# On the chain structure in the de Branges spaces

Yurii Belov, Alexander Borichev  Yurii Belov,
Department of Mathematics and Computer Sciences, St. Petersburg State
University, St. Petersburg, Russia
<EMAIL_ADDRESS>
x Alexander Borichev,
Aix-Marseille University, CNRS, Centrale Marseille, I2M, Marseille, France
<EMAIL_ADDRESS>
###### Abstract.

We study the indivisible intervals and the monotonicity of the growth of the
exponential type in the chains of de Branges subspaces in terms of the
spectral measure. We prove that for spectral measures supported on
$\mathbb{Z}$, there exist at most two subspaces of the same type, which then
bound an indivisible interval. Furthermore, in this case, we study possible
locations of the indivisible intervals.

###### Key words and phrases:

Hilbert spaces of entire functions, de Branges spaces, indivisible intervals,
exponential type.

###### 2020 Mathematics Subject Classification:

Primary: 46E22, Secondary: 30D15, 47B32

The first author was supported by a grant of the Government of the Russian
Federation for the state support of scientific research, carried out under the
supervision of leading scientists, agreement 075-15-2021-602 and Russian
Foundation for Basic Research grant 20-51-14001-ANF-a. The second author was
supported by the project ANR 18-CE40-0035.

##  1\. Introduction and main results

One of the main parts of the de Branges theory of Hilbert spaces of entire
functions is the study of the chains of the de Branges subspaces of a given de
Branges space. There are different equivalent definitions of de Branges spaces
(axiomatic approach, in terms of Hermite–Biehler functions, in terms of the
Hamiltonians, as the weighted Cauchy transforms, spaces isometrically imbedded
into $L^{2}$ with respect to a measure on the real line). For more information
on the de Branges theory see [10, 14, 15] and the references therein. For some
recent progress see, for example, [6, 16].

In this paper we use the weighted Cauchy transform definition of the de
Branges spaces. Let $T=\\{t_{n}\\}_{n\in\mathcal{N}}$ be a discrete subset of
the real line and let $\mu=\sum_{n\in\mathcal{N}}{\mu_{n}\delta_{t_{n}}}$ be a
positive measure such that

(1.1) $\sum_{n\in\mathcal{N}}\frac{\mu_{n}}{t^{2}_{n}+1}<\infty.$

Fix an entire function $A$ real on the real line with simple zeros at
$T=\operatorname{supp}\mu$ and define the corresponding de Branges space (in
the weighted Cauchy transform form) $\mathcal{HC}(\mu)$ as follows:

$\mathcal{HC}(\mu)=\mathcal{HC}(A,\mu)=\biggl{\\{}f(z)=A(z)\sum_{n\in\mathcal{N}}\frac{a_{n}\mu^{1/2}_{n}}{z-t_{n}}:\\{a_{n}\\}_{n\in\mathcal{N}}\in\ell^{2}\biggr{\\}},$
$\|f\|_{\mathcal{HC}(\mu)}=\|a_{n}\|_{\ell^{2}}.$

We study an important class of de Branges spaces corresponding to the so
called canonical systems on finite interval. Namely, given a $2\times 2$ real
summable a.e. positively semi-defined matrix function $H$ (Hamiltonian) on a
finite interval $[0,L]$, we consider the system

$JY^{\prime}(t)=zH(t)Y(t),\qquad t\in[0,L],\qquad
J=\biggl{(}\begin{array}[]{ccc}0&-1\\\ 1&0\end{array}\biggr{)},$

where $z\in\mathbb{C}$ is the so called spectral parameter, $Y$ is an
absolutely continuous column vector function such that $Y(0)^{T}=(0,1)$ and
$Y(L)^{T}=(A,B)$. The entire functions $A$ and $B$ are real on the real line
with simple real interlacing zeros. We define a measure $\mu$ supported on
$\mathcal{Z}_{A}$, where $\mathcal{Z}_{F}$ is the zero set of an entire
function $F$, with masses $B(t)/A^{\prime}(t)$, $t\in\mathcal{Z}_{A}$, and
associate to the canonical system with Hamiltonian $H$ the de Branges space
$\mathcal{HC}(A,\mu)$.

It is known that a de Branges space $\mathcal{HC}(A,\mu)$ corresponds to a
canonical system on a finite interval if and only if it is regular, that is,

$z\mapsto\frac{f(z)-f(w)}{z-w}\in\mathcal{HC}(A,\mu),$

whenever $f\in\mathcal{HC}(A,\mu)$, $w\in\mathbb{C}$. This is equivalent to
the condition that the functions in our space are in the Cartwright class and

(1.2)
$\sum_{n\in\mathcal{N}}\frac{1}{\mu_{n}(t^{2}_{n}+1)A^{\prime}(t_{n})^{2}}<\infty,$

and from now on we assume that this condition is satisfied.

By the de Branges theory, the de Branges subspaces of the space
$\mathcal{HC}(\mu)$ constitute a chain $\\{\mathcal{HC}(\mu_{s})\\}$ ordered
by inclusion. Every such $\mathcal{HC}(\mu_{s})$ corresponds to a point $s$ or
to a subinterval $(a,s]$ of $(0,L]$ and to the restriction of our Hamiltonian
$H$ to $[0,s]$. We set $(A_{s},B_{s})=Y(s)^{T}$ and we associate to the space
$\mathcal{HC}(\mu_{s})$ the corresponding entire function $A_{s}$. We have
$\mathcal{Z}_{A_{s}}=\operatorname{supp}\mu_{s}$. Given the de Branges space
$\mathcal{HC}(\mu)$ we denote the corresponding chain of the de Branges
subspaces by ${\rm Chain}(\mu)$. We are interested in the so called
$H$-indivisible intervals in this chain, that is the semi open maximal
intervals $I=(a,s]\subset[0,L]$ such that $H$ is a degenerate constant matrix
on $I$ and, hence, the subspaces $\mathcal{HC}(\mu_{s})$ coincide for $s\in
I$.

If the spectrum of $\mu$ is $\mathbb{Z}$, then one can easily verify (see
Section 2 below) that the chain does not contain indivisible end intervals
$(0,a]$ and $(a,L]$. It is of interest to study the indivisible intervals
inside the chain.

######  Theorem 1.

If $\operatorname{supp}(\mu)=\mathbb{Z}$, and the space $\mathcal{HC}(\mu)$ is
regular, then the corresponding chain ${\rm Chain}(\mu)$ can contain one
indivisible interval and cannot contain two contiguous indivisible intervals.

When the support of the measure $\mu$ is $\mathbb{Z}$, such chains may contain
infinitely many non-contiguous indivisible intervals, see Theorem 2 below. On
the other hand, these indivisible intervals are somewhat separated, see
Theorem 6.

Furthermore, if the support of the measure $\mu$ is similar enough to
$\mathbb{Z}$, then we get results analogous to Theorem 1, see Section 3.

###  1.1. Exponential type

Another important characteristic of a de Branges space $\mathcal{HC}(\mu)$ is
the exponential type ${\rm Type}(\mathcal{HC}(\mu))$, that is the exponential
type $t(A)$ of the function $A$.

######  Theorem 2.

Let $\Sigma$ be a countable subset of the interval $(0,\pi)$. There exists a
measure $\mu$ supported on $\mathbb{Z}$ such that the space
$\mathcal{HC}(\mu)$ is regular, and the corresponding chain ${\rm Chain}(\mu)$
contains indivisible intervals $J_{s}$ with ${\rm
Type}(\mathcal{HC}(\mu_{t}))=s$, for all $t\in J_{s}$, $s\in\Sigma$.

One can express the exponential type of a de Branges space $\mathcal{HC}(\mu)$
in terms of the Hamiltonian $H$ of the corresponding canonical system. Namely,
the Krein–de Branges formula (see e.g. [15, Theorem 11]) states that

${\rm Type}(\mathcal{HC}(\mu))=\int_{0}^{L}\sqrt{\det H(t)}\,dt.$

Let us recall here some key facts from the de Branges theory which are
necessary to formulate our results.

