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\title[Solvability criterion for multiple interpolation  problem]{Solvability criterion for multiple interpolation  problem  in  preimage of  convolution operator}


\author{V.V. Napalkov (jr.),  A.A. Nuyatov}

\address{Valerii Valentinovich Napalkov,
\newline\hphantom{iii} Institute of Mathematics,
\newline\hphantom{iii} Ufa Federal Research Center, RAS
\newline\hphantom{iii} Chernyshevsky str.  112,
\newline\hphantom{iii} 450008, Ufa, Russia}
\email{vnap@mail.ru}

\address{Andrey Alexandrovich Nuyatov,
\newline\hphantom{iii} Nizhny Novgorod State Technical University
\newline\hphantom{iii} named after R.E. Alekseev,
\newline\hphantom{iii} Minin str. 24,
\newline\hphantom{iii} 603155, Nizhny Novgorod, Russia}
\email{nuyatov1aa@rambler.ru}


\thanks{\sc V.V. Napalkov,  A.A. Nuyatov, Solvability criterion for multiple interpolation  problem  in  preimage of  convolution operator.}
\thanks{\copyright \ Napalkov V.V.,  Nuyatov A.A.  \ 2026}
\thanks{\rm The research by V.V. Napalkov (jr.) is supported by the Ministry of Science and Higher Education  in the framework of state task (code of scientific theme FMRS-2025-0010).}
\thanks{\it Submitted November 30,  2025.}

\maketitle {\small
\begin{quote}
\noindent{\bf Abstract.} In this paper we find a solvability criterion for the multiple interpolation problem in the preimage of a convolution operator and, consequently, a criterion for the solvability of   Abel~---~Goncharov problem in the same space. When the kernel of the operator serves as the preimage, uniqueness of the solution to these problems    holds provided  the set of interpolation nodes is the uniqueness set in the kernel of the convolution operator.
\medskip

\noindent{\bf Keywords:}  multiple interpolation, Abel  problem,
convolution operator, entire functions.

\medskip

\noindent{\bf Mathematics Subject Classification: }{46A13, 30D20}

\end{quote}

\vskip20pt

\section{Introduction}
Let $H(\CC)$ be the space of entire functions with the topology
of uniform convergence on compact sets; we denote the dual of the space
$H(\CC)$ by $H^*(\CC)$. We also introduce the notation
$$P_{\CC} = \{f(z)\in H(\CC): \exists\, c_1, c_2>0, |f(z)|\leq c_1
e^{c_2 |z|}\}.$$ With a function $\varphi(z)\in P_{\CC}$ we associate
a functional $F\in H^*(\CC)$ such that
$\widehat{F}(z)=\varphi(z)$, where $\widehat{F}(z)=\langle
F_{\lambda}, e^{\lambda z}\rangle$ is the Laplace transform
of the functional $F$.

 The convolution operator  acting from $H(\CC)$ to
$H(\CC)$  with the characteristic function $\varphi(z)\in P_{\CC}$
reads
\begin{equation}\label{N0}
M_{\varphi}[f](z)=\frac{1}{2\pi i}\int\limits_C f(z+t)\gamma(t)dt,
\end{equation}
where $C$ is a closed contour, $\gamma(t)$ is an analytic function on
$C$ and outside $C$, $\gamma(\infty)=0$.

We take and fix  an arbitrary  function $g_0(z)\in H(\CC)$ and consider the convolution equation
$$
M_{\varphi}[f](z)=g_0(z),
$$
We denote by $\IM^{-1}M_{\varphi}[f]$  the preimage of the convolution operator.
As   $g_0(z)\equiv 0$, the preimage of the convolution operator
becomes the kernel $\KER M_{\varphi}[f]$.


For an arbitrary function $\psi(z)\in H(\CC)$  we define the ideal in $H(\CC)$
$$
(\psi)=\{\psi(z)\cdot R(z):\  R(z)\in H(\CC)\}.
$$
\begin{definition}
The identities
\begin{align}\label{N1}
&H(\CC)=(\psi)+\IM^{-1}M_{\varphi},\\
\label{N1.1}
&H(\CC)=(\psi)\oplus \IM^{-1}M_{\varphi},
\end{align}
are called the Fisher representation and Fisher decomposition for the preimage of
$M_{\varphi}$ in $H(\CC)$, respectively.
\end{definition}

When identities (\ref{N1}) hold, each entire function can be represented, generally speaking, non--uniquely as
$$
f(z)=h(z)+g(z),\qquad g(z)\in \IM^{-1}M_{\varphi},\qquad h(z)\in (\psi).
$$

Let us provide an example when the characteristic function $\varphi(z)$
of the operator $M_{\varphi}$ is the polynomial $P(z)=z^2+1$, then the convolution operator
becomes a differential operator with constant
coefficients $$P\left(\frac{d}{dz}\right)=\frac{d^2}{d\,z^2}+1,$$
and as a function from the ideal we take the polynomial $Q(z)=z+1$.


