}}\,H_{n}^{\prime},$

say. Neither eigenvalue of $H_{n}^{\prime}$ is zero by Proposition 3.10.
Furthermore, eigenvalues of $H_{n}^{\prime}$ are calculated as

$\frac{1}{2}\left(a^{2}+b^{2}+c^{2}+d^{2}\pm\sqrt{(a^{2}+b^{2}+c^{2}+d^{2})^{2}-4(ad-
bc)^{2}}\right),$

and

$(a^{2}+b^{2}+c^{2}+d^{2})^{2}-4(ad-
bc)^{2}=((a-d)^{2}+(b+c)^{2})((a+d)^{2}+(b-c)^{2}\geq 0.$

Therefore, both eigenvalues of $H_{n}^{\prime}$ are positive, so
$H_{n}^{\prime}$ is positive definite.

On the other hand, by $P_{n}=(P_{1}^{\ast})^{-1}\cdots(P_{n}^{\ast})^{-1}$,
$P_{k}^{\ast}=\begin{bmatrix}\alpha_{k}^{\ast}&\beta_{k}^{\ast}\\\
\gamma_{k}^{\ast}&\delta_{k}^{\ast}\end{bmatrix}$, and (3.17), we obtain

$\frac{1}{\det
P_{n}}=\prod_{k=1}^{n}(|u_{k}(0)-1|^{2}-|v_{k}(k-1)|^{2})\left/\prod_{k=2}^{n}|u_{k-1}(0)-1|^{2}\right.,$

because the identity

$|u|^{2}\det\begin{bmatrix}\Re((z+w)/u)&\Im((z+w)/u)\\\
-\Im((z-w)/u)&\Re((z-w)/u)\end{bmatrix}=\det\begin{bmatrix}\Re(z+w)&\Im(z+w)\\\
-\Im(z-w)&\Re(z-w)\end{bmatrix}=|z|^{2}-|w|^{2}$

holds for general complex numbers $z,w,u$. Hence $H_{n}$ is written as

$H_{n}=-\left(\prod_{k=1}^{n}(|u_{k}(0)-1|^{2}-|v_{k}(k-1)|^{2})\right)\tilde{H}_{n}$

for some positive definite matrix $\tilde{H}_{n}$. Thus the proof is completed
if

$-\prod_{k=1}^{n}(|u_{k}(0)-1|^{2}-|v_{k}(k-1)|^{2})=\frac{2^{2n}\,|C_{L}|^{2(2n-1)}}{D_{n-1}(\mathcal{C})D_{n}(\mathcal{C})}\quad(1\leq
n\leq d)$ (3.24)

is proved. Applying Cramer’s rule to (3.19),

$\displaystyle u_{k}(0)\pm v_{k}(k-1)-1$ $\displaystyle=\frac{\det
L_{k}^{\pm}({\mathcal{C}};\mp 2\mathfrak{c};1)}{D_{k}(\mathcal{C})},$
$\displaystyle\overline{u_{k}(0)\pm v_{k}(k-1)-1}$ $\displaystyle=\frac{\det
L_{k}^{\pm}({\mathcal{C}};\mp 2\mathfrak{c};2k)}{D_{k}(\mathcal{C})}.$

Therefore,

$\displaystyle|u_{k}(0)-1|^{2}-|v_{k}(k-1)|^{2}=\frac{1}{2}\Bigl{(}(u_{k}(0)+v_{k}(k-1)-1)\overline{(u_{k}(0)-v_{k}(k-1)-1)})$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad+\overline{(u_{k}(0)+v_{k}(k-1)-1)}(u_{k}(0)-v_{k}(k-1)-1)\Bigr{)}$
$\displaystyle=\frac{1}{2D_{k}(\mathcal{C})^{2}}\Bigl{(}\det
L_{k}^{+}({\mathcal{C}};-2\mathfrak{c};1)\det
L_{k}^{-}({\mathcal{C}};2\mathfrak{c};2n)+\det
L_{k}^{+}({\mathcal{C}};-2\mathfrak{c};2n)\det
L_{k}^{-}({\mathcal{C}};2\mathfrak{c};1)\Bigr{)}.$

Using (3.22) on the right-hand side,

$\displaystyle|u_{k}(0)-1|^{2}-|v_{k}(k-1)|^{2}$
$\displaystyle=\begin{cases}\displaystyle{4\,|C_{L}|^{4}\,\frac{D_{k-2}(\mathcal{C})}{D_{k}(\mathcal{C})}},&k\geq
2,\\\\[10.0pt]
\displaystyle{-4\,|C_{L}|^{2}\,\frac{1}{D_{1}(\mathcal{C})}},&k=1.\end{cases}$

This implies (3.24). ∎

######  Proposition 3.13.

The pair of functions $(A(t,z),B(t,z))$ of (2.10) satisfies the boundary
condition (1.12).

###### Proof.

The first half of (1.12) follows from definition (2.10) by (2.8) and (2.9) for
$n=1$, since $A_{0}^{\ast}(0,z)=A(z)$ and $B_{0}^{\ast}(0,z)=B(z)$. We prove
the second half of (1.12). By definition (2.7) and Lemma 3.6,

$A_{d}^{\ast}(t,z)=C_{L}e^{i(L-t)z}+\overline{C_{L}}e^{-i(L-t)z},\quad-
iB_{d}^{\ast}(t,z)=C_{L}e^{i(L-t)z}-\overline{C_{L}}e^{-i(L-t)z}.$

Therefore,

$\lim_{t\to L}\begin{bmatrix}A(t,z)\\\
B(t,z)\end{bmatrix}=(P_{1}^{\ast})^{-1}\cdots(P_{d}^{\ast})^{-1}\begin{bmatrix}\Re(C_{L})\\\
\Im(C_{L})\end{bmatrix}$

for fixed $z\in\mathbb{C}$ by definition (2.10). In particular, the limit is
independent of $z$, but $A(t,0)$ and $B(t,0)$ are constant function of $t$ by
Proposition 3.9 and definitions (2.9) and (2.10), and hence $A(t,0)=A(0)$ and
$B(t,0)=B(0)$. ∎

Proof of Theorem 1.1. As a summary of the above results, we obtain the
following theorem which implies Theorem 1.1.