######  Theorem 3 ([15, Theorem 17]).

Given a positive measure $\nu$ on $\mathbb{R}$ such that

(1.3) $\int_{\mathbb{R}}\frac{d\nu(t)}{1+t^{2}}<\infty,$

there exists a chain of regular de Branges spaces $\mathcal{H}_{t,\nu}$,
$(t\in(0,\infty)$ or $t\in(0,L])$ such that $\mathcal{H}_{t,\nu}$ is
isometrically embedded in $L^{2}(\nu)$, the set
$\bigcup_{t}\mathcal{H}_{t,\nu}$ is dense in $L^{2}(\nu)$ and

$\mathcal{H}_{t_{1},\nu}\subset\mathcal{H}_{t_{2},\nu},\qquad t_{1}\leq
t_{2}.$

Every regular de Branges space isometrically embedded in $L^{2}(\nu)$ belongs
to this chain. Furthermore, if $L<\infty$, then $T=\operatorname{supp}\nu$ is
discrete, and for an entire function $A$ real on the real line with simple
zeros at $T$, we have $\mathcal{H}_{L,\nu}=\mathcal{HC}(A,\nu^{*})$, where
$\nu^{*}(\\{t\\})=1/(\nu(\\{t\\})(A^{\prime}(t))^{2})$,
$t\in\operatorname{supp}T$. Furthermore, $\mathcal{H}_{L,\nu}$ restricted to
$T$ is equal to $L^{2}(\nu)$.

For example if $\nu$ is the Lebesque measure, then the corresponding de
Branges chain consists of the Paley–Wiener spaces $\mathcal{PW}_{a}$,
$a\in(0,\infty)$.

The number $T=\sup_{t}({\rm Type}(\mathcal{H}_{t,\nu}))$ is called the
exponential type of the measure $\nu$. One of the fundamental question of
harmonic analysis is to determine $T$ via $\nu$, see [9], [13] and the
references therein. We are interested in the closely related question about
the regularity of growth of the exponential type.

######  Definition 4.

Given a positive measure $\nu$ on the real line satisfying (1.3), we say that
$\nu$ generates a thin chain if for any type $t>0$ there exists at most one
element $\mathcal{H}_{t,\nu}$ of the chain such that ${\rm
Type}(\mathcal{H}_{t,\nu})=t$.

In particular, Hamiltonians corresponding to thin chains satisfy the condition
$\det H\neq 0$ a.e. on any interval and vice versa.

Given a measure $\nu$ satisfying (1.3), we define the Hilbert transform of
$\nu$ by the formula

(1.4)
$\widetilde{\nu}(z)=\frac{1}{\pi}\int_{\mathbb{R}}\biggl{(}\frac{1}{z-t}+\frac{t}{t^{2}+1}\biggr{)}\,d\nu(t).$

The function $\widetilde{\nu}(z)$ is well defined for
$z\in\mathbb{C}\setminus\mathbb{R}$. Moreover, it is well-known that for
absolutely continuous measures $\nu$, $d\nu=w\,dt$, the Hilbert transform
$\widetilde{\nu}$ exists on $\mathbb{R}$ if we understand the right-hand side
of (1.4) in the principal value sense. The Hilbert transform naturally appears
in many problems of harmonic analysis.

If the weight $w$ has convergent logarithmic integral, then it is well-known
that the type of $w(x)dx$ is infinite (see [9], [13]). We are able to show
that under some additional regularity assumptions, the measure $w(x)dx$
generates a thin chain.

######  Theorem 5.

Let $w$ be a $C^{1}$ smooth positive function such that

$w\in
L^{1}\biggl{(}\frac{dx}{1+x^{2}}\biggr{)},\qquad\int_{\mathbb{R}}\frac{\log
w(x)}{1+x^{2}}\,dx>-\infty,\qquad(\widetilde{\log w})^{\prime}\in
L^{\infty}(\mathbb{R}).$

Then the measure $w(x)dx$ generates a thin chain.

Sometimes the chain generated by a measure is not thin, but “almost thin”,
that is, for every $t>0$ the chain contains at most two subspaces of type $t$.
In particular, this is the case if the support of the measure is $\mathbb{Z}$,
as shows the following result.

######  Theorem 6.

Let $\nu$ be a positive measure satisfying (1.3) such that
$\operatorname{supp}\nu=\mathbb{Z}$ and the corresponding chain of regular de
Branges spaces $\mathcal{H}_{t,\nu}$ is defined on a finite interval $(0,L]$.
Then for any $0<t_{2}<t_{1}\leq L$ such that ${\rm
Type}(\mathcal{H}_{t_{1},\nu})={\rm Type}(\mathcal{H}_{t_{2},\nu})$, we have

$\dim(\mathcal{H}_{t_{1},\nu}\ominus\mathcal{H}_{t_{2},\nu})\leq 1.$

Starting with a measure $\mu=\sum_{n\in\mathbb{Z}}\mu_{n}\delta_{n}$
satisfying the conditions of Theorem 1, we can define
$\nu=\sum_{n\in\mathbb{Z}}\mu^{-1}_{n}\delta_{n}$ satisfying the conditions of
Theorem 6 and such that $\mathcal{H}_{L,\nu}=\mathcal{HC}(\sin\pi z,\mu)$, see
Theorem 3. Thus, Theorem 6 extends the result of Theorem 1 on the absence of
contiguous indivisible intervals.

###  1.2. Notation and organization of the paper

In this text, $A\lesssim B$ means that $A\leq CB$ with a positive constant
$C$, $A\gtrsim B$ means $A\geq cB$ with a constant $c>0$, and $A\asymp B$
means that $A\lesssim B$ and $A\gtrsim B$ simultaneously.

Some function theoretic criteria for the existence of (contiguous) indivisible
intervals in a chain of the Branges spaces are given in Section 2. In Section
3 we consider the indivisible intervals in the chains associated with the de
Branges spaces represented as the Cauchy transforms with spectrum in
$\mathbb{Z}$ or some perturbations of $\mathbb{Z}$. In Section 4 we deal with
the de Branges subspaces of the same type in a chain. Section 5 describes some
properties of different isometric Cauchy transform representations for the de
Branges spaces.

##  2\. Indivisible intervals

In this section we deal with regular de Branges spaces. We start with some
equivalent conditions for the existence of an indivisible interval in a de
Branges chain.

######  Lemma 7.

Given a de Branges space $\mathcal{HC}(\mu)$, the following assertions are
equivalent:

  1. (i)

The chain ${\rm Chain}(\mu)$ contains an indivisible interval.

  2. (ii)

For some subspaces in the chain, we have
$\dim(\mathcal{HC}(\mu_{s})\ominus\mathcal{HC}(\mu_{a}))=1$.

  3. (iii)

For some subspace $\mathcal{HC}(A,\nu)=\mathcal{HC}(\mu_{s})$ in the chain, we
have $A\sum_{n\in\mathcal{N}}\nu_{n}/(\cdot-t_{n})\in\mathcal{HC}(A,\nu)$,
where $\nu=\sum_{n\in\mathcal{N}}\nu_{n}\delta_{t_{n}}$.

  4. (iv)

There exists a function $G$ in $\mathcal{HC}(\mu)$ real on the real line with
simple real zeros such that $G$ is orthogonal to $G/(\cdot-\lambda)$,
$\lambda\in\mathcal{Z}_{G}$.

  5. (v)

For some subspace $\mathcal{HC}(\mu_{s})$ in the chain, the measure $\mu_{s}$
is finite.

  6. (vi)

For some subspace $\mathcal{HC}(\mu_{s})$ in the chain, the domain of the
operator of multiplication by $z$ is not dense in $\mathcal{HC}(\mu_{s})$.

Under the conditions of (iv), the de Branges space $\mathcal{H}$ spanned by
$G$ and $G/(\cdot-\lambda)$, $\lambda\in\mathcal{Z}_{G}$, has exponential type
equal to that of $G$.

###### Proof.

For the implication (i)$\implies$(ii) see [15, Section 4.3]. The implication
(ii)$\implies$(i) is evident, see [10, Problem 86]. By [10, Theorem 29], we
obtain the implication (ii)$\implies$(iii). Next, taking
$G=A\sum_{n\in\mathcal{N}}\nu_{n}/(\cdot-t_{n})$, and using that $G\perp
G/(\cdot-\lambda)$, $\lambda\in\mathcal{Z}_{G}$, we obtain the implication
(iii)$\implies$(iv). The implication (iv)$\implies$(ii) is evident, because
the closed space spanned by $G/(\cdot-\lambda)$, $\lambda\in\mathcal{Z}_{G}$,
is a de Branges subspace of $\mathcal{HC}(\mu)$. The equivalence
(iii)$\iff$(v) follows because (iii) means that the sequence
$\\{\nu_{n}^{1/2}\\}_{n\in\mathcal{N}}$ is in $\ell^{2}$. The equivalence
(iii)$\iff$(vi) follows from [10, Theorem 29].