We consider the differential equation
$$
\frac{d^2f(z)}{d\,z^2}+f(z)=e^z,
$$
the function from the preimage is $$f(z)=c_1e^{i\,z}+c_2e^{-i\,z}+\frac{1}{2} e^z.$$
Then the Fisher representation reads
$$H(\CC)=(Q)+c_1e^{i\,z}+c_2e^{-i\,z}+\frac{1}{2} e^z.$$
This means that for each function $f(z)\in H(\CC)$ we have
$$
\frac{f(z)-(c_1e^{i\,z}+c_2e^{-i\,z}+\frac{1}{2} e^z)}{z+1}\in H(\CC),
$$
hence, as the zero of the polynomial $z+1$ we have the identity
$$
f(-1)=c_1e^{-i}+c_2e^{i}+\frac{1}{2} e^{-1}
$$
Hence, the function  $f(z)$ is defined but not uniquely.

As identities (\ref{N1.1}) hold, then such a representation of the entire
function is unique.

  The multiple interpolation problem
in $\IM^{-1}M_{\varphi}$ with nodes $\mu_j\in\CC$ being the zeroes of $\psi\in H(\CC)$,
with multiplicities
$q_j$, $j=0,1,2,\ldots$ is formulated as follows: given
an arbitrary sequence of complex numbers
$a^k_j$, $j=0,1,2,\ldots;$, $k=0,1,\ldots, q_k-1$, whether there exists a function $y\in\IM^{-1}M_{\varphi}$ such that
$$y^{(k)}(\mu_j)=a^k_j, \quad j=0,1,2,\ldots; \quad k=0,1,\ldots, q_k-1.$$


\begin{lemma}\label{lemma1}
The solvability of the multiple interpolation problem is equivalent to the validity of the  Fisher representation
$$
H(\CC)=(\psi)+\IM^{-1}M_{\varphi}.
$$
\end{lemma}

\begin{proof}
1. Let   Fisher representation (\ref{N1}) hold.   Mittag~---~Leffler theorem and the Weierstrass theorem on the existence of entire functions with given zeros imply the existence of a function $h(z)\in H(\CC)$ such that $h^{(k)}(\mu_j)=a^k_j,$ $j=0,1,2,\ldots;$ $k=0,1,\ldots, q_k-1$. By (\ref{N1})
$$
h(z)=y(z)+\psi(z)\cdot g(z), \qquad y(z)\in \IM^{-1}M_{\varphi},\quad g(z)\in H(\CC),
$$
and this implies the identities  $y^{(k)}(\mu_j)=a^k_j,$ $j=0,1,2,\ldots;$ $k=0,1,\ldots, q_k-1$.
We note that if (\ref{N1.1}) holds, then the function $y(z)$ is unique.

2. Let the multiple interpolation problem be solvable. The validity of  the Fisher representation holds can be written in an another way: for each function $h(z)\in H(\CC)$, there exists a function $y(z)\in \IM^{-1}M_{\varphi}$ such that
$$
\dfrac{h(z)-y(z)}{\psi(z)}\in H(\CC).
$$
The latter holds if $h(z)-y(z)$ vanishes at the zeros of the function $\psi(z)$ (otherwise the function $\dfrac{h(z)-y(z)}{\psi(z)}$ would be meromorphic).

Since  by the assumption the problem of multiple interpolation is solvable in the preimage of  convolution operator and there exists a function $h(z)\in H(\CC)$ such that $h^{(k)}(\mu_j)=a^k_j,$ $j=0,1,2,\ldots;$ $k=0,1,\ldots,q_k-1$,  we have
$$
h^{(k)}(\mu_j)=a^k_j= y^{(k)}(\mu_j), \quad j=0,1,2,\ldots;\quad k=0,1,\ldots, q_k-1,
$$
where $\mu_j$, $j=0,1,2,\ldots$  the zeros of the function $\psi(z)$.
Thus,  Fisher representation (\ref{N1}) holds. Note that if the problem of multiple interpolation in the preimage of  convolution operator has a unique solution, then (\ref{N1.1}) holds. The proof is complete.
\end{proof}

A particular case of the  considered problem under is
$$
y^{(k)}(\mu_j)=a_j^k,\quad j=0,1,2,\ldots;\quad k=\overline{0,j}.
$$
Hence,   the preimage of the convolution operator contains a function $y(z)$,  which for the sequence of complex numbers $a_0^0$, $a_1^1$, \ldots, $a_n^n$, \ldots  satisfies
$$
y^{(k)}(\mu_k)=a_k^k.
$$
Thus, we obtain the Abel problem   \cite{MathEncycl}  in the preimage of  convolution operator.

