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Sampling Theory, Orthogonal Sampling Formulas, 
Krein's Theory of Entire Operators, Reproducing Kernel Hilbert Spaces,de Branges Spaces.
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\begin{document}
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\title{Applications of M.\,G. Krein's Theory of Regular Symmetric
       Operators to Sampling Theory \thanks{Mathematics Subject
       Classification(2000): 41A05, 46E22, 47B25, 47N50, 47N99, 94A20.}
       \thanks{Keywords: Sampling Theory, Orthogonal Sampling Formulas,
       Krein's Theory of Entire Operators, Reproducing Kernel Hilbert Spaces,
       de Branges Spaces.}
       \thanks{Research partially supported by CONACYT under Project P42553�F.}}
       \author{\textbf{Luis O. Silva and Julio H. Toloza}
       \\[5mm]
       Instituto de Investigaciones en Matem\'aticas Aplicadas y en Sistemas\\
       Universidad Nacional Aut\'onoma de M\'exico\\
       Apartado Postal 20-726, M\'exico D.\,F. 01000
       \\[3mm]
       \texttt{silva@leibniz.iimas.unam.mx}\\
       \texttt{jtoloza@leibniz.iimas.unam.mx}} \date{} \maketitle
\begin{center}
\begin{minipage}{5in}
  \centerline{{\bf Abstract}} \bigskip
  The classical Kramer sampling theorem establishes general conditions
  that allow
  the reconstruction of functions by orthogonal sampling formulas. One major
  task in sampling theory is to find concrete, non trivial
  realizations of this theorem. In this paper we provide a new approach to
  this subject on the basis of the M. G. Krein's theory of simple
  regular symmetric operators, with deficiency indices $(1,1)$,
  for obtaining Kramer-type sampling formulas. We show that these
  formulas have the form of Lagrange interpolation series. Concerning
  the case of entire operators, we also characterize the space of functions
  reconstructible by our sampling formulas.
\end{minipage}
\end{center}
\newpage

\section{Introduction}
\label{sec:intro}

It has been argued recently that sampling theory might be the
bridge that would allow to reconcile the continuous nature of
physical fields with the need of discretization of space-time,
as required by a yet-to-be-formulated theory of quantum
gravity \cite{kempf1,kempf2,kempf3}.  This idea, which is
partially developed in \cite{kempf1} for the one-dimensional
case, introduces the use of simple regular symmetric operators
with deficiency indices $(1,1)$ to obtain orthogonal sampling
formulas.  It is remarkable that the class of operators under
consideration in \cite{kempf1} had already been studied in
detail by M. G.  Krein \cite{krein1,krein2,krein3} more than
60 years ago.

The conjunction of the ideas of \cite{kempf1} and Krein's
theory of symmetric operators with equal deficiency indices
suggests the means of treating sampling theory in a new and
straightforward way.
By introducing this new approach, we put in a
mathematically rigorous framework the ideas in \cite{kempf1} and
propose a general method for obtaining analytical Kramer-type
sampling formulas associated with the self-adjoint extensions of a
simple regular symmetric operator with deficiency indices $(1,1)$.

A seminal result in sampling theory is the
Whittaker-Shannon-Kotel'nikov sampling theorem
\cite{kotelnikov,shannon,whittaker}. This theorem states that
functions that belong to a Paley-Wiener space may be uniquely
reconstructed from their values at certain discrete sets of
points. A general approach to Whittaker-Shannon-Kotel'nikov
type formulas was developed by Kramer \cite{kramer} based on
the following result. Given a finite interval $I=[a,b]$, let
$K(y,x)\in L^2(I,dy)$ for all $x\in\mathbb{R}$.  Assume that
there exists a sequence $\{x_n\}_{n\in\mathbb{Z}}$ for which
$\{K(y,x_n)\}_{n\in\mathbb{Z}}$ is an orthogonal basis of
$L^2(I,dy)$.  Let $f$ be any function of the form
\[
f(x):=\inner{K(\cdot,x)}{g(\cdot)}_{L^2(I)}
\]
for some $g\in L^2(I,dy)$, where the brackets denotes the inner
product in $L^2(I,dy)$. Then $f$ can be reconstructed from its
samples $\{f(x_n)\}_{n\in\mathbb{Z}}$ by the formula
\[
f(x)=\sum_{n=-\infty}^\infty
        \frac{\inner{K(\cdot,x)}{K(\cdot,x_n)}_{L^2(I)}}
        {\norm{K(\cdot,x_n)}^2_{L^2(I)}}f(x_n),\qquad x\in\mathbb{R}.
\]
The search for concrete realizations of the Kramer sampling
theorem, which in some cases has proven to be a difficult
task, has motivated a large amount of literature.  See, for
instance, \cite{annaby1,garcia1,garcia2,garcia3,zayed1,zayed2}
and references therein. On the basis of the approach proposed
in the present work, we obtain Kramer-type analytic sampling
formulas for functions belonging to linear sets of analytic
functions determined by a given regular simple symmetric
operator.  A central idea in Krein's theory is the fact that
any simple symmetric operator with deficiency indices $(1,1)$,
acting in a certain Hilbert space $\cH$, defines a bijective
mapping from $\cH$ onto a space of scalar functions of one
complex variable having certain analytic properties. In this
space of functions one can introduce an inner product. In the
particular case when the starting point is an entire operator
(see Definition 2), the space of functions turns out to be a
Hilbert space of entire functions known as a de Branges space
\cite{debranges}.

Summing up, in this paper we present an original technique in
sampling theory based on Krein's theory of regular symmetric
operators.  Concretely, we prove an analytic sampling formula
valid for functions belonging to linear sets of analytic
functions associated to symmetric operators of the kind
already mentioned. We then focus our attention to entire
operators to provide a characterization of the corresponding
spaces of entire functions. As a byproduct, we also provide
rigorous proofs to some of the results formally obtained in
\cite{kempf1}. In a sense this work may be considered as
introductory of the theoretical framework.  Further rigorous
results related to \cite{kempf1}, as well as other
applications to sampling theory, will be discussed in
subsequent papers.