∎

The chain ${\rm Chain}(\mu)$ starts with an indivisible interval $(0,a]$ (or
several contiguous indivisible intervals
$(0,a_{1}],(a_{1},a_{2}],\ldots,(a_{k-1},a_{k}]$) if and only if
$1\in\mathcal{HC}(\mu)$ or, correspondingly,
$1,\ldots,z^{k-1}\in\mathcal{HC}(\mu)$) if and only if

$\sum_{n\in\mathcal{N}}\frac{1}{\mu_{n}A^{\prime}(t_{n})^{2}}<\infty$

or, correspondingly,

$\sum_{n\in\mathcal{N}}\frac{t^{2(k-1)}_{n}}{\mu_{n}A^{\prime}(t_{n})^{2}}<\infty.$

The chain ends with $k$ contiguous indivisible intervals
$(a_{1},a_{2}],(a_{2},a_{3}],\ldots,(a_{k},L]$ if and only if

$\sum_{n\in\mathcal{N}}\mu_{n}t_{n}^{2(k-1)}<\infty.$

Furthermore, the chain ${\rm Chain}(\mu)$ contains $k$ contiguous indivisible
intervals if and only if we can find an entire function $G$ real on the real
line with simple real zeros such that $G$ is orthogonal to
$G/(\cdot-\lambda)$, $\lambda\in\mathcal{Z}_{G}$, and
$z^{k-1}G\in\mathcal{HC}(\mu)$.

In this article, we are mainly interested in indivisible intervals inside the
chain.

The reproducing kernel $K_{t_{n}}$ of $\mathcal{HC}(\mu)$ at $t_{n}\in T$,

$\langle F,K_{t_{n}}\rangle=F(t_{n}),\qquad F\in\mathcal{HC}(\mu),$

is given by

$K_{t_{n}}(z)=\mu_{n}A^{\prime}(t_{n})\frac{A(z)}{z-t_{n}},\qquad
n\in\mathcal{N}.$

Therefore,

$\|K_{t_{n}}\|=\mu^{1/2}_{n}|A^{\prime}(t_{n})|,\qquad n\in\mathcal{N},$

and

$\frac{K_{t_{n}}(z)}{\|K_{t_{n}}\|}=\frac{\|K_{t_{n}}\|}{A^{\prime}(t_{n})}\frac{A(z)}{z-t_{n}},\qquad
n\in\mathcal{N}.$

######  Lemma 8.

If the chain ${\rm Chain}(\mu)$ contains an indivisible interval, and $A$ is
associated to $\mathcal{HC}(\mu)$, then there exist entire functions $S$ and
$G$ real on the real line such that $G\in\mathcal{HC}(\mu)$,
$S\in\ell^{2}(\mu)$, and

$\frac{GS}{A}=\sum_{n\in\mathcal{N}}\frac{a^{2}_{n}}{\cdot-t_{n}},$

where $a_{n}=\mu^{-1/2}_{n}G(t_{n})/A^{\prime}(t_{n})$.

In the opposite direction, if there exist two entire functions $S$ and $G$
real on the real line such that
$\lim_{|y|\rightarrow\infty}y^{k-1}G(iy)/A(iy)=0$ and

(2.1) $\displaystyle S$ $\displaystyle\in\ell^{2}(\mu),$ (2.2)
$\displaystyle\frac{G}{A^{\prime}}$ $\displaystyle\in\ell^{2}(1/\mu),$ (2.3)
$\displaystyle\frac{GS}{A}$
$\displaystyle=\sum_{n\in\mathcal{N}}\frac{c_{n}}{\cdot-t_{n}},$

where $\\{c_{n}\\}_{n\in\mathcal{N}}\in\ell^{1}$,
$\sum_{n\in\mathcal{N}}c_{n}\neq 0$, then the chain ${\rm Chain}(\mu)$
contains an indivisible interval corresponding to a subspace of exponential
type coinciding with that of $G$.

###### Proof.

Let $G\in\mathcal{HC}(A,\mu)$ be an entire function real on the real line with
simple real zeros, orthogonal to $G/(\cdot-\lambda)$,
$\lambda\in\mathcal{Z}_{G}$. Then we have

$G=A\sum_{n\in\mathcal{N}}\frac{a_{n}\mu^{1/2}_{n}}{\cdot-
t_{n}}=\sum_{n\in\mathcal{N}}a_{n}\frac{K_{t_{n}}}{\|K_{t_{n}}\|}\cdot\frac{A^{\prime}(t_{n})}{|A^{\prime}(t_{n})|},$

with real coefficients $a_{n}$. Then

$\displaystyle
G\perp\frac{G}{\lambda-\cdot},\qquad\lambda\in\mathcal{Z}_{G}\qquad\Leftrightarrow$
$\displaystyle\Bigl{\langle}\frac{G}{\lambda-\cdot},\sum_{n\in\mathcal{N}}a_{n}\frac{K_{t_{n}}(z)}{\|K_{t_{n}}\|}\cdot\frac{A^{\prime}(t_{n})}{|A^{\prime}(t_{n})|}\Bigr{\rangle}=0,\qquad\lambda\in\mathcal{Z}_{G}\quad\Leftrightarrow$
(2.4) $\displaystyle\sum_{n\in\mathcal{N}}\frac{a^{2}_{n}}{\lambda-
t_{n}}=0,\qquad\lambda\in\mathcal{Z}_{G},$

because

(2.5) $G(t_{n})=a_{n}A^{\prime}(t_{n})\mu^{1/2}_{n}.$

Next, (2.4) is equivalent to the existence of an entire function $S$ real on
the real line such that

(2.6) $\sum_{n\in\mathcal{N}}\frac{a^{2}_{n}}{\cdot-t_{n}}=\frac{GS}{A}.$

Comparing the residues on $T$, we obtain that

$a_{n}^{2}=\frac{G(t_{n})S(t_{n})}{A^{\prime}(t_{n})},\qquad n\in\mathcal{N},$

and hence,

(2.7) $S(t_{n})=a_{n}\mu^{-1/2}_{n},\qquad n\in\mathcal{N}.$

Finally, (2.5)–(2.7) yield (2.1)–(2.3) with $c_{n}\geq 0$,
$0<\sum_{n\in\mathcal{N}}c_{n}<\infty$.

In the opposite direction, suppose that we can find two entire function $S$
and $G$ real on the real line such that
$\lim_{|y|\rightarrow\infty}G(iy)/A(iy)=0$, and relations (2.1)–(2.3) hold
with $\\{c_{n}\\}_{n\in\mathcal{N}}\in\ell^{1}$,
$\sum_{n\in\mathcal{N}}c_{n}\neq 0$.

Set $b_{n}=S(t_{n})\mu^{1/2}_{n}$, $n\in\mathcal{N}$, and consider

$H=\sum_{n\in\mathcal{N}}b_{n}\frac{K_{t_{n}}}{\|K_{t_{n}}\|}\cdot\frac{A^{\prime}(t_{n})}{|A^{\prime}(t_{n})|}\in\mathcal{HC}(A,\mu).$

Since $G/A^{\prime}\in\ell^{2}(1/\mu)$ and
$\lim_{|y|\rightarrow\infty}G(iy)/A(iy)=0$, a result from the de Branges
theory [10, Theorem 26] yields that $G\in\mathcal{HC}(A,\mu)$.

Set

$a_{n}=\frac{G(t_{n})}{A^{\prime}(t_{n})\mu^{1/2}_{n}},\qquad
n\in\mathcal{N}.$

Then $a_{n}b_{n}=c_{n}$, $n\in\mathcal{N}$,

$\langle
G,H\rangle=\sum_{n\in\mathcal{N}}a_{n}b_{n}=\sum_{n\in\mathcal{N}}c_{n}\neq
0,$

and

$\Bigl{\langle}\frac{G}{\lambda-\cdot},H\Bigr{\rangle}=\sum_{n\in\mathcal{N}}\frac{G(t_{n})b_{n}}{(\lambda-
t_{n})\|K_{t_{n}}\|}\cdot\frac{A^{\prime}(t_{n})}{|A^{\prime}(t_{n})|}\\\
=\sum_{n\in\mathcal{N}}\frac{a_{n}b_{n}}{\lambda-
t_{n}}=\frac{GS}{A}(\lambda)=0,\qquad\lambda\in\mathcal{Z}_{G}.$

Thus
$G\not\in\operatorname{Span}\\{G/(\cdot-\lambda)\\}_{\lambda\in\mathcal{Z}_{G}}$,
and, by Lemma 7, we get an indivisible interval. ∎

If $S$ and $G$ in the formulation of Lemma 8 are not polynomials, then the
indivisible interval we obtain is inside the chain.