When the characteristic function of   convolution operator is a polynomial, the convolution operator becomes a finite order linear differential operator with constant coefficients. Therefore, an important particular consequence is that for a finite order homogeneous linear differential equation with constant coefficients  the multiple interpolation problem and the Abel~---~Goncharov problem are solvable. Furthermore, a differential--difference operator, an integro--differential operator, and a linear differential operator of infinite order with constant coefficients are also particular cases of the convolution operator, and, as a consequence, these problems are solvable for the corresponding homogeneous equations.

 In   \cite{NapalkovNuyatov}, the interpolation problem was solved for the case where
the nodes are simple and lie on the real axis. In
\cite{NapalkovNuyatovTMPh}, the interpolation problem in the kernel
of the convolution operator was solved with complex nodes. In   \cite{NapalkovNuyatovUMJ}, the problem of multiple interpolation in the kernel of  convolution operator was solved.
The aim of this paper is to find
conditions under which the problem of multiple interpolation in the preimage
of   convolution operator is solvable.

\vskip20pt

        \section{Preliminaries}
%-----------------------------------------------------

Along with the operator $M_{\varphi}[f](z)$, we introduce a linear
 operator  continuous in the topology of   space $H(\CC)$
\begin{equation}\label{N2.1}
M_{\varphi}[\psi(z)\cdot y(z)]+g_0(z):\ H(\CC)\rightarrow H(\CC),
\end{equation}
as above, $g_0(z)$ is a function from the image  of operator
$M_{\varphi}[f](z)$.

\begin{theorem}\label{T1}
Fisher representation in $H(\CC)$ holds
if and only if   operator (\ref{N2.1}) is surjective.
\end{theorem}

\begin{proof}
1. Let identity (\ref{N1}) hold and $g(z)\in H(\CC)$.
Since   convolution operator (\ref{N0}) is surjective
\cite{NapalkovBook},  for $g(z)$ there exists $f(z)\in H(\CC):\,
M_{\varphi}[f]=g(z)$. According to (\ref{N1}), the function $f(z)$ can
be represented as
$$f(z)=\psi(z)\cdot h(z)+u(z), \,
h(z)\in H(\CC),\,u(z)\in \IM^{-1}M_{\varphi},$$ this is why
$$M_{\varphi}[f]=M_{\varphi}[\psi(z)\cdot
h(z)]+M_{\varphi}[u(z)]=M_{\varphi}[\psi(z)\cdot
h(z)]+g_0(z)=g(z),$$ and we see that the operator
$M_{\varphi}[\psi(z)\cdot h(z)]+g_0(z)$ is surjective.

2. Let operator (\ref{N2.1}) be surjective. We need to prove identity  (\ref{N1}). Since operator (\ref{N2.1}) and the convolution operator
are surjective, then for each $f(z)\in H(\CC)$ there exist
functions $w_1(z)\in \IM^{-1}M_{\varphi}, y_1(z)\in H(\CC)$
satisfying the identity
$$
M_{\varphi}[f(z)]=M_{\varphi}[\psi(z)\cdot
y_1(z)]+g_0(z)=M_{\varphi}[\psi(z)\cdot y_1(z)]+M_{\varphi}[w_1(z)],
$$
and this is why
$$
f(z)=w_1(z)+\psi(z)\cdot y_1(z).
$$
This proves Fisher representation   (\ref{N1}). The proof is complete.
\end{proof}

We denote by $N_{\varphi}$ and $N_{\psi}$  the zero sets
of  functions $\varphi(z)$ and $\psi(z)$. Let us formulate the injectivity conditions
for the operator $M_{\varphi}[\psi\,\cdot\,]+g_0(z)$.


\begin{theorem}\label{T2}
If two conditions hold

\vskip5pt

\begin{enumerate}
	\item[1.] $\exists\, h_0(z)\in H(\CC): h_0(z)\in (\psi)\bigcap \{f\in H(\CC): M_{\varphi}[f]=g_0\};$