In relation with the articles on sampling theory mentioned
above, our results are close to those of \cite{garcia1}. There
are, of course, several differences. First, the sampling
theory developed in \cite{garcia1} is based upon the spectral
properties of a single self-adjoint operator (since
\cite{garcia1} considers only symmetric operators having zero
is their resolvent sets); as already mentioned, our sampling
theory is grounded on the family of self-adjoint extensions of
a simple regular symmetric operator with deficiency indices
$(1,1)$.  Second, we mainly use techniques of operator theory,
whereas \cite{garcia1} relies mostly on complex variable
methods complemented with some elementary results of
functional analysis. Finally, the results of \cite{garcia1} do
not fit easily in the framework proposed in \cite{kempf1} for
which our results indeed give some rigorous justification.

This paper is organized as follows. In Section 2, we introduce
some concepts of the theory of regular symmetric operators. In
Section 3, we state and prove a sampling formula. In Section
4, we provide a characterization of the Hilbert spaces
associated with the class of operators under consideration,
paying particular attention to the case of entire operators.
Finally, in Section 5, we discuss some examples.

\section{Preliminaries}
\label{sec:preliminaries}

Most of the mathematical background required in this work is
based on Krein's theory of entire operators as accounted in
the expository book \cite{R1466698}. In this section we review
some definitions and results from the theory of symmetric
operators and introduce the notation.

Let $\cH$ denote a separable Hilbert space whose inner product
$\inner{\cdot}{\cdot}$ will be assumed anti-linear in its
first argument.

\begin{definition}
  A closed symmetric operator $A$ with domain and range in
  $\cH$ is called:
\begin{enumerate}[(a)]
\item {\em simple} if it does not have non-trivial invariant
  subspaces on which $A$ is self-adjoint,
\item {\em regular} if every point in $\mathbb{C}$ is a point
  of regular type for $A$, that is, if the operator $(A-zI)$
  has bounded inverse for every $z\in\mathbb{C}$.
\end{enumerate}
\end{definition}

By \cite[Theorem 1.2.1]{R1466698}, an operator $A$ is simple if and only if
\[
\bigcap_{z:\im z\neq 0}\ran\left(A-zI\right)=\{0\}.
\]
Moreover, as shown in \cite[Propositions 1.3.3, 1.3.5 and
1.3.6]{R1466698}, a closed symmetric operator with finite
deficiency indices $(m,m)$ is regular if and only if some
(hence every) self-adjoint extension of $A$ within $\cal H$
has only discrete spectrum of multiplicity at most $m$.  Also,
\begin{equation}\label{def}
\text{dim}\left[\ran\left(A-zI\right)\right]^\perp=
\text{dim}\left[\Ker\left(A^*-\overline{z}I\right)\right] = m
\end{equation}
for all $z\in{\mathbb C}$. For this kind of operators it is
also known that, for given any $x\in\mathbb{R}$, there is
exactly one self-adjoint extension within $\cH$ such that $x$
is in its spectrum.

\begin{remark}
  In what follows, whenever we consider self-adjoint
  extensions of a symmetric operator $A$, we will always mean
  the self-adjoint restrictions of $A^*$, in other words, the
  self-adjoint extensions of $A$ within $\cH$ (cf. Naimark's
  theory on generalized self-adjoint extensions \cite[Appendix
  1]{MR1255973}).
\end{remark}

We will denote by $\text{Sym}^{(1,1)}_\text{R}({\cal H})$ the
class of regular, simple, closed symmetric operators, defined
on $\cH$, with deficiency indices $(1,1)$.

Let $A_\sharp$ be some self-adjoint extension of
$A\in\text{Sym}^{(1,1)}_\text{R}({\cal H})$. The generalized
Cayley transform is defined as
\[
\left(A_\sharp-wI\right)\left(A_\sharp-zI\right)^{-1}.
\]
for every $w\in\mathbb{C}$ and $z\in\mathbb{C}\setminus\Sp(A_\sharp)$.
This operator has several properties \cite[pag. 9]{R1466698}. We only
mention the following:
\begin{equation}\label{cayley2}
\left(A_\sharp-wI\right)\left(A_\sharp-zI\right)^{-1}:
\Ker\left(A^*-wI\right)\to\Ker\left(A^*-zI\right)
\end{equation}
one-to-one and onto.

A complex conjugation on $\cH$ is a bijective anti-linear
operator $C:\cH\to\cH$ such that $C^2=I$ and
$\inner{C\eta}{C\varphi}=\inner{\varphi}{\eta}$ for all
$\eta,\varphi\in\cH$.  A symmetric operator $A$ is said to be
real with respect to a complex conjugation $C$ if
$C\dom(A)\subseteq\dom(A)$ and $CA\varphi=AC\varphi$ for every
$\varphi\in\dom(A)$. Clearly, the condition
$C\dom(A)\subseteq\dom(A)$, along with $C^2=1$, implies that
$C\dom(A)=\dom(A)$. Consequently, $A^*$ is also real with
respect to $C$.


\section{Sampling formulae}
\label{sec:reg sym operators}

Let us consider an operator
$A\in\text{Sym}^{(1,1)}_\text{R}({\cal H})$.  Let $A_\sharp$
be some self-adjoint extension of $A$. Given $z_0\in{\mathbb
  C}\setminus\Sp(A_\sharp)$ and
$\psi_0\in\Ker\left(A^*-z_0I\right)$, define
\begin{equation}\label{psi}
\psi(z):=\left(A_\sharp-z_0I\right)\left(A_\sharp-zI\right)^{-1}\psi_0
        =\psi_0 + (z-z_0)\left(A_\sharp-zI\right)^{-1}\psi_0
\end{equation}
for every $z\in{\mathbb C}\setminus\Sp(A_\sharp)$.  This
vector-valued function is analytic in the resolvent set of
$A_\sharp$ which, by (\ref{cayley2}), takes values in
$\Ker\left(A^*-zI\right)$ when evaluated at $z$. Moreover,
$\psi(z)$ has simple poles at the points of
$\Sp(A_\sharp)=\Sp_\text{disc}(A_\sharp)$. By looking at the
first equality in (\ref{psi}), it is clear that the dependency
of $\psi(z)$ on $z_0$ and $\psi_0$ is rather unessential.
Indeed, for any other $z_0'\in{\mathbb
  C}\setminus\Sp(A_\sharp)$, we can take
$\psi_0'=\left(A_\sharp-z_0I\right)
\left(A_\sharp-z_0'I\right)^{-1}\psi_0$ in
$\Ker\left(A^*-z_0'I\right)$ and thus
$\psi(z)=\psi_0'+(z-z_0')\left(A_\sharp-zI\right)^{-1}\psi_0'$
by the first resolvent identity.