######  Lemma 9.

If the chain ${\rm Chain}(\mu)$ contains $k$ contiguous indivisible intervals,
and $A$ is associated to $\mathcal{HC}(\mu)$, then there exist entire
functions $S$ and $G$ real on the real line such that
$z^{k-1}G\in\mathcal{HC}(A,\mu)$,

$\displaystyle S$ $\displaystyle\in\ell^{2}(\mu),$ $\displaystyle\frac{GS}{A}$
$\displaystyle=\sum_{n\in\mathcal{N}}\frac{a^{2}_{n}}{\cdot-t_{n}},$

where $a_{n}=\mu^{-1/2}_{n}G(t_{n})/A^{\prime}(t_{n})$.

In the opposite direction, if there exist two entire functions $S$ and $G$
real on the real line such that
$\lim_{|y|\rightarrow\infty}y^{k-1}G(iy)/A(iy)=0$ and

$\displaystyle S$ $\displaystyle\in\ell^{2}(\mu),$
$\displaystyle\frac{z^{k-1}G}{A^{\prime}}$ $\displaystyle\in\ell^{2}(1/\mu),$
$\displaystyle\frac{GS}{A}$
$\displaystyle=\sum_{n\in\mathcal{N}}\frac{c_{n}}{\cdot-t_{n}},$

where $\\{c_{n}\\}_{n\in\mathcal{N}}\in\ell^{1}$,
$\sum_{n\in\mathcal{N}}c_{n}\neq 0$, then the chain ${\rm Chain}(\mu)$
contains $k$ contiguous indivisible intervals.

The proof is analogous to that of Lemma 8.

Again, if $S$ and $G$ are not polynomials, then the contiguous indivisible
intervals we obtain are inside the chain.

##  3\. The spectrum $\mathbb{Z}$ and its perturbations

Here, we start with the case when the spectrum $T$ of the de Branges space is
$\mathbb{Z}$, and, correspondingly, $A(z)=A_{0}(z)=\sin(\pi z)$.

###  3.1. Indivisible interval inside the chain; proof of Theorem 1

###### Proof of Theorem 1.

By Lemma 9 we know that the existence of $k$ contiguous intervals is
equivalent to the existence of two non-zero entire functions $S$ and $G$ real
on the real line and such that $z^{k-1}G\in\mathcal{HC}(A_{0},\mu)$,

(3.1) $\begin{cases}S\in\ell^{2}(\mu),\\\ z^{k-1}G\in\ell^{2}(1/\mu),\\\
\dfrac{GS}{A_{0}}=\sum\limits_{n\in\mathbb{Z}}\dfrac{c_{n}}{\cdot-n},\end{cases}$

$\\{c_{n}\\}_{n\in\mathbb{Z}}\in\ell^{1}$. Additionally, we could impose the
restriction $c_{n}\geq 0$, $n\in\mathbb{Z}$.

Now for $k=1$, choose entire functions $G$ and $S$ real on the real line such
that $G(z)S(z)=z^{-1}A_{0}(z)$,
$|G(x)|\asymp\operatorname{dist}(x,\mathcal{Z}_{G})(1+|x|)^{-1/2}$,
$|S(x)|\asymp\operatorname{dist}(x,\mathcal{Z}_{S})(1+|x|)^{-1/2}$. (For
example, we can take $G(z)=\prod_{n\geq
1}\bigl{(}1-\frac{z}{2n-1}\bigr{)}\bigl{(}1+\frac{z}{2n}\bigr{)}$). Then
$\lim_{|y|\to\infty}G(iy)/A_{0}(iy)=0$.

Set

$\mu_{n}=\begin{cases}|n|^{-1/2},\quad n\in\mathcal{Z}_{G},\\\ |n|^{1/2},\quad
n\in\mathcal{Z}_{S},\end{cases}$

and $\mu_{0}=1$. The measure $\mu=\sum_{n\in\mathbb{Z}}\mu_{n}\delta_{n}$
satisfies conditions (1.1) and (1.2). Furthermore, conditions (3.1) are
satisfied and the space $\mathcal{HC}(\mu)$ contains an indivisible interval
inside the chain.

In the opposite direction, suppose that there are two contiguous indivisible
intervals. Conditions (1.1), (1.2) and (3.1) imply that

$\displaystyle\sum_{n\in\mathbb{Z}}(S^{2}(n)+(1+|n|)^{-2})\mu_{n}<\infty,$
$\displaystyle\sum_{n\in\mathbb{Z}}(n^{2}G^{2}(n)+(1+|n|)^{-2})\mu^{-1}_{n}<\infty.$

Therefore,

$\sum_{n\in\mathbb{Z}}(|S(n)|+(1+|n|)^{-1})(n|G(n)|+(1+|n|)^{-1})<\infty.$

By the Cartwright theorem [8], $S$ and $G$ have strictly positive exponential
types. Since $t(S)+t(G)\leq\pi$, these exponential types are smaller than
$\pi$.

Since $G\in\ell^{1}(\mathbb{Z})$, we conclude that $G\in L^{1}(\mathbb{R})$
[8, Section 10.6]. In a similar way, since
$S(n)(1+|n|)^{-1}\in\ell^{1}(\mathbb{Z})$, we have $S(x)(1+|x|)^{-1}\in
L^{1}(\mathbb{R})$ [1, Theorem 3a]. Next we use that

$\frac{GS}{A_{0}}=\sum_{n\in\mathbb{Z}}\frac{a^{2}_{n}}{\cdot-n},\qquad
c_{n}\geq 0,\,n\in\mathbb{Z},\qquad\\{c_{n}\\}_{n\in\mathbb{Z}}\in\ell^{1}.$

Using a version of Boole’s lemma by Khrushchev–Vinogradov [11], we obtain that
$|G(x)S(x)|\asymp|x|^{-1}$ on a set $E\subset\mathbb{R}$ of infinite
logarithmic length. As a result,

$\infty=\biggl{(}\int_{E}\frac{dt}{1+|t|}\biggr{)}^{2}\leq\int_{E}|G(t)|\,dt\int_{E}\frac{|S(t)|\,dt}{1+|t|}\\\
\leq\int_{\mathbb{R}}|G(t)|\,dt\int_{\mathbb{R}}\frac{|S(t)|\,dt}{1+|t|}<\infty.$

This contradiction shows that no de Branges space with spectrum $\mathbb{Z}$
possesses two contiguous indivisible intervals. ∎

Next, we consider some situations where the spectrum of our de Branges space
is a perturbation of $\mathbb{Z}$.

######  Proposition 10.

Suppose that for some $\gamma\in\mathbb{R}$ and $T\subset\mathbb{R}$ we have

$|A(z)|\asymp\min(1,\operatorname{dist}(z,T))(1+|z|)^{\gamma}\exp(\pi|\Im
z|),\qquad z\in\mathbb{C}.$

If a de Branges space $\mathcal{HC}(A,\mu)$ is regular, then it can contain
$k$ contiguous indivisible intervals if and only if $k<2+\gamma$.

In the proof, we use a possibility to factorize such entire functions $A$ into
factors of precise asymptotics. For a similar arguments, see Lemmas 13 and 14
below.

###### Proof.

Let $1\leq k<2+\gamma$. Choose $\alpha\in(k-\gamma-1,1)$. Then choose
$\beta\in(-1-2\gamma,\min(1,1-2\alpha-2\gamma))$,
$\delta\in(\max(-1-2\gamma,2k-2\alpha-2\gamma-1),1)$ and define entire
functions $G$ and $S$ real on the real line such that $GS=A/(\cdot-\lambda)$
for some $\lambda\in T$,
$|G(t)|\asymp\operatorname{dist}(t,\mathcal{Z}_{G})(1+|t|)^{-\alpha}$,
$|S(t)|\asymp\operatorname{dist}(t,\mathcal{Z}_{S})(1+|t|)^{\gamma-1+\alpha}$,
$t\in\mathbb{R}$, and a measure $\mu=\sum_{t\in T}\mu_{t}\delta_{t}$ with
$\mu_{t}=(1+|t|)^{\beta}$, $t\in\mathcal{Z}_{G}$, $\mu_{t}=(1+|t|)^{\delta}$,
$t\in\mathcal{Z}_{S}$, $\mu_{\lambda}=1$. It remains to apply Lemma 9. In the
opposite direction, we argue by analogy with the proof of Theorem 1. ∎

######  Proposition 11.

Given $\beta>0$, set $T=\cup_{n\in\mathbb{Z}}\\{n,n+(2+|n|)^{-\beta}\\}$. If
the space $\mathcal{HC}(T,\mu)$ is regular, then $\beta<1$, and the
corresponding de Branges subspaces chain ${\rm Chain}(\mu)$ can contain an
indivisible interval and cannot contain two contiguous indivisible intervals.