######  Theorem 3.1.

Let $\mathcal{C}\in\mathbb{C}^{d+1}$ be as in (1.5) and define
$E=E_{\mathcal{C}}$ by (1.6). Suppose that $D_{d}(\mathcal{C})\not=0$. Then,

  1. (1)

$A(t,z)$ and $B(t,z)$ are well-defined and continuous on $[0,L)$ with respect
to $t$,

  2. (2)

$A(t,z)$ and $B(t,z)$ are continuously differentiable on $(r(n-1)/2,rn/2)$
with respect to $t$ for every $1\leq n\leq d$,

  3. (3)

the left-sided limit $\lim_{t\nearrow rn/2}(A(t,z),B(t,z))$ defines entire
functions of $z$ for every $1\leq n\leq d$,

  4. (4)

$A(t,z)$ and $B(t,z)$ have the forms (1.13).

  5. (5)

matrices $H_{n}$ of (1.10) are well-defined for all $1\leq n\leq d$ and
satisfy (1.14),

  6. (6)

the pair of functions $(A(t,z),B(t,z))$ defined in (2.10) satisfies the system
(1.3) associated with $H(t)$ defined in (1.10),

  7. (7)

the pair of functions $(A(t,z),B(t,z))$ satisfies the boundary condition
(1.12).

###### Proof.

(1), (2), and (3) are consequences of (2.9), (2.10), (3.14), and Proposition
3.10. For (4), we put
$P_{n}=\begin{pmatrix}\alpha_{n}^{\ast\ast}&\beta_{n}^{\ast\ast}\\\
\gamma_{n}^{\ast\ast}&\delta_{n}^{\ast\ast}\end{pmatrix}$ and

$a_{n}(k)=(\alpha_{n}^{\ast\ast}+i\beta_{n}^{\ast\ast})r_{n}(k),\quad
b_{n}(k)=(\gamma_{n}^{\ast\ast}+i\delta_{n}^{\ast\ast})r_{n}(k),$

where
$r_{n}(k)=p_{n}^{+}(k)+\overline{q_{n}^{+}(k)}=p_{n}^{-}(k)-\overline{q_{n}^{-}(k)}$
as in the proof of Proposition 3.9. Then, we have (1.13) by (2.9), (2.10),
(3.14), and the changing of index $k=(L-rj-r+1)/r$. (5) follows from
Propositions 3.10, 3.10, and 3.12. (6) is a consequence of Proposition 3.9,
(2.9), and (2.10). In fact,

$-\frac{d}{dt}\begin{bmatrix}A_{n}(t,z)\\\
B_{n}(t,z)\end{bmatrix}=z\begin{bmatrix}0&-1\\\
1&0\end{bmatrix}H_{n}\begin{bmatrix}A_{n}(t,z)\\\ B_{n}(t,z)\end{bmatrix}$

for every $r(n-1)/2\leq t<rn/2$ and $1\leq n\leq d$ by Proposition 3.9. This
implies (1.3) for $H(t)$ defined by (1.11). (7) is a consequence of
Proposition 3.13. ∎

##  4\. Proofs of Theorems 1.2 and 1.3

To prove Theorems 1.2 and 1.3, we prepare a proposition. The proof about it
below is the almost same as the argument in the literature on canonical
systems; for example, the proof of equation (2.4) and Lemma 2.1, and Step 1 of
the proof of Theorem 5.1 in Dym [3]. However, we purposely give the detailed
proof to confirm that the positive semidefiniteness of the Hamiltonian, which
is usually assumed in the theory of canonical systems, is not necessary for
the proof as well as [10, Proposition 5.1].

######  Proposition 4.1.

Let $H(t)$ and $(A,B)$ be as in Theorem 1.2, and write
$\displaystyle{H_{n}=\begin{bmatrix}\alpha_{n}&\beta_{n}\\\
\beta_{n}&\gamma_{n}\end{bmatrix}}$ for $1\leq n\leq d$. Then the solution
$(A(t,z),B(t,z))$ mentioned in Theorem 1.2 exists and it is represented as

$\begin{aligned} \begin{bmatrix}A(t,z)\\\
B(t,z)\end{bmatrix}&=\begin{bmatrix}\cos((rn/2-t)z)-\beta_{n}\sin((rn/2-t)z)&-\gamma_{n}\sin((rn/2-t)z)\\\
\alpha_{n}\sin((rn/2-t)z)&\cos((rn/2-t)z)+\beta_{n}\sin((rn/2-t)z)\end{bmatrix}\\\
&\quad\times\begin{bmatrix}\cos((r/2)z)-\beta_{n+1}\sin((r/2)z)&-\gamma_{n+1}\sin((r/2)z)\\\
\alpha_{n+1}\sin((r/2)z)&\cos((r/2)z)+\beta_{n+1}\sin((r/2)z)\end{bmatrix}\\\
&\quad\cdots\times\begin{bmatrix}\cos((r/2)z)-\beta_{d}\sin((r/2)z)&-\gamma_{d}\sin((r/2)z)\\\
\alpha_{d}\sin((r/2)z)&\cos((r/2)z)+\beta_{d}\sin((r/2)z)\end{bmatrix}\begin{bmatrix}A\\\
B\end{bmatrix}\end{aligned}$

(4.1)

for $r(n-1)/2\leq t<rn/2$ and $1\leq n\leq d$, where the product of quadratic
matrices on the right-hand side consists of only the first matrix if $n=d$. In
particular, for any $0\leq t<L$, there exists a quadratic matrix-valued
function $M(t,z)$ consisting of entire functions of $z$ such that

$\begin{bmatrix}A(t,z)\\\ B(t,z)\end{bmatrix}=M(t,z)\begin{bmatrix}A\\\
B\end{bmatrix},$ (4.2)

holds and $\det M(t,z)=1$.