Given $z_1\in{\mathbb C}\setminus\Sp(A_\sharp)$, let us choose
some $\mu\in\cH$ such that $\inner{\psi(\cc{z_1})}{\mu}\neq
0$.  The inner product $\inner{\psi(\cc{z})}{\mu}$ then
defines an analytic function in
$\mathbb{C}\setminus\Sp(A_\sharp)$, having zeroes at a
countable set $S_\mu$ devoid of accumulation points in
$\mathbb{C}$. The set $S_\mu$ is defined as the subset of
$\mathbb{C}$ for which $\cH$ can not be written as the direct
sum of $\ran (A-zI)$ and $\text{Span}\{\mu\}$. Following
Krein, we call the element $\mu$ a {\em gauge} \cite{krein3}.

Let us define
\begin{equation}\label{thetrick}
\xi(z):=\frac{\psi(\cc{z})}{\inner{\mu}{\psi(\cc{z})}}\,
\end{equation}
for all $z\in\mathbb{C}\setminus S_\mu$.

\begin{lemma}
  The vector-valued function $\xi(z)$ is analytic in
  $\mathbb{C}\setminus S_\mu$ and has simple poles at points
  of $S_\mu$.  Moreover, it does not depend on the
  self-adjoint extension of $A$ used to define $\psi(z)$.
\end{lemma}
\begin{proof}
  The first statement holds by rather obvious reasons, notice
  only that the poles of $\inner{\psi(\cc{z})}{\mu}$ coincide
  with those of $\psi(\cc{z})$. Let us pay attention to the
  second statement. Consider two self-adjoint extensions
  $A_\sharp$ and $A_\sharp'$. We have
\[
\psi(z)=\left(A_\sharp-z_0I\right)\left(A_\sharp-zI\right)^{-1}\psi_0\quad
\text{ and }\quad
\psi'(z)=\left(A'_\sharp-z_0I\right)\left(A'_\sharp-zI\right)^{-1}\psi_0.
\]
Since both $\psi(z)$ and $\psi'(z)$ belong to
$\Ker\left(A^*-zI\right)$ and the dimension of this subspace
is always equal to one, it follows that $\psi'(z)=g(z)\psi(z)$
for every $z\not\in\Sp(A_\sharp)\cup\Sp(A'_\sharp)$, where
$g(z)$ is a scalar function. Inserting this identity into
(\ref{thetrick}) yields $\xi(z)=\xi'(z)$.
\end{proof}

For every $z\in{\mathbb C}\setminus S_\mu$ we have the decomposition
$
{\cal H} = \ran\left(A-zI\right)\dot{+}\,\text{Span}\left\{\mu\right\},
$
in which case every element $\varphi\in\cH$ can be written as
\[
\varphi = \left[\varphi-\widehat{\varphi}(z)\mu\right] +
\widehat{\varphi}(z)\mu,
\]
where $\varphi-\widehat{\varphi}(z)\in\ran\left(A-zI\right)$.
A simple computation shows that
the non-orthogonal projection $\widehat{\varphi}(z)$ is given by
\begin{equation}\label{transform}
\widehat{\varphi}(z)
        :=\frac{\inner{\psi(\cc{z})}{\varphi}}{\inner{\psi(\cc{z})}{\mu}}
        =\inner{\xi(z)}{\varphi}
\end{equation}
whenever $z\in{\mathbb C}\setminus S_\mu$; it is otherwise not defined.
Indeed, the function $\widehat{\varphi}(z)$ is analytic in
$\mathbb{C}\setminus S_\mu$ and meromorphic in $\mathbb{C}$ for every
$\varphi\in\cH$. We note that $\widehat{\mu}(z)\equiv 1$.

Let us denote the linear map
$\varphi\mapsto\widehat{\varphi}(z)$ as $\Phi_\mu$ and the
linear space of functions given by (\ref{transform}) as $\Phi_\mu{\cal
  H}$. Since the operator $A$ is simple, it follows that
$\Phi_\mu$ is injective and, therefore, is an isomorphism from
$\cal H$ onto $\Phi_\mu{\cal H}$ \cite[Theorem 1.2.2]{R1466698}.
Moreover, $\Phi_\mu$ transforms $A$ into the multiplication
operator on $\Phi_\mu{\cal H}$, that is,
\[
\widehat{\left(A\varphi\right)}(z) =
\left(\Phi_\mu A\Phi_\mu^{-1}\widehat{\varphi}\right)(z)=
z\widehat{\varphi}(z),\quad \varphi\in\dom(A).
\]