\vskip5pt

	\item[2.] $N_{\psi}$ is the uniqueness set in $\KER M_{\varphi}$,
\end{enumerate}
then the operator  $M_{\varphi}[\psi\cdot ]+g_0(z)$ is injective.
\end{theorem}

\begin{proof} According to Assertion~1, $\exists\hat{h}_0\in
H(\CC): h_0(z)=\hat{h}_0(z)\cdot\psi(z)$, and
$$
M_{\varphi}[\psi\cdot f ]+g_0(z)=M_{\varphi}[\psi\cdot f
]+M_{\varphi}[h_0]=M_{\varphi}[\psi(f+\hat{h}_0)].
$$
We consider the equation
$$
M_{\varphi}[\psi(z)(f(z)+\hat{h}_0(z))]=0.
$$
The function $u(z)=\psi(z)(f(z)+\hat{h}_0(z))$ vanishes at the points
of the set $N_{\psi}$, which by assumption is the uniqueness set
in $\KER M_{\varphi}$, therefore $u(z)\equiv 0$. Since for a linear operator the injectivity is equivalent
to the triviality of its kernel, the operator $M_{\varphi}[\psi\,\cdot\, ]+g_0(z)$ is injective. The proof is complete.
\end{proof}

\begin{corollary}\label{corT2}
If $N_{\psi}$ is the uniqueness set in $\KER M_{\varphi}$,
then the operator $M_{\varphi}[\psi\,\cdot\,]$ is injective.
\end{corollary}

By Theorem~\ref{T1}, the surjectivity of   operator $M_{\varphi}[\psi\,\cdot\,]+g_0(z)$
is equivalent to the Fisher representation
$$H(\CC)=(\psi)+ \IM^{-1}M_{\varphi}.$$
It is obvious that for the   Fisher decomposition requites the injectivity of the operator $M_{\varphi}[\psi\,\cdot\,]+g_0(z)$, and this  injectivity is equivalent to
$$
(\psi)\cap \{f\in H(\CC): M_{\varphi}[f]=g_0\}=\{0\}
$$
since the direct sum of two linear subspaces is possible only
if the subspaces intersect at zero. Therefore, by
Theorem~\ref{T2}, generally speaking, the Fisher decomposition does not exist, but
the Fisher representation  exists, i.e., the interpolation problem in
the preimage of   convolution operator is solvable  but not uniquely.
Note that by Corollary \ref{corT2}, in the case where
$N_{\psi}$ is the uniqueness set in $\KER M_{\varphi}$ the
 decomposition exists
$$H(\CC)=(\psi)\oplus \KER M_{\varphi}.$$
%-----------------------------------------------------------------------------

\vskip20pt

\section{Convolution operator in space $P_{\mathds{C}}$}
%-----------------------------------------------------------------------------
According to the results of   \cite{Dickson}, \cite{Muggle}
the function $\psi\in H(\CC)$ generates
a surjective convolution operator $M_{\psi}: P_{\CC}\rightarrow P_{\CC}$
in the space $P_{\CC}$
\begin{equation}\label{N3.1}
M_{\psi}[f](z)=\frac{1}{2\pi i}\int\limits_C \psi(t)\gamma(t)e^{z
t}\,dt,
\end{equation}
where $\gamma(t)$ is the function associated with $f(z)$ in the Borel sense and $C$ is a closed contour enclosing all singular points of $\gamma(t)$.

We arbitrarily choose and fix the function $G_0(z)\in P_{\CC}$, and
consider the equation
$$
M_{\psi}[f]=G_0(z).
$$

Let $(\varphi)=\{\varphi(z)\cdot G(z):G(z)\in P_{\CC}\}$ be an ideal in the space $P_{\CC}$, and $\IM^{-1}M_{\psi}[f]$ be the preimage of  operator (\ref{N3.1}).
\begin{definition} Identities
\begin{align*}
&P_{\CC}=(\varphi)+\IM^{-1}M_{\psi},\\
&P_{\CC}=(\varphi)\oplus \IM^{-1}M_{\psi},
\end{align*}
are call the Fisher  representation and Fisher decomposition for the preimage of $M_{\psi}$ in $P_{\CC}$, respectively.
\end{definition}

We introduce  the linear  continuous operator
\begin{equation}\label{N5}
M_{\psi}[\varphi(z)\cdot y(z)]+G_0(z):\ P_{\CC}\rightarrow P_{\CC}.
\end{equation}

Let us formulate an analogue of   Theorem~\ref{T1} for operator (\ref{N5}).


\begin{theorem}\label{T3}
  Fisher representation holds in $P_{\CC}$ if and only if
 operator (\ref{N5}) is surjective.
\end{theorem}

The proof of Theorem~\ref{T3} is similar to the proof of Theorem~\ref{T1}  since for each $g(z)\in P_{\CC}$ there exists $f(z)\in
P_{\CC}: M_{\psi}[f](z)=g(z)$.