###### Proof.

By definition, $H(t)$ is integrable on $[t_{0},t_{1}]$ for any $0\leq
t_{0}<t_{1}<L$. Hence,

$\begin{aligned} \begin{bmatrix}A(t_{0},z)\\\
B(t_{0},z)\end{bmatrix}&=\left[I+z\int_{t_{0}}^{t_{1}}J(s_{1})\,ds_{1}+z^{2}\int_{t_{0}}^{t_{1}}\int_{s_{1}}^{t_{1}}J(s_{1})J(s_{2})\,ds_{2}ds_{1}\right.\\\
&\qquad\qquad\left.+z^{3}\int_{t_{0}}^{t_{1}}\int_{s_{1}}^{t_{1}}\int_{s_{2}}^{t_{1}}J(s_{1})J(s_{2})J(s_{3})\,ds_{3}ds_{2}ds_{1}+\cdots\right]\begin{bmatrix}A(t_{1},z)\\\
B(t_{1},z)\end{bmatrix},\end{aligned}$

(4.3)

where $I=I_{2}$ and $\displaystyle{J(t)=\begin{bmatrix}0&-1\\\
1&0\end{bmatrix}H(t)}$. Taking
$C=\max\\{|\alpha_{n}|,\,|\beta_{n}|,\,|\gamma_{n}|~{}:~{}1\leq n\leq d\\}$
and by using the formula

$\int_{t_{0}}^{t_{1}}\int_{s_{1}}^{t_{1}}\int_{s_{2}}^{t_{1}}\cdots\int_{s_{k-1}}^{t_{1}}1\,ds_{k}\cdots
ds_{2}ds_{1}=\frac{1}{k!}(t_{1}-t_{0})^{k},$

we obtain

$\left|\left[\int_{t_{0}}^{t_{1}}\int_{s_{1}}^{t_{1}}\int_{s_{2}}^{t_{1}}\cdots\int_{s_{k-1}}^{t_{1}}J(s_{1})\cdots
J(s_{k})\,ds_{k}\cdots ds_{2}ds_{1}\right]_{ij}\right|\leq
2^{k-1}C^{k}\frac{1}{k!}(t_{1}-t_{0})^{k}$

for every $1\leq i,j\leq 2$, where $[M]_{ij}$ means the $(i,j)$-entry of a
matrix $M$. This estimate implies that the right-hand side of (4.3) converges
absolutely and uniformly if $z$ lies in a bounded region. Suppose that
$H(t)=\Bigl{[}\begin{smallmatrix}\alpha&\beta\\\
\beta&\gamma\end{smallmatrix}\Bigr{]}$ (a constant matrix) with
$\alpha\gamma-\beta^{2}=1$ for $t_{0}\leq s\leq t_{1}$. Then the series of
integrals in (4.3) is calculated as

$\begin{bmatrix}\cos((t_{1}-t_{0})z)-\beta\sin((t_{1}-t_{0})z)&-\gamma\sin((t_{1}-t_{0})z)\\\
\alpha\sin((t_{1}-t_{0})z)&\cos((t_{1}-t_{0})z)+\beta\sin((t_{1}-t_{0})z)\end{bmatrix}.$

Hence we have

$\begin{aligned} \,&\begin{bmatrix}A(t_{0},z)\\\
B(t_{0},z)\end{bmatrix}=\begin{bmatrix}\cos((t_{1}-t_{0})z)-\beta\sin((t_{1}-t_{0})z)&-\gamma\sin((t_{1}-t_{0})z)\\\
\alpha\sin((t_{1}-t_{0})z)&\cos((t_{1}-t_{0})z)+\beta\sin((t_{1}-t_{0})z)\end{bmatrix}\begin{bmatrix}A(t_{1},z)\\\
B(t_{1},z)\end{bmatrix}.\end{aligned}$

Therefore, we obtain (4.1) for $t\geq r(d-1)/2$ by taking the limit $t_{1}\to
L$. Also, the determinant of the matrix on the right-hand side is

$\displaystyle\det$
$\displaystyle\begin{bmatrix}\cos((t_{1}-t_{0})z)-\beta\sin((t_{1}-t_{0})z)&-\gamma\sin((t_{1}-t)z)\\\
\alpha\sin((t_{1}-t_{0})z)&\cos((t_{1}-t_{0})z)+\beta\sin((t_{1}-t_{0})z)\end{bmatrix}$
(4.4)
$\displaystyle=\cos^{2}((t_{1}-t_{0})z)+(\alpha\gamma-\beta^{2})\sin^{2}((t_{1}-t_{0})z)=1.$

Following the above case, applying (4.3) to $r(d-2)/2\leq t_{0}<r(d-1)/2$ and
$t_{1}=r(d-1)/2$ and using the result for $t\geq r(d-1)/2$, we obtain (4.1)
for $t\geq r(d-2)/2$. By repeating this process, (4.1) is obtained for all
$0\leq t<L$. ∎

###  4.1. Proof of Theorem 1.2.

To prove (1.13), we put

$\displaystyle{\begin{bmatrix}M_{11}^{n}(z)&M_{12}^{n}(z)\\\
M_{21}^{n}(z)&M_{22}^{n}(z)\end{bmatrix}=\prod_{k=n+1}^{d}\begin{bmatrix}\cos((r/2)z)-\beta_{k}\sin((r/2)z)&-\gamma_{k}\sin((r/2)z)\\\
\alpha_{k}\sin((r/2)z)&\cos((r/2)z)+\beta_{k}\sin((r/2)z)\end{bmatrix}}$