###### Proof.

Set

$A(z)=\sin(\pi
z)\cdot\prod_{n\in\mathbb{Z}}\biggl{(}1-\frac{z}{n+(2+|n|)^{-\beta}}\biggr{)}.$

Then

$|A^{\prime}(t)|\asymp(1+|t|)^{-\beta},\qquad t\in T.$

Since the space $\mathcal{HC}(T,\mu)$ is regular, by (1.1) and (1.2) we obtain
that

$\sum_{t\in T}(1+|t|)^{\beta-2}<\infty,$

and, hence, $\beta<1$.

Next, if ${\rm Chain}(\mu)$ contains two contiguous indivisible intervals,
then, as in the proof of Theorem 1, relations (1.1), (1.2), and (3.1) imply
that

$\displaystyle\sum_{t\in T}(S^{2}(t)+(1+|t|)^{-2})\mu_{t}$
$\displaystyle<\infty,$ $\displaystyle\sum_{t\in
T}(t^{2\beta+2}G^{2}(t)+(1+|t|)^{2\beta-2})\mu^{-1}_{t}$
$\displaystyle<\infty,$

and we conclude as in the proof of Theorem 1.

If now $\beta\in(0,1)$, let us verify that ${\rm Chain}(\mu)$ can contain an
indivisible interval. Given $n\in\mathbb{Z}$, set $n^{*}=n+(2+|n|)^{-\beta}$.
Choose $\Lambda\subset\mathbb{Z}$ such that $0\not\in\Lambda$ and

$S(x)=\prod_{n\in\Lambda}\Bigl{(}1-\frac{x}{n}\Bigr{)}\asymp(1+|x|)^{-(\beta+1)/2},\qquad
x\in\mathbb{Z}\setminus\Lambda,$

and set $G(z)=A(z)/(zS(z))$. Denote $\Lambda^{*}=\\{n^{*}:n\in\Lambda\\}$. We
have

$\displaystyle|S(t)|$
$\displaystyle\asymp\begin{cases}(1+|t|)^{-(3\beta+1)/2},\qquad
t\in\Lambda^{*},\\\ (1+|t|)^{-(\beta+1)/2},\qquad t\in
T\setminus(\Lambda\cup\Lambda^{*}),\end{cases}$ $\displaystyle|G(t)|$
$\displaystyle\asymp(1+|t|)^{-(\beta+1)/2},\qquad t\in\Lambda\cup\\{0\\}.$

Now, we set

$\mu_{t}=\begin{cases}(1+|t|)^{(\beta+1)/2},\qquad t\in\Lambda\cup\\{0\\},\\\
(1+|t|)^{(3\beta-1)/2},\qquad t\in T\setminus(\Lambda\cup\\{0\\}),\end{cases}$

A direct calculation shows that the measure $\mu=\sum_{t\in
T}\mu_{t}\delta_{t}$, satisfies conditions (1.1) and (1.2). Furthermore,
$\lim_{|y|\to\infty}G(iy)/A(iy)=0$, $S\in\ell^{2}(\mu)$,
$G/A^{\prime}\in\ell^{2}(1/\mu)$, and we conclude by applying Lemma 8. ∎

###  3.2. Infinite number of indivisible intervals. Proof of Theorem 2

In the following result we consider lacunary canonical products constructed by
rapidly growing zeros $\\{z_{k}\\}_{k\geq 1}$, $|z_{k+1}/z_{k}|\geq q>1$,
$k\geq 1$.

######  Proposition 12.

Let $U$ be a lacunary canonical product $\Lambda\subset\mathbb{R}$,
$\operatorname{dist}(\Lambda,\mathbb{Z})>0$, $T=\Lambda\cup\mathbb{Z}$, and
let $A(z)=\sin(\pi z)U(z)$. Then we can find a measure $\mu$ on $T$ such that
the corresponding space $\mathcal{HC}(A,T)$ contains a two sided sequence of
infinitely many contiguous indivisible intervals.

###### Proof.

Here we just choose entire functions $G$ and $S$ real on the real line such
that $zG(z)S(z)=A(z)$,
$|G(t)|\asymp\operatorname{dist}(t,\mathcal{Z}_{G})\psi(t)(1+|t|)^{-1/2}$,
$|S(t)|\asymp\operatorname{dist}(t,\mathcal{Z}_{S})\psi(t)(1+|t|)^{-1/2}$,
where $\psi(t)=1+\max_{|z|=t}|U(z)|^{1/2}$, $t\in\mathbb{R}$. Then set
$\mu_{t}=\psi(t)^{-3}$, $t\in\mathcal{Z}_{G}$, $\mu_{t}=1$,
$t\in\mathcal{Z}_{A}\setminus\mathcal{Z}_{G}$. Then we obtain that
$z^{k}S\in\ell^{2}(\mu)$, $z^{k}G/A^{\prime}\in\ell^{2}(1/\mu)$, for any
$k\geq 0$ and apply a natural analog of Lemma 9. ∎

Next we need some standard information on the asymptotics of canonical
products associated with very regular sequences on the real line.

Given a countable symmetric $\Lambda\subset\mathbb{R}\setminus\\{0\\}$ of
finite linear density, we set

$\mathcal{C}_{\Lambda}(z)=\prod_{t\in\Lambda}\Bigl{(}1-\frac{z}{t}\Bigr{)}=\prod_{t\in\Lambda_{+}}\Bigl{(}1-\frac{z^{2}}{t^{2}}\Bigr{)},$

where $\Lambda_{+}=\Lambda\cap\mathbb{R}_{+}$. Denote by $n_{\Lambda}$ the
counting function of $\Lambda$,

$n_{\Lambda}(x)=\operatorname{card}\bigl{(}\Lambda\cap(0,x]\bigr{)},\qquad
x>0.$

We say that a symmetric $\Lambda\subset\mathbb{R}\setminus\\{0\\}$ has strong
linear asymptotics $a_{\Lambda}x+b_{\Lambda}$ if the function

$\psi_{\Lambda}(t)=\int_{0}^{t}\bigl{(}n_{\Lambda}(x)-\lfloor
a_{\Lambda}x+b_{\Lambda}\rfloor\cdot\mathbf{1}_{[1,\infty)}(x)\bigr{)}\,dx$

is bounded on $(0,\infty)$. We say that $\Lambda\subset\mathbb{R}$ is
uniformly discrete if
$\inf\bigl{\\{}|t_{1}-t_{2}|:t_{1},t_{2}\in\Lambda,\,t_{1}\not=t_{2}\bigr{\\}}>0$.

######  Lemma 13.

Given an infinite symmetric uniformly discrete
$\Lambda\subset\mathbb{R}\setminus\\{0\\}$ with strong linear asymptotics
$ax+b$, we have

$|\mathcal{C}_{\Lambda}(z)|\asymp\min(1,\operatorname{dist}(z,\Lambda))(1+|z|)^{-1-2b}e^{a\pi|\Im
z|},\qquad z\in\mathbb{C}.$

###### Proof.

Since the linear density of $\lambda$ is $a$, we need only to verify that

$|\mathcal{C}_{\Lambda}(x+i)|\asymp(1+|x|)^{-1-2b},\qquad x\in\mathbb{R}.$

It is easily seen that for $\Lambda_{a,b}=\\{\pm(n-b)/a\\}_{n\geq 1}$ (with
trivial modifications for small $n$) we have $n_{\Lambda_{a,b}}(x)=\lfloor
ax+b\rfloor\cdot\mathbf{1}_{[1,\infty)}(x)$,