\begin{theorem}
If two conditions hold

\vskip5pt

\begin{enumerate}
	\item[1.]  $\exists\, h_0(z)\in P_{\CC}: h_0(z)\in (\varphi)\bigcap \{f\in P_{\CC}: M_{\psi}[f]=G_0\};$

\vskip5pt

\item[2.] $N_{\varphi}$ is the uniqueness set in $\KER M_{\psi}$,

\vskip5pt
\end{enumerate}
then the operator $M_{\psi}[\varphi\cdot ]+G_0(z)$ is injective.
\end{theorem}

The proof is similar to the proof of Theorem~\ref{T2}.

\begin{corollary}
If $N_{\varphi}$ is a uniqueness set in $\KER M_{\psi}$, then the operator $M_{\psi}[\varphi\,\cdot\,]$ is injective.
\end{corollary}
By analogy with the space $H(\CC)$, generally speaking, there is uniqueness in the interpolation problem in the preimage of operator $M_{\psi}$, but
under certain conditions the decomposition
$$
P_{\CC}=(\varphi)\oplus \KER M_{\psi}
$$
holds. We rewrite  operator (\ref{N2.1}) as
$$
M_{\varphi}[\psi\cdot f+f_0].
$$
Here $\psi$ and $f_0$ are fixed
entire functions, and $f$ ``ranges'' over the entire space $H(\CC)$.
Since the operator $M_{\varphi}[\psi\cdot f+f_0]$ linearly and
continuously maps $H(\CC)$ into $H(\CC)$, the adjoint operator
$M^*_{\varphi}[\psi\cdot f+f_0]$ linearly and continuously maps
the space $H^*(\CC)$ into $H^*(\CC)$. Since the spaces
$H^*(\CC)$ and $P_{\CC}$ are topologically isomorphic
\cite{NapalkovBook}, the operator $M^*_{\varphi}[\psi\cdot f+f_0]$
generates a linear  continuous operator acting from $P_{\CC}$
to $P_{\CC}$ by rule: for $G(z)\in P_{\CC}$
\begin{equation}%\label{N5}
\widehat{M}^*_{\varphi}[G]=M_{\psi}[\varphi\cdot G],
\end{equation}
where $M_{\psi}$ is an operator of   form (\ref{N3.1}). Since the space
$H(\CC)$ is a Fr\'echet space, we apply the result of
\cite{DieudonneSchwartz} (or \cite[Thm. 1.2]{Brauder}) and  obtain the following theorem.
\begin{theorem}\label{T5}
The statements hold

\vskip5pt

\begin{enumerate}
  \item  $\IM (M_{\varphi}[\psi\,\cdot\,]+g_0)$ is closed
in $H(\CC)$ if and only if $\IM M_{\psi}[\varphi\,\cdot\,]$ is closed
in $P_{\CC}$;

\vskip5pt

  \item $\IM(M_{\varphi}[\psi\,\cdot\,]+g_0)$ is everywhere dense in
$H(\CC)$ if and only if
$M_{\psi}[\varphi\,\cdot\,]$ is
injective;

\vskip5pt

  \item $M_{\varphi}[\psi\,\cdot\,]+g_0$ is injective
if and only if $\IM M_{\psi}[\varphi\,\cdot\,]$ is dense in $P_{\CC}$.
\end{enumerate}
\end{theorem}


By Theorems~\ref{T1}, \ref{T3} and~\ref{T5}
we obtain the next statement.
\begin{corollary}\label{C3}
$$H(\CC)=(\psi)\oplus \KER M_{\varphi}\Leftrightarrow
P_{\CC}=(\varphi)\oplus \KER M_{\psi},$$ where  $\KER M_{\psi}$ is the kernel of operator (\ref{N3.1}).
\end{corollary}

\vskip20pt

\section{Sufficiency sets}
%-----------------------------------------------------------

The topology $\tau_{\mathds C}$ of the space $P_{\mathds C}$ is defined as the
inductive limit of normed weighted spaces
$$B_n=\{\varphi(\lambda)\in P_{\mathds C}:\|\varphi\|_n=\sup
\limits_{\lambda\in\mathds
C}|\varphi(\lambda)|e^{-n|\lambda|}<\infty \},\quad n\in{\bf N}. $$