\begin{proposition}\label{interpolation}
  Assume $S_\mu\cap\mathbb{R}=\emptyset$.  Let $\{x_n\}$ be
  the spectrum of any self-adjoint extension $A_\sharp$ of
  $A$.  Then, for any analytic function $f(z)$ that belongs to
  $\Phi_\mu{\cal H}$, we have
\begin{equation}\label{sampling}
f(z) = \sum_{x_n\in\Sp(A_\sharp)}
      \frac{\inner{\xi(z)}{\xi(x_n)}}{\norm{\xi(x_n)}^2} f(x_n),
\end{equation}
The convergence in (\ref{sampling}) is uniform over
compact subsets of ${\mathbb C}\setminus S_\mu$.
\end{proposition}
\begin{proof}
  Fix some arbitrary self-adjoint extension $A_\sharp$ of $A$.
  Take another self-adjoint extension $A_\sharp'\ne A_\sharp$
  to define $\psi(z)$, that is,
$\psi(z)=\left(A_\sharp'-z_0I\right)\left(A_\sharp'-zI\right)^{-1}\psi_0$.
Arrange the elements of  $\Sp\left(A_\sharp\right)$ in a
  sequence  $\{x_n\}_{n\in M}$, where $M$ is a countable
  indexing set, and let $\eta_n$ be an eigenstate of
  $A_\sharp$ corresponding to $x_n$, i.\,e.,
  $A_\sharp\eta_n=x_n\eta_n$. Since $A^*\supset A_\sharp$,
  it follows that $\eta_m\in\Ker\left(A^*-x_mI\right)$, where
  furthermore
  $\text{dim}\left[\Ker\left(A^*-x_mI\right)\right]=1$. On the
  other hand, since $x_m$ is not a pole of $\psi(z)$,
  $\psi(x_m)$ is well defined and belongs also to
  $\Ker\left(A^*-x_mI\right)$. Therefore, up to a factor,
  $\eta_m=\xi(x_m)$.

  Pick a sequence $\{M_k\}_{k\in\mathbb{N}}$ of subsets of $M$
  such that $M_k\subset M_{k+1}$ and $\cup_kM_k=M$. Consider
  any analytic function $f(z)\in\Phi_\mu{\cal H}$.  Since $\Phi_\mu$
  is injective, there exists a unique $\varphi\in{\cal H}$
  such that $\widehat{\varphi}(z)=f(z)$.  Clearly,
\[
\left|\widehat{\varphi}(z)-
\sum_{n\in M_k}\frac{\inner{\xi(z)}{\xi(x_n)}}{\norm{\xi(x_n)}^2}
                     \inner{\xi(x_n)}{\varphi}\right|
        \le \norm{\xi(z)}\norm{\varphi-\sum_{n\in M_k}\frac{1}
                     {\norm{\xi(x_n)}^2}
        \inner{\xi(x_n)}{\varphi}\xi(x_n)}.
\]
The second factor in the r.\,h.\,s. of this inequality does
not depend on $z$ and obviously tends to zero as
$k\to\infty$. Since $\xi(z)$ is analytic on ${\mathbb
  C}\setminus S_\mu$, the uniform convergence of
(\ref{sampling}) has been proven.
\end{proof}

\begin{remark}
  Krein asserts that, for any operator in
  $\text{Sym}^{(1,1)}_\text{R}({\cal H})$, one can always
  choose $\mu$ so that $S_\mu\cap\mathbb{R}=\emptyset$
  \cite[Theorem 8]{krein2}.
\end{remark}

We notice that the sequence $\{x_n\}$ could be replaced by any
sequence $\{z_n\}$ of complex numbers for which
$\left\{\xi(z_n)\right\}$ is a sequence of orthogonal elements
that span $\cal H$ (for a related discussion, see
\cite{garcia2}).  In our case, the question of whether such a
sequence exists or not is answered to the affirmative by
invoking the self-adjoint extensions of the operator $A$.

Below we show that the interpolation formula (\ref{sampling})
is indeed a Lagrange interpolation series.

\begin{proposition}
  Under the hypotheses of Proposition~\ref{interpolation},
  there exists a complex function $G(z)$, analytic in
  $\mathbb{C}\setminus S_\mu$ (hence $\mathbb{R}$) and having
  simple zeroes at $\{x_n\}_{n\in I}=\Sp(A_\sharp)$, such that
\[
f(z) = \sum_{n\in M}
                \frac{G(z)}{(z-x_n)G'(x_n)} f(x_n),
\]
for every $f(z)\in\Phi_\mu{\cal H}$.
\end{proposition}
\begin{proof}
  Given $\{x_n\}_{n\in M}=\Sp(A_\sharp)$, set
  $\psi(z)=\left(A_\sharp-z_0I\right)\left(A_\sharp-zI\right)^{-1}\psi_0$,
  for some $z_0:\im z_0\neq 0$ and $\psi_0\in\Ker\left(A^*-z_0
    I\right)$. Define
\[
G(z):=\frac{1}{\inner{\psi(\cc{z})}{\mu}}.
\]
This function has simples zeroes at the poles of
$\psi(\cc{z})$, that is, at points of the set $\{x_n\}_{n\in
  M}$. Also, $\xi(z)=\cc{G(z)}\psi(\cc{z})$. Moreover, we can
write
\[
\psi(\cc{z})=\frac{\eta(\cc{z})}{\cc{z}-x_n}\quad\text{ and }\quad
G(z)=(\cc{z}-x_n)F(\cc{z}),
\]
where $\eta(z):=
(\cc{z}-x_n)\left(A_\sharp-z_0I\right)\left(A_\sharp-zI\right)^{-1}\psi_0$
is analytic at $z=x_n$, $\eta(x_n)\neq 0$, and $F(x_n)=G'(x_n)$. Thus,
a straightforward computation shows that
\[
\frac{\inner{\xi(z)}{\xi(x_n)}}{\norm{\xi(x_n)}^2} =
        \frac{G(z)}{(z-x_n)G'(x_n)}
        \frac{\inner{\eta(\cc{z})}{\eta(x_n)}}{\norm{\eta(x_n)}^2}
\]
so it remains to verify that the last factor above equals one.
By the Cauchy integral formula, we have
\begin{align*}
\eta(x_n)
        &=\frac{1}{2\pi i}\oint_{|w-x_n|=\epsilon}\frac{\eta(w)}{w-x_n}\,dw\\
        &=\frac{1}{2\pi i}\oint_{|w-x_n|=\epsilon}
         \left(A_\sharp-z_0I\right)\left(A_\sharp-wI\right)^{-1}\psi_0\,dw\\
        &=-\left(A_\sharp-z_0I\right)\left(
          \frac{1}{2\pi i}\oint_{|w-x_n|=\epsilon}
          \left(wI-A_\sharp\right)^{-1}dw\right)\psi_0\\
        &=-\left(x_n-z_0\right)P_n\psi_0,
\end{align*}
where $P_n$ denotes the orthoprojector onto the eigenspace associated to $x_n$.
Therefore,
\begin{align*}
\inner{\eta(\cc{z})}{\eta(x_n)}
        &=-\left(x_n-z_0\right)\inner{\eta(\cc{z})}{P_n\psi_0}\\
        &=-\left(x_n-z_0\right)\left(z-x_n\right)
          \inner{P_n\left(A_\sharp-z_0I\right)
          \left(A_\sharp-zI\right)^{-1}\psi_0}{\psi_0}\\
        &=\left|x_n-z_0\right|^2\inner{P_n\psi_0}{\psi_0}.
\end{align*}
Finally, it is clear that $\inner{\eta(x_n)}{\eta(x_n)}
=\left|x_n-z_0\right|^2\inner{P_n\psi_0}{\psi_0}$.
\end{proof}