(4.5)

for $1\leq n\leq d$. Then (4.1) implies

$\begin{aligned}
A(t,z)&=\cos((rn/2-t)z)\Bigl{[}AM_{11}^{n}(z)+BM_{12}^{n}(z)\Bigr{]}\\\
&\quad-\sin((rn/2-t)z)\Bigl{[}A(\beta_{n}M_{11}^{n}(z)+\gamma_{n}M_{21}^{n}(z))+B(\beta_{n}M_{12}^{n}(z)+\gamma_{n}M_{22}^{n}(z))\Bigr{]},\\\
B(t,z)&=\cos((rn/2-t)z)\Bigl{[}AM_{21}^{n}(z)+BM_{22}^{n}(z)\Bigr{]}\\\
&\quad+\sin((rn/2-t)z)\Bigl{[}A(\alpha_{n}M_{11}^{n}(z)+\beta_{n}M_{21}^{n}(z))+B(\alpha_{n}M_{12}^{n}(z)+\beta_{n}M_{22}^{n}(z))\Bigr{]}\end{aligned}$

(4.6)

for $r(n-1)/2\leq t<rn/2$ and $1\leq n\leq d$. Putting $X=e^{i(r/2)z}$,
$X^{\ast}=e^{i(rn/2-t)z}$, $Y=X^{-1}$, and $Y^{\ast}=(X^{\ast})^{-1}$, we
obtain

$M_{rs}^{n}(z)=\sum_{\nu=1}^{d-n}\left[N_{rs}^{n}(\nu)X^{\nu}Y^{d-n-\nu}+\overline{N_{rs}^{n}(\nu)}X^{d-n-\nu}Y^{\nu}\right]$
(4.7)

for $r,s\in\\{1,2\\}$ by induction for $n\geq 1$, where $N_{rs}^{n}(\nu)$ are
complex numbers depending only on the set $\\{H_{n}\\}_{1\leq n\leq d}$, and

$\cos((rn/2-t)z)=\frac{1}{2}(X^{\ast}+Y^{\ast}),\quad\sin((rn/2-t)z)=-\frac{i}{2}(X^{\ast}-Y^{\ast}).$
(4.8)

Substituting (4.7) and (4.8) into (4.6) and then carrying out a simple
calculation, we obtain (1.13).

By (1.3), $A(t,0)$ and $B(t,0)$ are constant function of $t$. Hence
$E(t,0)=A(t,0)-iB(t,0)=A-iB$ by the boundary condition at $t=L$. Suppose that
$E(0,z_{0})=0$ for some real number $z_{0}$. Then $A(0,z_{0})=B(0,z_{0})=0$,
and thus it should be $(A,B)=(0,0)$ by (4.2). It is a contradiction. Hence
$E(0,z)$ has no real zeros. $\Box$

###  4.2. Proof of Theorem 1.3.

From (4.7) the leading term of

$\begin{bmatrix}M_{11}^{d-n}(z)&M_{12}^{d-n}(z)\\\
M_{21}^{d-n}(z)&M_{22}^{d-n}(z)\end{bmatrix}\begin{bmatrix}A\\\
B\end{bmatrix}$

with respect to $X$ and $Y$ is written as

$\begin{bmatrix}P_{n}X^{n}+\overline{P_{n}}Y^{n}\\\
Q_{n}X^{n}+\overline{Q_{n}}Y^{n}\end{bmatrix}$

for some complex numbers $P_{n}$ and $Q_{n}$. Because

$\begin{aligned} \begin{bmatrix}M_{11}^{d-n-1}(z)&M_{12}^{d-n-1}(z)\\\
M_{21}^{d-n-1}(z)&M_{22}^{d-n-1}(z)\end{bmatrix}&=\begin{bmatrix}\frac{X+Y}{2}+i\beta_{d-n}\frac{X-Y}{2}&i\gamma_{d-n}\frac{X-Y}{2}\\\
-i\alpha_{d-n}\frac{X-Y}{2}&\frac{X+Y}{2}-i\beta_{d-n}\frac{X-Y}{2}\end{bmatrix}\begin{bmatrix}M_{11}^{d-n}(z)&M_{12}^{d-n}(z)\\\
M_{21}^{d-n}(z)&M_{22}^{d-n}(z)\end{bmatrix},\end{aligned}$

we have

$\displaystyle\begin{bmatrix}P_{n}\\\ Q_{n}\end{bmatrix}$
$\displaystyle=\frac{1}{2}\begin{bmatrix}1+i\beta_{d-n+1}&i\gamma_{d-n+1}\\\
-i\alpha_{d-n+1}&1-i\beta_{d-n+1}\end{bmatrix}\begin{bmatrix}P_{n-1}\\\
Q_{n-1}\end{bmatrix}$ $\displaystyle=\frac{1}{2}\left(\begin{bmatrix}1&0\\\
0&1\end{bmatrix}-i\begin{bmatrix}0&-1\\\
1&0\end{bmatrix}\begin{bmatrix}\alpha_{d-n+1}&\beta_{d-n+1}\\\
\beta_{d-n+1}&\gamma_{d-n+1}\end{bmatrix}\right)\begin{bmatrix}P_{n-1}\\\
Q_{n-1}\end{bmatrix}.$

The leading term of $E_{d-n+1}(t,z)=A_{d-n+1}(t,z)-iB_{d-n+1}(t,z)$ with
$t=r(d-n)/2$ is

$(P_{n}X^{n}+\overline{P_{n}}Y^{n})-i(Q_{n}X^{n}+\overline{Q_{n}}Y^{n}).$

Therefore, the eq105ficient of $X^{n}$ (resp. $Y^{n}$) is zero if
$(P_{n},\,Q_{n})$ is proportional to $(1,\,-i)$ (resp. $(1,\,i)$), and both
are zero if $(P_{n},Q_{n})=(0,0)$. Applying this to $n=d$ gives the desired
conclusion.