Let $S\subset\mathds C$ be the uniqueness set in $P_{\mathds C}$. Then in $P_{\mathds C}$ we can introduce the topology $\tau_S$ of the inductive limit
of spaces
$$ B_{n,S}=\{\varphi(\lambda)\in P_{\mathds C}:\|\varphi\|_{n,S}=\sup
\limits_{\lambda \in S}|\varphi(\lambda)|e^{-n|\lambda|}<\infty
\},\quad n\in{\bf N}.
$$
In what follows, we need convergence to zero in the topology
$\tau_{\CC}$   \cite{SebashtianISilva}: let $f_m$ be a countable
sequence of functions from $P_{\CC}$, then $f_m\rightarrow0$
for $m\rightarrow \infty$ in the topology $\tau_{\CC}$ if and only if
there exist numbers $\sigma>0$ and $M>0$ such that

\vskip5pt

\begin{enumerate}
\item[(a.1)] $|f_m(z)|\leq Me^{\sigma |z|},$ $\forall m\in\mathds{N},$ $\forall z\in \CC$;

    \vskip5pt

\item[(b.1)] for each compact set $K_{\CC}\subset \CC$ we have $|f_m(z)|\rightrightarrows 0 $ as  $m\rightarrow\infty$, $z\in K_{\CC}$.
\end{enumerate}

Let us introduce the concept of sufficiency of a set $S\subset \CC$ in $U\subset P_{\CC}$ with the topology induced from $P_{\CC}$.

\begin{definition}
 We  say that $S$ is a sufficient set on $U$  the conditions

\vskip5pt
 \begin{enumerate}
 \item[(a.2)]  for each sequence of functions $q_k(z)\in U$ there exist numbers $\sigma>0$ and $M>0$ such that  $|q_k(z)|\leq M e^{\sigma |z|}, $ $ \forall k\in \mathds{N},$ $ \forall z\in S$;

     \vskip5pt
 \item[(b.2)] for each compact set  $K_S\subset S$ we have $ |q_k(z)|\rightrightarrows 0$ as $k\rightarrow \infty$, $z\in K_S;$
 \end{enumerate}

 \vskip5pt

\noindent imply the convergence of this sequence on $U.$
\end{definition}
Conditions (a.2) and (b.2) define convergence to zero in the topology $\tau_S$.


 The next theorem was proved in
  \cite[Thm. 6]{NapalkovNuyatov}.

\begin{theorem}
If $N_{\varphi}$ is a sufficient set in $\KER
M_{\psi}$, then the operator $M_{\psi}[\varphi\,\cdot\,]$ is injective and
$\IM M_{\psi}[\varphi\,\cdot\,]$ is closed in $P_{\CC}$.
\end{theorem}


Since by Theorem~\ref{T5} the two conditions: $M_{\psi}[\varphi\,\cdot\,]$ is injective and
$\IM M_{\psi}[\varphi\,\cdot\,]$ is closed in $P_{\CC}$  are equivalent to the surjectivity of operator
$M_{\varphi}[\psi\,\cdot\,]+g_0$, we arrive at the next theorem.

\begin{theorem}\label{T7}
Let $\varphi(z)\in P_{\CC}, \psi(z)\in H(\CC)$. If
$N_{\varphi}$ is a sufficient set in $\KER
M_{\psi}$, then the Fisher representation
$$H(\CC)=(\psi)+ \IM^{-1}\,M_{\varphi}$$ holds. Moreover, if
$N_{\psi}$ is a uniqueness set in $\KER M_{\varphi}$, then
the Fisher decomposition
$$H(\CC)=(\psi)\oplus \KER M_{\varphi}$$
holds.
\end{theorem}

Let $N_{\varphi}=\{\lambda_k\}_{k=1}^{+\infty}$
be the zero set of   function
$\varphi\in P_{\mathds C}$, each zero is repeated as many times as its multiplicity (to avoid cumbersome notations, in what follows, by $\lambda_{\tilde{k}}$, $\tilde{k}=1,2\ldots$ we  mean some subsequence of the sequence
$\{\lambda_k\}_{k=1}^{+\infty}$); $N_{\psi}=\{\mu_k\}_{k=1}^{+\infty}$ is the set of zeros of the function $\psi\in H(\mathds C)$, each zero is repeated as many times as its multiplicity; by $q_j$ denote the multiplicity of zero $\mu_j$; $\tilde{N}_{\psi}$ is an infinite set,
which consists of all distinct zeros of the function $\psi\in H(\mathds
C)$.

\begin{theorem}[\cite{NapalkovNuyatovUMJ}]
 Let for some fixed $\alpha\in[0,+\infty)$ there exists a number $\beta\in[0,+\infty)$ such that
$\alpha\cdot\beta<1$  and the following conditions are satisfied:

\vskip5pt

\begin{itemize}
  \item[(a)]$N_{\varphi}\subset
D_{\alpha}=\{z\in\mathds C: |\IM z|\leq\alpha \RE z\}$ and there exists a subsequence $\lambda_{\tilde{k}}$ such that
$$
\RE(\lambda_{\tilde{k}})<\RE(\lambda_{\tilde{k}+1}),\qquad \tilde{k}\in{\mathds N}.
$$

\vskip5pt

  \item[(b)] $N_{\psi}\subset D_{\beta}=\{z\in\mathds C: |\IM z|\leq\beta \RE z\}$, and the elements of  set $\tilde{N}_{\psi}$ satisfy
  \begin{equation*}
\RE(\mu_k)<\frac{1-\alpha\beta}{1+\alpha\beta}\RE(\mu_{k+1}),\quad k\in{\mathds
N}.
\end{equation*}
\end{itemize}
\vskip5pt

Then the set
$N_{\varphi}$ is sufficient in $\KER M_{\psi}$.
\end{theorem}

In \cite{MerzlyakovPopenov} there is an
  example, which  shows that there exist convolution operators
in the considered class such that the problem of interpolation by functions from the
kernel of the convolution operator with arbitrary complex interpolation nodes $\mu_j$, $j=1,2,\ldots$, is, generally speaking, unsolvable. Suppose that
the set of interpolation nodes contains the points $\mu_1\in \mathds R$,
$\mu_2=\mu_1+i\in \CC$, and $\varphi(z)=1-e^{iz}$. Then the problem of
simple interpolation by entire functions from $\KER M_{\varphi}=\{f\in
H(\CC):f(z)=f(z+i)\}$ is unsolvable.

In this example, $\lambda_n=2\pi n\in \mathds R$. All functions from the $\KER M_{\varphi}$ kernel are periodic with period $i$,
therefore, it is not possible to specify arbitrary interpolation
data at the nodes $\mu_1\in \mathds R$, $\mu_2=\mu_1+i\in \CC$.

%-----------------------------------------------------------------------------------------------------------------------------

\vskip20pt

\section{Main result}


Let $N_{\varphi}=\{\lambda_k\}_{k=1}^{s\leq \infty}$ be the zero
set of $\varphi\in P_{\CC}$, $N_{\psi}=\{\mu_k\}_{k=1}^{m\leq \infty}$ be the zero
set of $\psi(z)\in H(\CC)$, each zero is repeated taking into account its multiplicity. Using the result of \cite{Dickson}, \cite{Muggle}, we note that $\KER M_{\psi}$ consists of quasi--polynomials with exponents from the set $N_{\psi}$, i.e., for each $r(z)\in \KER M_{\psi}$ we can write
\begin{equation}\label{N3}
r(z)=\sum\limits_{i=1}^n\sum\limits_{j=0}^{q_i-1}  C_{ij}z^j
e^{\mu_i\cdot z},
\end{equation}
where $q_i$ is the multiplicity of zero $\mu_i$, all coefficients $C_{ij}$,
$i=\overline{1,n}$, $j=\overline{0,q_i-1}$ are non--zero.


\begin{remark}
It is important that identity (\ref{N3}) involves only   monomials
with the exponents $\mu_i\in N_{\psi}$, $i=\overline{1,n}$, belonging to the conjugate diagram $r(z)$.
\end{remark}

 We arrange $\mu_i$, $i=\overline{1,n}$ in order of increasing absolute values. Since there are   finitely many terms in (\ref{N3}), we have
 $$
\left|\mu_i\right|\leq C_{\mu}, \quad \exists\, C_{\mu}>0, \quad i=\overline{1,n}.
$$