$|\mathcal{C}_{\Lambda_{a,b}}(x+i)|\asymp(1+|x|)^{-1-2b},\qquad
x\in\mathbb{R}.$

Therefore, we need only to check that the function $W$,

$W(x)=\log\Bigl{|}\frac{\mathcal{C}_{\Lambda}(x+i)}{\mathcal{C}_{\Lambda_{a,b}}(x+i)}\Bigr{|}=\int_{0}^{\infty}\log\Bigl{|}1-\Bigl{(}\frac{x+i}{t}\Bigr{)}^{2}\Bigr{|}\,\bigl{(}dn_{\Lambda}(t)-dn_{\Lambda_{a,b}}(t)\bigr{)},$

is bounded on the real line. Integrating by parts twice and using that
$n_{\Lambda}-n_{\Lambda_{a,b}}$ and $\psi_{\Lambda}$ are bounded, we obtain
that

$W(x)=\int_{0}^{\infty}\psi_{\Lambda}(t)\cdot\Re\,\Bigl{[}\frac{2}{t^{2}}-\frac{1}{(t-x+i)^{2}}-\frac{1}{(t+x+i)^{2}}\Bigr{]}\,dt.$

The function in the right hand side is bounded because $\psi_{\Lambda}$ are
bounded. ∎

######  Lemma 14.

Let $\Lambda_{1}\subset\Lambda_{2}$ be two symmetric subsets of
$\mathbb{R}\setminus\\{0\\}$ with strong linear asymptotics, correspondingly,
$a_{1}x+b_{1}$ and $a_{2}x+b_{2}$. Given $a\in(a_{1},a_{2})$ and
$b\in(b_{1},b_{2})$, there exists a symmetric subset $\Lambda$ of
$\mathbb{R}\setminus\\{0\\}$ with linear asymptotics $ax+b$ such that
$\Lambda_{1}\subset\Lambda\subset\Lambda_{2}$.

###### Proof.

By observation. ∎

###### Proof of Theorem 2.

We consider just the case of infinite $\Sigma$. The other case is much
simpler. Let $\Sigma=\\{\pi s_{k}\\}_{k\geq 1}$. By induction in $k\geq 1$, we
construct a disjoint system of intervals $(a_{k},b_{k})\subset(0,1)$ such that
$s_{k}<s_{m}\implies b_{k}<a_{m}$, $k,m\geq 1$. Set $r_{k}=(a_{k}+b_{k})/2$,
$k\geq 1$. Also by induction in $k\geq 1$, we construct, using Lemma 14,
symmetric sets $\Lambda_{k}\subset\mathbb{Z}\setminus\\{0\\}$ with strong
linear asymptotics $s_{k}x-(1+r_{k})/6$ such that
$s_{k}<s_{m}\implies\Lambda_{k}\subset\Lambda_{m}$, $k,m\geq 1$.

Set $G_{k}=\mathcal{C}_{\Lambda_{k}}$, $S_{k}=\sin(\pi z)/(zG_{k}(z))$, $k\geq
1$. For every $k\geq 1$, by Lemma 13 we have

(3.2) $\displaystyle|G_{k}(n)|$
$\displaystyle\asymp(1+|n|)^{(r_{k}-2)/3},\qquad
n\in\mathbb{Z}\setminus\Lambda_{k},$ (3.3) $\displaystyle|S_{k}(n)|$
$\displaystyle\asymp(1+|n|)^{-(r_{k}+1)/3},\qquad n\in\Lambda_{k}.$

Given $n\in\mathbb{Z}\setminus\\{0\\}$, we set

$s(n)=\sup\,\bigl{\\{}s_{k}:k\geq 1,\,n\not\in\Lambda_{k}\bigr{\\}}.$

By construction, if $s_{k}>s(n)$, then $n\in\Lambda_{k}$, and if $s_{k}<s(n)$,
then $n\not\in\Lambda_{k}$. If $s(n)=s_{m}\in\Sigma$ and $n\in\Lambda_{m}$,
then we set $u(n)=a_{m}$, otherwise set

$u(n)=\sup\,\bigl{\\{}b_{k}:s_{k}<s(n)\bigr{\\}}.$

Now, we set $\mu_{0}=1$,

$\mu_{n}=(1+|n|)^{(2u(n)-1)/3},\qquad n\in\mathbb{Z}\setminus\\{0\\}.$

Then the measure $\mu=\sum_{n\in\mathbb{Z}}\mu_{n}\delta_{n}$ satisfies
conditions (1.1) and (1.2).

Fix $k\geq 1$. To prove the existence of an indivisible interval corresponding
to the exponential type $s_{k}$, by Lemma 8, we need only to verify that
$S_{k}\in\ell^{2}(\mu)$ and $G_{k}\in\ell^{2}(1/\mu)$.

If $n\in\Lambda_{k}$, then $s(n)\leq s_{k}$ and $u(n)\leq a_{k}$. Therefore,
by (3.3), we have

$\sum_{n\in\Lambda_{k}}|S_{k}(n)|^{2}\mu_{n}\asymp\sum_{n\in\Lambda_{k}}(1+|n|)^{-(2r_{k}+2)/3}(1+|n|)^{(2u(n)-1)/3}\\\
\leq\sum_{n\in\mathbb{Z}}(1+|n|)^{-(2r_{k}+2)/3+(2a_{k}-1)/3}=\sum_{n\in\mathbb{Z}}(1+|n|)^{-1+2(a_{k}-r_{k})/3}<\infty.$

If $n\in\mathbb{Z}\setminus(\Lambda_{k}\cup\\{0\\})$, then $s(n)\geq s_{k}$
and $u(n)\geq b_{k}$. Therefore, by (3.2), we have

$\\!\\!\sum_{n\in\mathbb{Z}\setminus(\Lambda_{k}\cup\\{0\\})}|G_{k}(n)|^{2}\mu^{-1}_{n}\asymp\sum_{n\in\mathbb{Z}\setminus(\Lambda_{k}\cup\\{0\\})}(1+|n|)^{(2r_{k}-4)/3}(1+|n|)^{(1-2u(n))/3}\\\
\leq\sum_{n\in\mathbb{Z}}(1+|n|)^{(2r_{k}-4)/3+(1-2b_{k})/3}=\sum_{n\in\mathbb{Z}}(1+|n|)^{-1+2(r_{k}-b_{k})/3}<\infty.$

This completes the proof. ∎

##  4\. The same type subspaces

###  4.1. Regularity of growth of exponential type. Proof of Theorem 5

The proof of Theorem 5 is based on a combination of an atomization result in
[5] and some fact on the completeness of mixed systems in the Paley–Wiener
spaces from [2].

###### Proof of Theorem 5.

We start with the following simple fact. If two positive weights are
comparable, that is $w_{1}(x)\asymp w_{2}(x)$, then the chains of the de
Branges subspaces are the same, i.e. the de Branges subspaces from different
chains coincide as sets with equivalent norms. Therefore, it is sufficient to
consider any weight comparable to $w$.

Now we apply Theorem 2.6 from [5] (with sufficiently large $\sigma>0$) and
construct an entire function $H$ of finite exponential type $b$ with simple
zeros such that

$|H(x)|^{2}\asymp w(x),\qquad x\in\mathbb{R}.$

It remains to prove that the measure $|H(x)|^{2}\,dm$ generates a thin chain.
Assume the contrary. Then there exist two different de Branges spaces
$\mathcal{H}_{1}$, $\mathcal{H}_{2}$ from the chain of the same exponential
type. Let us fix some non-trivial function $F_{1}$ from
$\mathcal{H}_{1}\ominus\mathcal{H}_{2}$. Let $F_{2}$ be an $A$-function
corresponding to $\mathcal{H}_{2}$ such that
$\mathcal{Z}_{F_{2}}\cap\mathcal{Z}_{G}=\emptyset$. Set $a=t(F_{1})=t(F_{2})$.
We have

$F_{1}\perp\frac{F_{2}(z)}{z-\lambda},\qquad\lambda\in\mathcal{Z}_{F_{2}},$

where $\perp$ means orthogonality in $\mathcal{H}_{1}$. We recall that the
space $\mathcal{H}_{1}$ is isometrically embedded in $L^{2}(|H|^{2}\,dm)$.
Hence,

(4.1)
$\int_{\mathbb{R}}\frac{F_{1}(x)H(x)\overline{F_{2}(x)H(x)}\,dx}{x-\lambda}=0,\qquad\lambda\in\mathcal{Z}_{F_{2}}.$

Since $F_{1}H,F_{2}H/(\cdot-\lambda)\in L^{2}(\mathbb{R})$,
$\lambda\in\mathcal{Z}_{F_{2}}$, the functions
$F_{1}G,F_{2}H/(\cdot-\lambda)$, $\lambda\in\mathcal{Z}_{F_{2}}$, belong to
the Paley–Wiener space $\mathcal{P}W_{a+b}$. Thus, equation (4.1) can be
considered as orthogonality of some vectors from $\mathcal{P}W_{a+b}$. Denote
by $k_{\lambda}$ the reproducing kernel in the space $\mathcal{P}W_{a+b}$ at
the point $\lambda\in\mathbb{C}$. From (4.1) we obtain that the system