\begin{remark}
Notice that the function $G(z)$ is defined up to a factor. In particular,
one may adjust it so that $G'(x_k)=1$, where $x_k$ is a fixed
eigenvalue of $A_\sharp$. Thus, a computation like the one in the proof above
shows that
\begin{equation}\label{gz}
G(z)=(z-x_k)\frac{\inner{\xi(z)}{\xi(x_k)}}{\norm{\xi(x_k)}^2}.
\end{equation}
This identity may be useful in some applications; see Example 2 below.
\end{remark}

\section{Spaces of analytic functions}
\label{sec:entire operators}

In this section we characterize the set of functions given by the
mapping $\Phi_\mu$

Let ${\cal R}\subset{\cal H}$ be the linear space of elements
$\varphi$ for which $\widehat{\varphi}(z)$ is analytic on
$\mathbb{R}$. As a consequence of \cite[Corollary 1.2.1]{R1466698},
it follows that
\begin{equation}\label{isometry}
\inner{\varphi}{\eta}
     = \int_{-\infty}^\infty\cc{\widehat{\varphi}(x)}\,\widehat{\eta}(x)\,
          dm(x)
\end{equation}
for any $\varphi,\eta\in{\cal R}$ and
$m(x)=\inner{E_x\mu}{\mu}$, where $E_x$ is any spectral
function of the operator $A$. That is, $\Phi_\mu{\cal R}$ is a
linear space of analytic functions in ${\mathbb C}\setminus
S_\mu$ such that their restriction to $\mathbb{R}$ belong to
$L^2(\mathbb{R},dm)$; in short,
\[
\left.\Phi_\mu{\cal R}\right|_\mathbb{R}\subset L^2(\mathbb{R},dm).
\]
Moreover, in this restricted sense $\Phi_\mu$ is an isometry
from $\cal R$ into $L^2(\mathbb{R},dm)$.

The following theorem is due to Krein \cite[Theorem 3]{krein2}

\begin{theorem}[Krein]\label{krein}
  For $A\in\text{Sym}^{(1,1)}_\text{R}({\cal H})$, assume that
  $\cc{\cal R}={\cal H}$. Consider a distribution function
  $m(x)=\inner{E_x\mu}{\mu}$, where $E_x$ is a spectral
  function of $A$.  Then the map $\Phi$ generates a bijective
  isometry from $\cal H$ onto $L^2(\mathbb{R},dm)$ if and only
  if $E_x$ is orthogonal.
\end{theorem}

This theorem deserves some comments. For entire operators,
$\Phi_\mu\cH$ is a linear subset of entire functions. When
$E_x$ is orthogonal it occurs that
\[
\left.\Phi_\mu\cH\right|_\mathbb{R}=L^2(\mathbb{R},dm)
\]
in the usual sense of equivalence classes. Thus, every
function in $L^2(\mathbb{R},dm)$ is, up to a set of measure
zero with respect to $m(x)$, the restriction to $\mathbb{R}$
of a unique entire function that is the image under $\Phi_\mu$
of one and only element belonging to $\cal H$.  In passing, we
notice that every orthogonal spectral function of the operator
$A$ is the restriction to $\dom(A)$ of the spectral function
of some of its self-adjoint extensions within $\cal H$.  Since
these self-adjoint extensions have only discrete spectrum, the
inner product in $L^2(\mathbb{R},dm)$, with
$m(x)=\inner{E_x\mu}{\mu}$, reduces to an expression like
\[
\int_{-\infty}^\infty\cc{f(x)}g(x)\,dm(x)=\sum_{k}c_k\cc{f(x_k)}g(x_k)
\]
whenever $E_x$ is orthogonal. That is, the equivalence classes in these
spaces are quite broad.

The following corollary is partly a straightforward consequence of
Theorem~\ref{krein}. Notice that
$S_\mu\cap\mathbb{R}=\emptyset$ implies $\cal{R}=\cH$.

\begin{corollary}\label{hachesombrero}
  Let $A\in\text{Sym}^{(1,1)}_\text{R}({\cal H})$ and choose a
  gauge $\mu$ for this operator. Assume
  that $S_\mu\cap\mathbb{R}=\emptyset$. Let $E_x$ be one of its
  orthogonal spectral functions. Then the linear space of
  functions $\widehat{\cH}_\mu:=\Phi_\mu{\cH}$, equipped with
  the inner product
\begin{equation}\label{innerprod}
\inner{f(\cdot)}{g(\cdot)}:=\int_{-\infty}^\infty\cc{f(x)}g(x)\,dm(x)
\qquad \text{where }m(x)=\inner{E_x\mu}{\mu},
\end{equation}
is a reproducing kernel Hilbert space, with reproducing kernel
$k(z,w):=\inner{\xi(z)}{\xi(w)}$.
\end{corollary}
\begin{proof}
We only verify the last statement. Given
$f(z)=\inner{\xi(z)}{\varphi}\in\widehat{\cH}_\mu$, we have
\[
\inner{k(\cdot,w)}{f(\cdot)}=\int_{-\infty}^\infty\cc{k(x,w)}f(x)\,dm(x)
                            =\inner{\xi(w)}{\varphi}=f(w),
\]
where the second equality follows from (\ref{isometry}).
\end{proof}

Notice that the linear space $\widehat{\cH}_\mu$ depends only
on the choice of gauge $\mu$.  By
Corollary~\ref{hachesombrero}, for those gauges that obeys
$S_\mu\cap\mathbb{R}=\emptyset$, $\widehat{\cH}_\mu$ may be
endowed with different Hilbert space structures, one for each
orthogonal spectral function of the operator $A$. By
(\ref{isometry}), all these Hilbert spaces are however
isometrically equivalent.