The latter half of the theorem is a consequence of Schur-Cohn test and
Theorems 1.1 and 1.2, since $H$ of Theorem 1.2 must be equal to $H$ of Theorem
1.1 defined for $E_{f}(z)=e^{irdz/2}f(e^{-irz})$ by Proposition 4.1. $\Box$

##  5\. Inductive construction

To state the result, we introduce special matrices $\mathfrak{P}_{n}(H)$ and
$\mathfrak{Q}_{n}$ as follows. For $n=0$, we define

$\mathfrak{P}_{0}=\mathfrak{P}_{0}(H)=\left[\begin{array}[]{cc|cc}1&0&0&0\\\
0&1&0&0\\\ \hline\cr 0&0&1&0\\\
0&0&0&1\end{array}\right],\quad\mathfrak{Q}_{0}=\left[\begin{array}[]{cc|cc|cc|cc}1&0&1&0&0&0&0&0\\\
0&1&0&1&0&0&0&0\\\ \hline\cr 0&0&0&0&1&0&1&0\\\
0&0&0&0&0&1&0&1\end{array}\right].$

For $n\geq 1$ and $\displaystyle{H=\begin{bmatrix}\alpha&\beta\\\
\beta&\gamma\end{bmatrix}}$, we define

$\mathfrak{P}_{k}(H)=\left[\begin{array}[]{c|c|c|c}I_{k+2,k+1}^{+}&I_{k+2,k+1}^{-}&\mathbf{0}&\mathbf{0}\\\
\hline\cr\mathbf{0}&\mathbf{0}&I_{k+2,k+1}^{+}&I_{k+2,k+1}^{-}\\\
\hline\cr(1-i\beta)\cdot\mathbf{0}I_{k}&\mathbf{0}_{k,k+1}&(-i\gamma)\cdot\mathbf{0}I_{k}&\mathbf{0}_{k,k+1}\\\
\mathbf{0}_{k,k+1}&(-i\alpha)\cdot
I_{k}\mathbf{0}&\mathbf{0}_{k,k+1}&(1-i\beta)\cdot I_{k}\mathbf{0}\\\
\end{array}\right],$
$\mathfrak{Q}_{k}=\left[\begin{array}[]{c|c|c|c}I_{k+2}&I_{k+2}&\mathbf{0}&\mathbf{0}\\\
\hline\cr\mathbf{0}&\mathbf{0}&I_{k+2}&I_{k+2}\\\
\hline\cr\mathbf{0}_{2k,k+2}&\mathbf{0}_{2k,k+2}&\mathbf{0}_{2k,k+2}&\mathbf{0}_{2k,k+2}\\\
\end{array}\right],$

where

$I_{k+2,k+1}^{+}=\begin{bmatrix}I_{k+1}\\\
\mathbf{0}_{1,k+1}\end{bmatrix},\quad
I_{k+2,k+1}^{-}=\begin{bmatrix}\mathbf{0}_{1,k+1}\\\
I_{k+1}\end{bmatrix},\quad\mathbf{0}I_{k}=\begin{bmatrix}\mathbf{0}_{k,1}&I_{k}\end{bmatrix},\quad
I_{k}\mathbf{0}=\begin{bmatrix}I_{k}&\mathbf{0}_{k,1}\end{bmatrix}.$

The matrices $\mathfrak{P}_{k}(H)$ are invertible if $\det H=1$, because

$\det\mathfrak{P}_{k}(H)=\det\begin{bmatrix}1-i\beta&i\gamma\\\
i\alpha&1-i\beta\end{bmatrix}^{k}=(\alpha\gamma-(\beta+i)^{2})^{k}.$

Using these matrices, an inductive formula for coefficient of $A_{n}(t,z)$ and
$B_{n}(t,z)$ is described as follows.

######  Proposition 5.1.

Let $\mathcal{C}\in\mathbb{C}^{d+1}$ be as in (1.5) and define
$E=E_{\mathcal{C}}$ by (1.6). Suppose that $D_{d}(\mathcal{C})\not=0$. Put
$r_{n}(k)=p_{n}^{+}(k)+\overline{q_{n}^{+}(k)}=p_{n}^{-}(k)-\overline{q_{n}^{-}(k)}$,
$P_{n}=\begin{bmatrix}\alpha_{n}^{\ast\ast}&\beta_{n}^{\ast\ast}\\\
\gamma_{n}^{\ast\ast}&\delta_{n}^{\ast\ast}\end{bmatrix}$, and

$a_{n}(k)=(\alpha_{n}^{\ast\ast}+i\beta_{n}^{\ast\ast})r_{n}(k),\quad
b_{n}(k)=(\gamma_{n}^{\ast\ast}+i\delta_{n}^{\ast\ast})r_{n}(k),$

as in Proposition 3.9 and its proof. For $0\leq n\leq d$, define the column
vectors $A_{n}^{\ast\ast}$ and $B_{n}^{\ast\ast}$ of length $d-n+1$ by

$\displaystyle A_{n}^{\ast\ast}$ $\displaystyle=~{}{}^{\rm
t}\\!\begin{bmatrix}a_{n}(\tfrac{d-r+1}{2})&a_{n}(\tfrac{d-r+1}{2}-1)&\cdots&a_{n}(-\tfrac{d+r-1}{2}+n)\end{bmatrix},$
$\displaystyle B_{n}^{\ast\ast}$ $\displaystyle=~{}{}^{\rm
t}\\!\begin{bmatrix}b_{n}(\tfrac{d-r+1}{2})&b_{n}(\tfrac{d-r+1}{2}-1)&\cdots&b_{n}(-\tfrac{d+r-1}{2}+n)\end{bmatrix}$

for $1\leq n\leq d$ and

$A_{0}^{\ast\ast}=B_{0}^{\ast\ast}=\frac{1}{2}~{}{}^{\rm
t}\\!\begin{bmatrix}C_{(d-r+1)/2}&C_{(d-r+1)/2-1}&\cdots&C_{-(d+r-1)/2}\end{bmatrix}.$
(5.1)

Define the column vectors $\Omega_{n}$ of length $4(d-n+1)$ by

$\Omega_{n}=\begin{bmatrix}A_{n}^{\ast\ast}\\\
J_{d-n+1}\overline{A_{n}^{\ast\ast}}\\\ B_{n}^{\ast\ast}\\\
J_{d-n+1}\overline{B_{n}^{\ast\ast}}\end{bmatrix}\quad(1\leq n\leq
d),\quad\Omega_{0}=\begin{bmatrix}A_{0}^{\ast\ast}\\\
J_{d+1}\overline{A_{0}^{\ast\ast}}\\\ B_{0}^{\ast\ast}\\\
J_{d+1}\overline{B_{0}^{\ast\ast}}\end{bmatrix}.$ (5.2)

Then, vectors $\Omega_{n}$ satisfies the linear relation

$\mathfrak{P}_{d-(n+1)}(H_{n+1})\Omega_{n+1}=\mathfrak{Q}_{d-(n+1)}\Omega_{n}$
(5.3)

for every $0\leq n\leq d-1$, where $H_{n}$ is of (2.11).