$\bigl{\\{}k_{\lambda}\bigr{\\}}_{\lambda\in\mathcal{Z}_{H}}\cup\biggl{\\{}\frac{F_{2}H}{\cdot-\lambda}\biggr{\\}}_{\lambda\in\mathcal{Z}_{F_{2}}}$

is not complete in $\mathcal{P}W_{a+b}$. This contradicts the following lemma:

######  Lemma 15.

Let $T=T_{1}T_{2}$ be an entire function in the Paley–Wiener space
$\mathcal{PW}_{\pi}$ with the conjugate indicator diagram $[-\pi i,\pi i]$ and
with simple zeroes. Then the mixed system

$\bigl{\\{}k_{\lambda}\bigr{\\}}_{\lambda\in\mathcal{Z}_{T_{1}}}\cup\biggl{\\{}\frac{T_{1}T_{2}}{\cdot-\lambda}\biggr{\\}}_{\lambda\in\mathcal{Z}_{T_{2}}}$

is always complete in $\mathcal{PW}_{\pi}$.

This lemma follows immediately from [2, Proposition 2.1]. For other versions
of this result see [3, 4]. ∎

###  4.2. Spectrum $\mathbb{Z}$. Proof of Theorem 6

The proof of Theorem 6 is based on a combination of Theorem 1, some results on
the classical Pólya problem, and the second Beurling–Malliavin theorem.

###### Proof of Theorem 6.

First of all, since $\mathcal{H}=\mathcal{H}_{L,\nu}=\mathcal{H}(\sin(\pi
z),\mu)$ is regular and the support of $\nu$ and $\mu$ is $\mathbb{Z}$, by the
Cartwright theorem [8], the space $\mathcal{H}$ contains no entire functions
of zero exponential type except $0$.

Set $\mathcal{H}_{1}=\mathcal{H}_{t_{1},\nu}$,
$\mathcal{H}_{2}=\mathcal{H}_{t_{2},\nu}$. Using Theorem 1 we obtain that if
$\dim(\mathcal{H}_{1}\ominus\mathcal{H}_{2})<\infty$, then
$\dim(\mathcal{H}_{1}\ominus\mathcal{H}_{2})\leq 1$. Thus, it remains to
consider the case $\dim(\mathcal{H}_{1}\ominus\mathcal{H}_{2})=\infty$, ${\rm
Type}(\mathcal{H}_{1})={\rm Type}(\mathcal{H}_{2})$.

Choose a function $F\in\mathcal{H}_{1}\setminus\\{0\\}$ such that
$F\perp\mathcal{H}_{2}$. Let $A_{2}$ be an $A$-function corresponding to the
space $\mathcal{H}_{2}$ such that
$\mathcal{Z}_{A_{2}}\cap\mathbb{Z}=\emptyset$. We have

(4.2) $F\perp\frac{A_{2}}{\cdot-s_{n}},\qquad s_{n}\in\mathcal{Z}_{A_{2}}.$

Now, relation (4.2) is equivalent to the interpolation formula

(4.3)
$\sum_{n\in\mathbb{Z}}\frac{F(n)A_{2}(n)}{\mu_{n}(z-n)}=\frac{A_{2}(z)S(z)}{\sin(\pi
z)},$

for some entire function $S$. Since there exists infinitely many linear
independent functions $F$ satisfying (4.3) we can assume that the functions
$F$ and $S$ have at least $100$ common non-integer zeroes
$\lambda_{1},\ldots,\lambda_{100}$. Set
$P(z)=\prod_{k=1}^{100}(z-\lambda_{k})$. From (4.3) we conclude that

$\sum_{n\in\mathbb{Z}}\frac{F(n)A_{2}(n)}{(n-\lambda_{1})\mu_{n}(z-n)}=\frac{A_{2}(z)S(z)}{(z-\lambda_{1})\sin(\pi
z)},$

and then, by induction,

$\sum_{n\in\mathbb{Z}}\frac{F(n)A_{2}(n)}{P(n)\mu_{n}(z-n)}=\frac{A_{2}(z)S(z)}{P(z)\sin(\pi
z)}.$

Hence,

(4.4) $\frac{F}{P}\perp\frac{A_{2}}{\cdot-s_{n}},\qquad
s_{n}\in\mathcal{Z}_{A_{2}}.$

Thus, we can assume that our function $F$ satisfies the inequality
$|F(x)|\leq|x|^{-100}$, and is real on the real line. Using Lemmas 17,18 we
find such a representation $\mathcal{HC}(T,\gamma)$ of our space that the
zeroes of $F$ on the real line are away from the support
$(x_{n})_{n\in\mathbb{Z}}$ of
$\gamma=\sum_{n\in\mathbb{Z}}\gamma_{n}\delta_{x_{n}}$:

(4.5) $\text{the set\
}\mathcal{Z}_{F}\cap\bigcup_{n\in\mathbb{Z}\setminus\\{0\\}}\bigl{[}x_{n}-|n|^{-10},x_{n}+|n|^{-10}\bigr{]}\text{\
is bounded}.$

By (4.4), using this representation $\mathcal{HC}(T,\nu)$, we obtain that

$\sum_{n\in\mathbb{Z}}\frac{F(x_{n})A_{2}(x_{n})}{\gamma_{n}(\cdot-
x_{n})}=\frac{A_{2}U}{T}$

for some entire function $U$. Moreover, $t(A_{2})+t(U)\leq t(T)=1$ and
$0<t(F)\leq t(A_{2})$. By comparing residues we obtain

$\displaystyle U(x_{n})=F(x_{n})\gamma_{n}^{-1}T^{\prime}(x_{n}),$ (4.6)
$\displaystyle|U(x_{n})|\lesssim(1+|x_{n}|)^{-10},\qquad n\in\mathbb{Z}.$

Hence,

$\frac{UF}{T}=\sum_{n\in\mathbb{Z}}\frac{U(x_{n})F(x_{n})}{T^{\prime}(x_{n})(\cdot-
x_{n})}+R=\sum_{n\in\mathbb{Z}}\frac{F^{2}(n)}{\gamma_{n}(\cdot-x_{n})}+R,$

for some entire function $R$ of zero exponential type which is real on the
real line.

Case 1. $R$ is a polynomial. Then the zeroes of the product $UF$ are
sufficiently close to the support of $\gamma$, which contradicts to (4.5).

Case 2. $R$ is a transcendental entire function of zero exponential type. The
product $UF$ has at least one zero on every interval $(n,n+1)$. By (4.5) and
(4.6), $R$ is bounded on $\Sigma=\mathcal{Z}_{F}\cap\mathbb{R}$.

As in [3], we use now some information on the classical Pólya problem and the
second Beurling–Malliavin theorem.

A sequence $X=\\{x_{n}\\}\subset\mathbb{R}$ is a Pólya sequence if any entire
function of zero exponential type which is bounded on $X$ is a constant. We
say that a disjoint sequence of intervals $\\{I_{n}\\}$ on the real line is a
long sequence of intervals if

$\sum_{n}\frac{|I_{n}|^{2}}{1+\operatorname{dist}^{2}(0,I_{n})}=+\infty.$

Since $\Sigma$ is not a Pólya sequence and is a union of two separated
sequences, a theorem by Mitkovski–Poltoratski [12] (see also the discussion in
[3]) gives that there exists a long sequence of intervals $\\{I_{n}\\}$ such
that

$\frac{\operatorname{card}(\Sigma\cap I_{n})}{|I_{n}|}\to 0.$

Therefore, if $\Sigma^{\prime}=\mathcal{Z}_{U}\cap\mathbb{R}$, then

$\frac{\operatorname{card}(\Sigma^{\prime}\cap I_{n})}{|I_{n}|}\to 1.$

By the second Beurling–Malliavin theorem [7], we obtain that $t(U)\geq 1$,
and, hence, $t(F)$=0. This contradiction completes the proof. ∎

##  5\. Isometric Cauchy transform representations for de Branges spaces

We start with two standard results. For reader’s convenience we formulate them
here and give the proofs.