Irrespective of any Hilbert space structure, $\widehat{\cH}_\mu$
possesses the following properties.

\begin{proposition}
  \label{prop-cond-1}
Let $w$ be a non real zero of $f(z)\in\widehat\cH_\mu$. Then the function
$g(z):=f(z)(z-\cc{w})(z-w)^{-1}$ is also in $\widehat\cH_\mu$ and
$\norm{g(\cdot)}=\norm{f(\cdot)}$.
\end{proposition}
\begin{proof}
  Suppose that $w$ is a non real zero of $f(z)$. Since
$f(z)=\inner{\xi(z)}{\varphi}$ for some $\varphi\in{\cal H}$, it follows
that $\varphi$ is orthogonal to $\psi(\cc{w})$ and therefore
$\varphi\in\ran\left(A-wI\right)$. Note that
\begin{align*}
f(z)
   &=\inner{\xi(z)}{\left(A-wI\right)\left(A-wI\right)^{-1}\varphi}\\
   &=\inner{\left[A^*-\cc{z}+\left(\cc{z}-\cc{w}\right)I\right]\xi(z)}
               {\left(A-wI\right)^{-1}\varphi}\\
   &=\left(z-w\right)\inner{\xi(z)}{\left(A-wI\right)^{-1}\varphi}.
\end{align*}
In these computations we have used that
$\xi(z)\in\Ker\left(A^*-\cc{z}\right)$.  Moreover,
\begin{align*}
\inner{\xi(z)}{\left(A-\cc{w}I\right)\left(A-wI\right)^{-1}\varphi}
    &=\inner{\left[A^*-\cc{z}I+\left(\cc{z}-w\right)I\right]\xi(z)}
                {\left(A-wI\right)^{-1}\varphi}\\
    &=\left(z-\cc{w}\right)\inner{\xi(z)}{\left(A-wI\right)^{-1}\varphi}\\
    &=\frac{z-\cc{w}}{z-w}f(z).
\end{align*}
Then $f(z)\left(z-\cc{w}\right)\left(z-w\right)^{-1}\in\widehat{\cH}_\mu$.
The equality of norms follows from the fact that the Cayley transform
$\left(A-\cc{w}I\right)\left(A-wI\right)^{-1}$ is an isometry.
\end{proof}
\begin{proposition}
  \label{prop-cond-2}
  The evaluation functional, defined by $f(\cdot)\mapsto
  f(z)$, is continuous.
\end{proposition}
\begin{proof}
Let $f(z),g(z)\in\widehat{\cH}_\mu$.
Then $f(z)=\inner{\xi(z)}{\varphi}$ and $g(z)=\inner{\xi(z)}{\eta}$
for some $\varphi,\eta\in\cH$, and furthermore
\[
\left|f(w)-g(w)\right|=\left|\inner{\xi(w)}{\varphi-\eta}\right|
        \le \norm{\xi(w)}\norm{\varphi-\eta}
        = \norm{\xi(w)}\norm{f(\cdot)-g(\cdot)}.
\]
In other words, this result follows from the fact that $\widehat{\cH}_\mu$ is a
reproducing kernel Hilbert space.
\end{proof}
\begin{definition}
  An operator $A\in\text{Sym}^{(1,1)}_\text{R}({\cal H})$ is
  called {\em entire} if there exists a so-called {\em entire
    gauge} $\mu\in\cH$ such that $\widehat{\varphi}(z)$ is an
  entire function for every $\varphi\in\cH$. Equivalently, $A$
  is entire if ${\cal
    H}=\ran\left(A-zI\right)\dot{+}\,\text{Span}\left\{\mu\right\}$
  for all $z\in\mathbb{C}$.
\end{definition}

Notice that if $A$ is entire, $\xi(z)$ is a vector-valued
entire function and $\widehat\cH_\mu$ is a Hilbert space of entire
functions.

\begin{definition}\label{def:de-branges}
A Hilbert space of entire functions is called a {\em de Branges space} if,
for every $f(z)$ in that space, the following conditions holds:
\begin{enumerate}[(i)]
\item for every non real zero $w$ of $f(z)$, the function
        $f(z)(z-\cc{w})(z-w)^{-1}$ belongs to the Hilbert space
        and has the same norm as $f(z)$,
\item the function $f^*(z):=\cc{f(\cc{z})}$ belongs to the Hilbert space
        and also has the same norm as $f(z)$;
\end{enumerate}
and furthermore,
\begin{enumerate}[(i)]
\item[(iii)] for every $w:\im w\neq 0$, the linear functional
        $f(\cdot)\mapsto f(w)$  is continuous.
\end{enumerate}
\end{definition}

There is an extensive literature concerning the properties of de Branges
spaces. We refer to \cite{debranges} for more details.