###### Proof.

By Lemma 3.8 and (3.7), we have

$(\mathsf{I}\pm\mathsf{J}){\mathsf{E}}(\phi_{n}^{\pm}+X(0))=\sum_{k=-(d+r-1)/2+n}^{(d-r+1)/2}\Bigl{(}r_{n}(k)X(k)\pm\overline{r_{n}(k)}Y(k)\Bigr{)}.$

Therefore,

$\displaystyle
A_{n}(t,z)=\sum_{k=-(d+r-1)/2+n}^{(d-r+1)/2}\Bigl{(}a_{n}(k)X(k)+\overline{a_{n}(k)}Y(k)\Bigr{)},$
$\displaystyle
B_{n}(t,z)=\sum_{k=-(d+r-1)/2+n}^{(d-r+1)/2}\Bigl{(}b_{n}(k)X(k)+\overline{b_{n}(k)}Y(k)\Bigr{)}$

by (2.9) and (2.10). Evaluating these for $n$ and $n+1$ at $t=rn/2$ noting
$Y(k)=X(n-k-r+1)$,

$\displaystyle A_{n}(rn/2,z)$
$\displaystyle=\sum_{k=-(d+r-1)/2+n}^{(d-r+1)/2}\Bigl{[}a_{n}(k)+\overline{a_{n}(n-k-r+1)}\Bigr{]}X(k),$
(5.4) $\displaystyle B_{n}(rn/2,z)$
$\displaystyle=\sum_{k=-(d+r-1)/2+n}^{(d-r+1)/2}\Bigl{[}b_{n}(k)+\overline{b_{n}(n-k-r+1)}\Bigr{]}X(k)$

and

$\displaystyle A_{n+1}(rn/2,z)$
$\displaystyle=a_{n+1}((d-r+1)/2)\,X((d-r+1)/2)$ (5.5)
$\displaystyle\quad+\sum_{k=-(d+r-1)/2+n+1}^{(d-r+1)/2-1}\Bigl{[}a_{n+1}(k)+\overline{a_{n+1}(n-k-r+1)})\Bigr{]}\,X(k)$
$\displaystyle\quad+\overline{a_{n+1}((d-r+1)/2)}\,X(-(d+r-1)/2+n),$
$\displaystyle B_{n+1}(rn/2,z)$
$\displaystyle=b_{n+1}((d-r+1)/2)\,X((d-r+1)/2)$
$\displaystyle\quad+\sum_{k=-(d+r-1)/2+n+1}^{(d-r+1)/2-1}\Bigl{[}b_{n+1}(k)+\overline{b_{n+1}(n-k-r+1)})\Bigr{]}\,X(k)$
$\displaystyle\quad+\overline{b_{n+1}((d-r+1)/2)}\,X(-(d+r-1)/2+n).$

On the other hand, by Proposition 3.9,

$\displaystyle\frac{1}{z}\frac{d}{dt}A_{n}(t,z)$
$\displaystyle=\beta_{n}A_{n}(t,z)+\gamma_{n}B_{n}(t,z),$ (5.6)
$\displaystyle-\frac{1}{z}\frac{d}{dt}B_{n}(t,z)$
$\displaystyle=\alpha_{n}A_{n}(t,z)+\beta_{n}B_{n}(t,z),$

where $H_{n}=\begin{bmatrix}\alpha_{n}&\beta_{n}\\\
\beta_{n}&\gamma_{n}\end{bmatrix}$. Because
$\displaystyle{\frac{d}{dt}X(k)=-izX(k)}$ and
$\displaystyle{\frac{d}{dt}Y(k)=izY(k),}$ the left-hand sides are

$\displaystyle\frac{1}{z}\frac{d}{dt}A_{n}(t,z)$
$\displaystyle=\sum_{k=-(d+r-1)/2+n}^{(d-r+1)/2}\Bigl{(}-ia_{n}(k)X(k)+i\overline{a_{n}(k)}Y(k)\Bigr{)},$
$\displaystyle\frac{1}{z}\frac{d}{dt}B_{n}(t,z)$
$\displaystyle=\sum_{k=-(d+r-1)/2+n}^{(d-r+1)/2}\Bigl{(}-ib_{n}(k)X(k)+i\overline{b_{n}(k)}Y(k)\Bigr{)}.$

Therefore, by comparing both sides of (5.6), we obtain

$(1-i\beta_{n})a_{n}(k)-i\gamma_{n}b_{n}(k)=0,\quad(1+i\beta_{n})b_{n}(k)+i\alpha_{n}a_{n}(k)=0$
(5.7)

for $-(d+r-1)/2+n\leq k\leq(d-r+1)/2$.