######  Lemma 16.

Given a de Branges space $\mathcal{HC}(A,\mu)$, its reproducing kernel is

$K_{w}(z)=A(z)\overline{A(w)}\sum_{n\in\mathcal{N}}\frac{\mu_{n}}{(z-t_{n})(\overline{w}-t_{n})}.$

If $w_{1},w_{2}\not\in\operatorname{supp}\mu$, then

(5.1)
$K_{w_{1}}(w_{2})=A(w_{2})\overline{A(w_{1})}\frac{\psi(w_{2})-\overline{\psi(w_{1})}}{\overline{w_{1}}-w_{2}},$

where

$\psi(z)=\sum_{n\in\mathcal{N}}\mu_{n}\Bigl{(}\frac{1}{z-t_{n}}+\frac{t_{n}}{t^{2}_{n}+1}\Bigr{)}.$

###### Proof.

By observation. ∎

######  Lemma 17.

Given a regular de Branges space $\mathcal{H}=\mathcal{HC}(A_{0},\mu)$,
$A_{0}(z)=\sin(\pi z)$, and $u\in\mathbb{R}$, set

$T=A_{0}\cdot(\psi-u).$

Then $T$ is an entire function of exponential type real on the real line, with
the conjugate indicator diagram $[-\pi i,\pi i]$. For every $n\in\mathbb{Z}$,
$T$ has exactly one simple zero $x_{n}$ on $(n,n+1)$, $\psi(x_{n})=u$, and
$\mathcal{Z}_{T}=\\{x_{n}\\}_{n\in\mathbb{Z}}$.

Next, $T\not\in\mathcal{H}$, $\\{K_{x_{n}}\\}_{n\in\mathbb{Z}}$ is an
orthogonal basis in $\mathcal{H}$, and

$\mathcal{H}=\mathcal{HC}(T,\nu),$

where

$\nu=\sum_{n\in\mathbb{Z}}\|T/(\cdot-
x_{n})\|^{-2}_{\mathcal{H}}\delta_{x_{n}}.$

###### Proof.

Since the zeros of $T$ and $A_{0}$ are interlacing, $T$ is of exponential type
with the conjugate indicator diagram $[-\pi i,\pi i]$, and

(5.2) $|A_{0}(iy)|=O(|yT(iy)|),\qquad|y|\to\infty.$

If $T=A_{0}\cdot(\psi-u)\in\mathcal{H}$, then

$A_{0}(z)\bigl{(}\psi(z)-u\bigr{)}=A_{0}(z)\sum_{n\in\mathbb{Z}}\frac{a_{n}\mu^{1/2}_{n}}{z-n}$

for some sequence $\\{a_{n}\\}_{n\in\mathbb{Z}}\in\ell^{2}$. Comparing the
values at the integer points we obtain that $a_{n}=\mu^{1/2}_{n}$,
$n\in\mathbb{Z}$, and, hence, $\sum_{n\in\mathbb{Z}}\mu_{n}<\infty$, which
contradicts to (1.2).

By formula (5.1), $\\{K_{x_{n}}\\}_{n\in\mathbb{Z}}$ is an orthogonal system
in $\mathcal{H}$. If $F\in\mathcal{H}\setminus\\{0\\}$ is orthogonal to
$\\{K_{x_{n}}\\}_{n\in\mathbb{Z}}$, then $F=TS$ for some entire function $S$.
By (5.2), we obtain that $|S(iy)|=O(|y|)$, $|y|\to\infty$. Furthermore,

$S=\frac{F}{A_{0}}\cdot\frac{A_{0}}{T}$

is of zero exponential type. Therefore, $S$ is a polynomial of order at most
$1$. Dividing $F$ by $S$ we obtain that $T\in\mathcal{H}$, which is
impossible. Thus, $\\{K_{x_{n}}\\}_{n\in\mathbb{Z}}$ is an orthogonal basis in
$\mathcal{H}$.

Denote

$\nu_{n}=\|T/(\cdot-x_{n})\|^{-2}_{\mathcal{H}}.$

Since $\\{T/(\cdot-x_{n})\\}_{n\in\mathbb{Z}}$ is a biorthogonal system to
$\\{K_{x_{n}}\\}_{n\in\mathbb{Z}}$, it is an orthogonal basis in
$\mathcal{H}$. Hence, for every $f\in\mathcal{H}$ we have

$f(z)=T(z)\sum_{n\in\mathbb{Z}}\frac{1}{z-x_{n}}\cdot\frac{\langle f,T/(\cdot-
x_{n})\rangle}{\|T/(\cdot-
x_{n})\|^{2}_{\mathcal{H}}}=T(z)\sum_{n\in\mathbb{Z}}\frac{a_{n}\nu_{n}^{1/2}}{z-x_{n}},$

where $a_{n}=\langle f,T/(\cdot-x_{n})\rangle\nu_{n}^{1/2}$, $n\in\mathbb{Z}$,
and $\\{a_{n}\\}_{n\in\mathbb{Z}}\in\ell^{2}$. Therefore,

$\mathcal{H}\subset\mathcal{HC}(T,\nu),$

where $\nu=\sum_{n\in\mathbb{Z}}\nu_{n}\delta_{x_{n}}$, and the inclusion is
isometric. Finally, again since $\\{T/(\cdot-x_{n})\\}_{n\in\mathbb{Z}}$ is an
orthogonal basis in $\mathcal{H}$, we have $\mathcal{H}=\mathcal{HC}(T,\nu)$.
∎

Next we show that for every subset $\Lambda$ of $\mathbb{R}$ of finite upper
linear density, we can find an isometric representation of our space with
respect to a measure somewhat separated from $\Lambda$.

######  Lemma 18.

In the conditions of Lemma 17, given a sequence of points $\\{y_{k}\\}_{k\geq
1}$ of finite upper linear density, we can find $u\in\mathbb{R}$ such that the
intersection

$\\{y_{k}\\}_{k\geq
1}\cap\bigcup_{n\in\mathbb{Z}\setminus\\{0\\}}\bigl{[}x_{n}-|n|^{-10},x_{n}+|n|^{-10}\bigr{]}$

is bounded.

###### Proof.

Set $h=\arctan\psi$. We have

$|h^{\prime}(t)|=\frac{\sum_{n\in\mathbb{Z}}\frac{\mu_{n}}{(t-n)^{2}}}{\bigl{(}\sum_{n\in\mathbb{Z}}\mu_{n}\bigl{(}\frac{1}{t-n}+\frac{n}{n^{2}+1}\bigr{)}\bigr{)}^{2}+1}.$

Let $t\in(m,m+1)$. Without loss of regularity we can assume that $m\geq 1$,
$s=t-m\leq 1/2$. Since

$\sum_{n\in\mathbb{Z}}\frac{\mu_{n}}{n^{2}+1}<\infty,$

we have

$\displaystyle\sum_{n\in\mathbb{Z}}\frac{\mu_{n}}{(t-n)^{2}}$
$\displaystyle\lesssim m^{2}+\mu_{m}s^{-2},$
$\displaystyle\sum_{n\in\mathbb{Z}}\mu_{n}\Bigl{(}\frac{1}{t-n}+\frac{n}{n^{2}+1}\Bigr{)}$
$\displaystyle\geq\mu_{m}s^{-1}-O(m^{2}).$

Hence, since

$\sum_{n\in\mathbb{Z}}\frac{1}{(n^{2}+1)\mu_{n}}<\infty,$

we obtain

(5.3) $|h^{\prime}(t)|=O(t^{6}),\qquad|t|\to\infty.$

Given $k\geq 1$, choose $n$ such that $y_{k}\in[n,n+1)$ and denote by
$\ell_{k}$ the length of the set

$J_{k}=h\bigl{(}[y_{k}-2|n|^{-10},y_{k}+2|n|^{-10}]\bigr{)}.$

By (5.3), $\sum_{k\geq 1}\ell_{k}<\infty$. Therefore, we can find
$u\in\mathbb{R}$ which belongs to at most finitely many sets $J_{k}$. Then for
sufficiently large $k$,

$\tan u\not\in\psi\bigl{(}[y_{k}-2|n|^{-10},y_{k}+2|n|^{-10}]\bigr{)}.$

If $\psi(s)=\tan u$, $s\in(n,n+1)$, then
$s\not\in\bigl{[}y_{k}-2|n|^{-10},y_{k}+2|n|^{-10}\bigr{]}$ and, hence,
$y_{k}\not\in\bigl{[}s-|n|^{-10},s+|n|^{-10}\bigr{]}$. The same is true for
$s\in(n-1,n)\cup(n+1,n+2)$. This completes the proof. ∎

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