It can be shown that
$\widehat{\cal H}_\mu$ is a de Branges space for certain choices of the
entire gauge $\mu$.
There are some evidence indicating that Krein noticed this fact
\cite[pag. 209]{R1466698}. Also, some hints supporting this assertion
has been given by de Branges himself
\cite{debranges1}. We however could not find any formal proof of this
statement. Thus, for the sake of completeness and for the lack of a
proper reference, we provide a proof below. In any case, we do not claim
any originality regarding this matter.

\begin{remark}
\label{rem:real-entire-gauge}
As a consequence of \cite[Lemma 2.7.1]{R1466698}, given any
self-adjoint extension $A_\sharp$ of an operator
$A\in\text{Sym}^{(1,1)}_\text{R}({\cal H})$, one can always
find a complex conjugation $C$ for which $A_\sharp$ is real.
If follows from the proof of the cited lemma that
$C\psi(z)=\psi(\cc{z})$ when $\psi(z)$ is written in terms of
the real self-adjoint extension.  Moreover, by \cite[Theorem
2.7.1]{R1466698}, the operator $A$ is also real with respect
to $C$. Since by \cite[Corollary 2.5]{MR1627806} all the
self-adjoint extensions of $A$ are real, it follows that
$C\psi(z)=\psi(\cc{z})$ for every realization of $\psi(z)$.

If furthermore $A$ is entire, then an entire gauge $\mu$ may
be chosen such that $C\mu=\mu$ (see \cite[Theorem 1]{krein1} and
also \cite[Section 2.7.7]{R1466698}).
\end{remark}

\begin{proposition}
  Assume that an entire operator $A$ is real with respect to
  some complex conjugation $C$ and let $\mu$ be a real entire
  gauge. Then the associated Hilbert space $\widehat{\cal
    H}_\mu$ is a de Branges space.
\end{proposition}
\begin{proof}
In view of Propositions~\ref{prop-cond-1} and
\ref{prop-cond-2} we only have to verify (ii).
By Remark~\ref{rem:real-entire-gauge} we know that
$C\psi(\cc{z})=\psi(z)$ thence $C\xi(z)=\xi(\cc{z})$.
Now consider any $f(z)=\inner{\xi(z)}{\varphi}$.  Clearly
$f^*(z):=\inner{\xi(z)}{C\varphi}$ also belongs to
$\widehat{\cH}_\mu$.  Furthermore,
\[
f^*(z)=\inner{\xi(z)}{C\varphi}=\cc{\inner{C\xi(z)}{\varphi}}
      =\cc{\inner{\xi(\cc{z})}{\varphi}}=\cc{f(\cc{z})}.
\]
Since $C$ is an isometry, the equality of norms follows.
\end{proof}

Notice that we only have used the simultaneous reality of the
entire operator and its entire gauge $\mu$ in showing that
$\widehat{\cH}_\mu$ obeys (ii) of
Definition~\ref{def:de-branges}.  Indeed, this condition is
also necessary.

\begin{proposition}
\label{prop-necessary}
If $\widehat{\cH}_\mu$ is a de Branges space there is a
complex conjugation $C$ with respect to which both $A$ and
$\mu$ are real.
\end{proposition}
\begin{proof}
  Let $\widehat{C}:\widehat{\cH}_\mu\to\widehat{\cH}_\mu$ be
  defined by $(\widehat{C}f)(z)=\cc{f(\cc{z})}$, for every
  $f(z)\in\widehat{\cal H}$.  Clearly, $\widehat{C}$ is a
  complex conjugation. Moreover, the self-adjoint
  multiplication operator $\widehat{A}$, defined on the
  maximal domain in $\widehat{\cH}$ by
  $(\widehat{A}f)(z)=zf(z)$, is real with respect to
  $\widehat{C}$. Now define
  $C:=\Phi_\mu^{-1}\widehat{C}\Phi_\mu$. By construction, $C$
  is a complex conjugation in $\cH$. Since
  $(\widehat{C}\widehat{\mu})(z)\equiv
  1\equiv\widehat{\mu}(z)$, it follows that $\mu$ is real with
  respect to $C$. Finally, it not difficult to see that
  $\Phi_\mu^{-1}\widehat{A}\Phi_\mu$ is a self-adjoint
  extension of $A$ real with respect to $C$. The
  considerations of Remark~\ref{rem:real-entire-gauge}
  complete the proof.
\end{proof}

\section{Examples}

\begin{example}
Consider the semi-infinite Jacobi matrix
\begin{equation}
  \label{eq:jm}
  \begin{pmatrix}
  q_1 & b_1 & 0  &  0  &  \cdots \\[1mm]
  b_1 & q_2 & b_2 & 0 & \cdots \\[1mm]
  0  &  b_2  & q_3  & b_3 &  \\
  0 & 0 & b_3 & q_4 & \ddots\\
  \vdots & \vdots &  & \ddots & \ddots
  \end{pmatrix}\,,
\end{equation}
where $b_k>0$ and $q_k\in\mathbb{R}$ for $k\in\mathbb{N}$. Fix
an orthonormal basis $\{\delta_k\}_{k\in\mathbb{N}}$ in
$\cH$. Let $J$ be the operator in $\cH$ whose matrix
representation with respect to $\{\delta_k\}_{k\in\mathbb{N}}$
is (\ref{eq:jm}). Thus, $J$ is the minimal closed operator
satisfying
\begin{equation*}
  \langle\delta_n,J\delta_n\rangle=q_n\,,\quad
 \langle\delta_{n+1},J\delta_n\rangle=\langle\delta_n,J\delta_{n+1}\rangle
  =b_n\,,\quad\forall n\in\mathbb{N}\,.
\end{equation*}
(Consult \cite{MR1255973} for a discussion on matrix
representation of unbounded symmetric operators.) It is well
known that $J$ may have only deficiency indices $(1,1)$ or
$(0,0)$ \cite{MR0184042}. A classical result is that if $J$
has deficiency indices $(1,1)$, then the orthogonal
polynomials of the first kind $P_k(z)$ associated with
(\ref{eq:jm}) are such that
\begin{equation*}
  \sum_{k=0}^\infty\abs{P_k(z)}^2<\infty
\end{equation*}
uniformly in any compact domain of the complex plane
\cite{MR0184042} .
Therefore, for any $z\in\mathbb{C}$,
$\pi(z)=\sum_{k=1}^\infty P_{k-1}(z)\delta_k$ is in $\cH$.
By construction, $\pi(z)$ is in the one-dimensional space $\Ker
(J^*-zI)$. It is also known that, when the deficiency indices are (1,1),
$J$ is an entire operator and $\delta_1$ is an entire gauge for
$J$ \cite[Section 3.1.1 and Theorem 3.1.2]{R1466698}.