For $4(d-n)$ complex numbers
$\\{a_{n+1}(k),\overline{a_{n+1}(k)},b_{n+1}(k),\overline{b_{n+1}(k)}\\}_{k=-(d+r-1)/2+n+1}^{(d-r+1)/2}$,
we obtain $2(d-n+1)$ linear equations by comparing coefficient of $X(k)$ for
$-(d+r-1)/2+n\leq k\leq(d-r+1)/2$ in equalities
$A_{n+1}(rn/2,z)=A_{n}(rn/2,z)$ and $B_{n+1}(rn/2,z)=B_{n}(rn/2,z)$ by using
(5.4) and (5.5). In addition, we obtain $2(d-n-1)$ linear equations from
differential equations (5.6) by using (5.7) for $-(d+r-1)/2+n+1\leq
k\leq(d-r+1)/2-1$. In total, we obtain $4(d-n)$ linear equations, which is
expressed in the form of (5.3). ∎

The pair of functions $(A(t,z),B(t,z))$ of (2.10) is written as

$\displaystyle A(t,z)$
$\displaystyle=\frac{1}{2}\alpha_{n}^{\ast\ast}\cdot\begin{bmatrix}F(t,z)&F(t,z)\end{bmatrix}\cdot(I+\mathfrak{J})\mathfrak{E}_{0}(\mathfrak{E}_{0}+\mathfrak{E}_{n}^{\sharp})^{-1}\mathfrak{E}_{0}\chi$
$\displaystyle\quad+\frac{i}{2}\beta_{n}^{\ast\ast}\cdot\begin{bmatrix}F(t,z)&-F(t,z)\end{bmatrix}\cdot(I-\mathfrak{J})\mathfrak{E}_{0}(\mathfrak{E}_{0}-\mathfrak{E}_{n}^{\sharp})^{-1}\mathfrak{E}_{0}\chi$
$\displaystyle B(t,z)$
$\displaystyle=\frac{1}{2}\gamma_{n}^{\ast\ast}\cdot\begin{bmatrix}F(t,z)&F(t,z)\end{bmatrix}\cdot(I+\mathfrak{J})\mathfrak{E}_{0}(\mathfrak{E}_{0}+\mathfrak{E}_{n}^{\sharp})^{-1}\mathfrak{E}_{0}\chi$
$\displaystyle\quad+\frac{i}{2}\delta_{n}^{\ast\ast}\cdot\begin{bmatrix}F(t,z)&-F(t,z)\end{bmatrix}\cdot(I-\mathfrak{J})\mathfrak{E}_{0}(\mathfrak{E}_{0}-\mathfrak{E}_{n}^{\sharp})^{-1}\mathfrak{E}_{0}\chi$

for $r(n-1)/2\leq t<rn/2$ by (2.9), (3.4), (3.13), and (3.14). These formulas
are explicit but it involves the complexity of calculating
$P_{n}=\begin{bmatrix}\alpha_{n}^{\ast\ast}&\beta_{n}^{\ast\ast}\\\
\gamma_{n}^{\ast\ast}&\delta_{n}^{\ast\ast}\end{bmatrix}$. In contrast, the
following method, based on Proposition 5.1, is often useful for computing the
triple $(H(t),A(t,z),B(t,z))$.

######  Theorem 5.1.

Let $\widetilde{\Omega}_{0}$ be a column vector of length $4(d+1)$. Define
column vectors $\widetilde{\Omega}_{n}$ $(1\leq n\leq d)$ of length $4(d-n+1)$
inductively as follows:

$\displaystyle\widetilde{a}_{n+1}:$
$\displaystyle=\widetilde{\Omega}_{n}(1)+\widetilde{\Omega}_{n}(d-n+2),$ (5.8)
$\displaystyle\widetilde{b}_{n+1}:$
$\displaystyle=\widetilde{\Omega}_{n}(2(d-n+1)+1)+\widetilde{\Omega}_{n}(3(d-n+1)+1),$
$\displaystyle\widetilde{\alpha}_{n+1}$
$\displaystyle:=\frac{|\widetilde{b}_{n+1}|^{2}}{\Re(\,\widetilde{a}_{n+1}\overline{(i\widetilde{b}_{n+1})}\,)},\quad\widetilde{\beta}_{n+1}:=\frac{\Im(\,\widetilde{a}_{n+1}\overline{(i\widetilde{b}_{n+1})}\,)}{\Re(\widetilde{a}_{n+1}\overline{(i\widetilde{b}_{n+1})}\,)},$
$\displaystyle\widetilde{\gamma}_{n+1}$
$\displaystyle:=\frac{|\widetilde{a}_{n+1}|^{2}}{\Re(\,\widetilde{a}_{n+1}\overline{(i\widetilde{b}_{n+1})}\,)},$
$\widetilde{H}_{n+1}:=\begin{bmatrix}\widetilde{\alpha}_{n+1}&\widetilde{\beta}_{n+1}\\\
\widetilde{\beta}_{n+1}&\widetilde{\gamma}_{n+1}\end{bmatrix},$ (5.9)
$\widetilde{\Omega}_{n+1}:=(\mathfrak{P}_{d-(n+1)}(\widetilde{H}_{n+1}))^{-1}\mathfrak{Q}_{d-(n+1)}\,\widetilde{\Omega}_{n},$
(5.10)

where $\mathfrak{P}_{0}(\tilde{H}_{0}):=\mathfrak{P}_{0}$ and $v(j)$ means the
$j$-th component of a column vector $v$.

Suppose that $\widetilde{\Omega}_{0}$ is the vector defined by (5.1) and (5.2)
for a vector $\mathcal{C}\in\mathbb{C}^{d+1}$ as in (1.5) such that
$D_{d}(\mathcal{C})\not=0$. Then $\widetilde{H}_{n}$ and
$\widetilde{\Omega}_{n}$ are well-defined as functions of $\mathcal{C}$ for
every $1\leq n\leq d$, and

$H_{n}=\widetilde{H}_{n},\quad\Omega_{n}=\widetilde{\Omega}_{n},$

where $H_{n}$ and $\Omega_{n}$ are defined in (2.11) and (5.2), respectively.