Let us find $\xi(z)$ for the operator $J$. Taking into account
(\ref{thetrick}), $\inner{\delta_1}{\xi(z)}=1$ and
$\inner{\delta_1}{\pi(\cc{z})}=1$ for all $z\in\mathbb{C}$.
Then, since both $\pi(\cc{z})$ and $\xi(z)$ are in $\Ker(J^*-\cc{z}I)$ and
$\delta_1$ is entire, $\pi(\cc{z})=\xi(z)$ for all
$z\in\mathbb{C}$. Thus, for any $\varphi$ in $\cH$ we have
$\varphi=\sum_{k=1}^\infty\varphi_k\delta_k$ and
$\widehat{\varphi}(z)\in\widehat\cH_{\delta_1}$ is then given by
\begin{equation*}
  \widehat\varphi(z):=\langle \pi(\cc{z}),\varphi\rangle=\sum_{k=1}^\infty
    P_k(z)\varphi_k\,,\quad z\in\mathbb{C}\,.
\end{equation*}

Clearly, if $\widehat\varphi(z)\in\widehat\cH_{\delta_1}$,
$\cc{\widehat\varphi(\cc{z})}\in\widehat\cH_{\delta_1}$.
Whence, in virtue of Propositions~\ref{prop-cond-1} and
\ref{prop-cond-2}, our space $\widehat\cH_{\delta_1}$ is a de
Branges space and then, by Proposition~\ref{prop-necessary},
$\delta_1$ is real with respect to
$C=\Phi_{\delta_1}^{-1}\widehat{C}\Phi_{\delta_1}$
($\widehat{C}$ is the conjugation in $\widehat\cH_{\delta_1}$
given in the proof of Proposition~\ref{prop-necessary}).

Taking into account that $\norm{\pi(\cc{z})}\equiv 1$, formula
(\ref{sampling}) is written in this case as
\begin{equation*}
  \begin{split}
f(z)&=\sum_{x_n\in \Sp\left(J_\sharp\right)}
    \langle \pi(\cc{z})\pi(x_n)\rangle f(x_n)\\
&=\sum_{x_n\in \Sp\left(J_\sharp\right)}f(x_n)
\sum_{k=1}^\infty P_k(z)
  P_k(x_n)\,,\quad z\in\mathbb{C}\,,
\end{split}
\end{equation*}
where $J_\sharp$ is certain self-adjoint extension of $J$.

In a different setting, sampling formulas obtained on the basis of
Jacobi operators have been studied before \cite{garcia00,garcia0}.
\end{example}

\begin{example}
The entire operator used here has been taken from
\cite{R1466698} and is a particular case of an example
given by Krein in \cite{krein4}.

Consider a non-decreasing bounded function $s(t)$ such that
\begin{equation*}
  s(-\infty)=0\quad\text{and}\quad s(t-0)=s(t)\,.
\end{equation*}
Fix a function defined for any $x$ in the real interval
$(-a,a)$ by
\begin{equation*}
  F(x):=\int_{-\infty}^\infty e^{ixt}ds(t)\,.
\end{equation*}
In the linear space $\widetilde{\cal L}$ of continuous
functions in $[0,a)$ vanishing in some left neighborhood of
$a$, we define a sesquilinear form as follows
\begin{equation}
\label{eq:quasiinner-product}
  (g,f):=\int_{0}^a\int_{0}^a
  F(x-t)f(x)\cc{g(t)}dxdt\,.
\end{equation}
This form is a quasi-scalar product, i.\,e., the existence of
elements $f$ in $\widetilde{\cal L}$ such that
$f\ne 0$ and nevertheless $(f,f)=0$ is not
excluded.

Denote by $\cal D$ the set of  continuously differentiable functions
$f\in\widetilde{\cal L}$ such that $f(0)=0$ and define in
$\cal D$ the differential operator $\widetilde{A}$ by the rule
$\widetilde{A}f:=i\partial_xf$. It is not difficult to show that
$(g,\widetilde{A}f)=(\widetilde{A}g,f)$ and $\cal D$ is
quasi-dense in $\widetilde{\cal L}$. Now, proceeding as in
\cite[Section 2.8.2]{R1466698}, one defines the space $\cal{L}$
as follows
\begin{equation*}
  {\cal L}=\widetilde{\cal{L}}\setminus\widetilde{0}\,,
  \qquad\widetilde{0}=\{f\in\widetilde{\cal L}: (f,f)=0\}\,.
\end{equation*}
In $\cal{L}$ we define an inner product by
\begin{equation}
  \label{eq:inner-product}
  \inner{\eta}{\varphi}:=(g,f)\,,
\end{equation}
where $\varphi$ and $\eta$ are equivalence classes containing
$f$ and $g$, respectively. Let $\cH$ be the completion of
$\cal{L}$ and consider in it the operator $A$ such that, for
the equivalence class $\varphi$ containing $f\in\cal D$,
$A\varphi$ is the equivalence class containing
$\widetilde{A}f$. It can be shown
that $A$ is an entire operator and that
\begin{equation*}
  \widehat{\varphi}(z)=\inner{\xi(z)}{\varphi}=\int_{0}^ae^{izt}f(t)dt\,,
\end{equation*}
where $f\in\varphi$.  This identity, together with
(\ref{eq:quasiinner-product}) and (\ref{eq:inner-product}),
determines $\xi(z)$ completely \cite[Section 3.2.2]{R1466698}.

Notice that, in this example, the entire gauge associated to
$\xi(z)$ remains unknown. This is not however an issue since
the sampling kernel can be computed anyway by resorting to
expression (\ref{gz}).
\end{example}

\begin{acknowledgments}
We express our gratitude to Miguel Ballesteros for drawing our
attention to the results of \cite{kempf1}.
\end{acknowledgments}


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