###### Proof.

Solving (5.7) for fixed $k$,

$\displaystyle\alpha_{n}$
$\displaystyle=\frac{|b_{n}(k)|^{2}}{\Re(\,a_{n}(k)\overline{(ib_{n}(k))}\,)},\quad\beta_{n}=\frac{\Im(\,a_{n}(k)\overline{(ib_{n}(k))}\,)}{\Re(\,a_{n}(k)\overline{(ib_{n}(k))}\,)},\quad\gamma_{n}=\frac{|a_{n}(k)|^{2}}{\Re(\,a_{n}(k)\overline{(ib_{n}(k))}\,)}.$

Therefore, $H_{n}$ and $\Omega_{n}$ of (2.11) and (5.2) satisfy (5.9) and
(5.10) by the definitions of $\mathfrak{P}_{k}(H_{k})$, $\mathfrak{Q}_{k}$,
and (5.3). Therefore, $H_{n}\not\equiv 0$ as a function of $\mathcal{C}$ for
every $1\leq n\leq d$ by Theorem 1.1, since all roots of the derivative of the
cyclotomic polynomial of degree $d+1$ lie inside the unit circle. Hence, the
invertibility of $\mathfrak{P}_{k}(H_{k})$ implies that
$\widetilde{\Omega}_{1},\widetilde{\Omega}_{2},\cdots,\widetilde{\Omega}_{d}$
and $\widetilde{H}_{1},\widetilde{H}_{2},\cdots,\widetilde{H}_{d}$ are
uniquely determined from the initial vector $\widetilde{\Omega}_{0}$.
Therefore, $\Omega_{n}=\widetilde{\Omega}_{n}$ and $H_{n}=\widetilde{H}_{n}$
for every $1\leq n\leq d$ if $\tilde{\Omega}_{0}=\Omega_{0}$. ∎

By definition of the matrices $\mathfrak{P}_{k}(H_{k})$, in (5.3),
$\Omega_{n+1}(1)$, $\Omega_{n+1}(2(d-n))$, $\Omega_{n+1}(2(d-n)+1)$, and
$\Omega_{n+1}(4(d-n))$ are determined from $\Omega_{n}$ independent of
$H_{n+1}$. Hence, we can define $\Omega_{n}$ by taking

$\Omega_{n}^{\prime}=\mathfrak{P}_{d-n}(H_{n})^{-1}\mathfrak{Q}_{d-n}\,\Omega_{n-1}$

for $\Omega_{n-1}$ and then substituting
$H_{n}=\begin{bmatrix}\alpha_{n}&\beta_{n}\\\
\beta_{n}&\gamma_{n}\end{bmatrix}$ defined by

$\displaystyle\alpha_{n}$
$\displaystyle=\frac{|\Omega_{n+1}^{\prime}(2(d-n)+1)|^{2}}{\Re(\,\Omega_{n+1}^{\prime}(1)\overline{(i\Omega_{n+1}^{\prime}(2(d-n)+1))}\,)},\quad\beta_{n}=\frac{\Im(\,\Omega_{n+1}^{\prime}(1)\overline{(i\Omega_{n+1}^{\prime}(2(d-n)+1))}\,)}{\Re(\,\Omega_{n+1}^{\prime}(1)\overline{(i\Omega_{n+1}^{\prime}(2(d-n)+1))}\,)},$
$\displaystyle\gamma_{n}$
$\displaystyle=\frac{|\Omega_{n+1}^{\prime}(1)|^{2}}{\Re(\,\Omega_{n+1}^{\prime}(1)\overline{(i\Omega_{n+1}^{\prime}(2(d-n)+1))}\,)}.$

into $H_{n}$ of $\Omega_{n}^{\prime}$. In this way we can inductively obtain
vectors $\Omega_{1},\dots,\Omega_{n}$ and quadratic real symmetric matrices
$H_{1},\dots,H_{d}$ starting with the initial vector $\Omega_{0}$.

## References

  * [1] L. de Branges,  Hilbert spaces of entire functions,  Prentice-Hall, Inc., Englewood Cliffs, N.J.,  1968\.
  * [2] A. Cohn,  Über die Anzahl der Wurzeln einer algebraischen Gleichung in einem Kreise,  Math. Z. 14 (1922), no. 1, 110–148.
  * [3] H. Dym,  An introduction to de Branges spaces of entire functions with applications to differential equations of the Sturm-Liouville type,  Advances in Math. 5 (1970),395–471.
  * [4] I. Gohberg (ed.)  I. Schur methods in operator theory and signal processing,  Operator Theory: Advances and Applications, 18.  Birkhäuser Verlag, Basel,  1986\.
  * [5] B. Ja. Levin,  Distribution of zeros of entire functions,  Revised edition, Translations of Mathematical Monographs, 5,  American Mathematical Society, Providence, R.I.,  1980\.
  * [6] M. Marden,  Geometry of polynomials, Second edition,  Mathematical Surveys, No. 3,
American Mathematical Society, Providence, R.I.,  1966\.

  * [7] Q. I. Rahman, G. Schmeisser,  Analytic theory of polynomials,  London Mathematical Society Monographs. New Series, 26,  The Clarendon Press, Oxford University Press, Oxford,  2002\.
  * [8] I. Schur,  Über Potenzreihen, die im Innern des Einheitskreises beschränkt sind,  J. Reine Angew. Math. 147 (1917), 205–232.
  * [9] I. Schur,  Über Potenzreihen, die im Innern des Einheitskreises beschränkt sind,  J. Reine Angew. Math. 148 (1918), 122–145.
  * [10] M. Suzuki,  An inverse problem for a class of canonical systems and its applications to self-reciprocal polynomials,  J. Anal. Math. 136 (2018), no. 1, 273–340.
  * [11] M. Suzuki,  An inverse problem for a class of canonical systems having Hamiltonians of determinant one,  J. Funct. Anal. 279 (2020), no. 12, 108699.
  * [12] M. Suzuki,  Chains of reproducing kernel Hilbert spaces generated by unimodular functions,
https://arxiv.org/abs/2012.11121.

Department of Mathematics, School of Science,
Tokyo Institute of Technology
2-12-1 Ookayama, Meguro-ku, Tokyo 152-8551, JAPAN
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