\begin{theorem}\label{mainTh}
Let a finite set of complex numbers $\widetilde{N}_{\psi}=\{\mu_i\}_{i=1}^n$ be the set of zeros
$\psi(z)$ located in the conjugate diagram of   function $r(z)\in \KER M_{\psi}[\varphi\,\cdot\,]$, i.e.
$$r(z)=\sum\limits_{i=1}^n\sum\limits_{j=0}^{q_i-1}  C_{ij}z^j
e^{\mu_i\cdot z}.$$
The set $N_{\varphi}$ is  sufficient  in
$\KER M_{\psi}[\varphi\,\cdot\,]$ if and only if
there exist $Q=\sum\limits_{i=1}^n q_i$ elements of the set $N_{\varphi}$ such that
$$
D_{\mu\lambda}=\det\left(\lambda^j_p e^{\mu_i\lambda_p}\right)\neq 0, \quad  i=\overline{1,n}, \quad j=\overline{0,q_i-1}, \quad p=1,\ldots,Q.$$
\end{theorem}
\begin{proof} 1. Let $D_{\mu\lambda}\neq 0$. We take a sequence in the space $\KER M_{\psi}$
$$
r_m(z)=\sum\limits_{i=1}^n\sum\limits_{j=0}^{q_i-1}  C_{ij}(m)z^j e^{\mu_i\cdot z}, \quad m=1,2,\ldots
$$
According to the assumptions of  theorem, there exist $Q$ elements of the set
$N_{\varphi}$ such that
$$
D_{\mu\lambda}=\det\left(\lambda^j_p e^{\mu_i\lambda_p}\right)\neq
0,\quad i=\overline{1,n}, \quad j=\overline{0,q_i-1}, \quad p=\overline{1,Q},
$$
then the coefficients $C_{ij}(m)$  solve the system
$$
\sum\limits_{i=1}^n\sum\limits_{j=0}^{q_i-1}  C_{ij}(m)\lambda_p^j
e^{\mu_i\cdot \lambda_p}=r_m(\lambda_p), \quad p=\overline{1,Q},
$$
and by Cramer  rule they are expressed by formulas
$$
C_{ij}(m)=\frac{\Delta_{l(j,i)}}{D_{\mu\lambda}},
$$
where $\Delta_{l(j,i)}$ is the determinant of the matrix obtained from the matrix
$A=\left(\lambda^j_p e^{\mu_i\lambda_p}\right)$ by replacing the
$l(j,i)$th column by a column of free terms, while
$$
\Delta_{l(j,i)}=\sum\limits_{p=1}^{Q}(-1)^{l(j,i)+p}\det
A_{l(j,i),p}r_m(\lambda_p).
$$

We arrange
$\lambda_p$, $p=\overline{1,Q}$ in order of increasing absolute values, then
$$
\left|\lambda_p\right|\leq C_{\lambda}, \quad \exists\, C_{\lambda}>0, \quad
p=\overline{1,Q}.
$$


Let us show that all
coefficients $C_{ij}(m)$, $i=\overline{1,n}$, $j=\overline{0,q_i-1},$
are bounded. We are going to estimate the determinants $\Delta_{l(j,i)}$; in order to do this, by using (a.2), we estimate $r_m(\lambda_p)$, $p=\overline{1,Q}$:
$$
\left|r_m(\lambda_p)\right|\leq M
e^{\sigma\left|\lambda_p\right|}\leq M e^{\sigma\cdot
C_{\lambda}},
$$
and also estimate $\det A_{l(j,i),p}$ from above
\begin{align*}
\left|\det A_{l(j,i),p}\right|& \leq (Q-1)!\left|\lambda_{Q-1}\right|^{Q-1}e^{(Q-1)\RE(\mu_n\cdot\lambda_{Q-1})}\\
&\leq(Q-1)!\left|C_{\lambda}\right|^{Q-1}e^{(Q-1)C_{\mu}C_{\lambda}}.
\end{align*}
Then $C_{ij}(m)$ satisfies the inequality
$$
\left|C_{ij}(m)\right|\leq C:=\frac{Q!\cdot M}{D_{\mu\lambda}}\cdot
\left|C_{\lambda}\right|^{Q-1}e^{(Q-1)C_{\mu}C_{\lambda}+\sigma
C_{\lambda}}.
$$

Let us prove that $N_{\varphi}$ is a uniqueness set in
$\KER M_{\psi}$. In order to do this, we need to show that the identity
$r_m(\lambda_p)=0$ implies the identity $r_m(z)\equiv 0$. We consider
the system
$$
\sum\limits_{i=1}^n\sum\limits_{j=0}^{q_i-1}  C_{ij}(m)\lambda_p^j e^{\mu_i\cdot \lambda_p}=0.
$$
According to the assumptions of theorem, the determinant of this system is nonzero.
Expressing the coefficients $C_{ij}(m)$ by the  Cramer  rule, as it has been done above, and replacing the $l(j,i)$ column in the determinant $\Delta_{l(j,i)}$ with the column of constant terms, we obtain
$C_{ij}(m)=0$, and therefore $r_m(z)\equiv 0$.

Let us prove that $$\lim\limits_{m\rightarrow\infty} C_{ij}(m)=0,\qquad i=\overline{1,n},\qquad j=\overline{0,q_i-1}.$$
The above arguing implies that for $C_{ij}(m)$, by the relation
$\left|r_m(\lambda_p)\right|\rightarrow 0$, $m\rightarrow\infty,$
the estimate holds
$$
\left|C_{ij}(m)\right|\leq\frac{Q!}{D_{\mu\lambda}}C_{\lambda}^{Q-1} e^{(Q-1)C_{\mu}C_{\lambda}}\cdot
\left|r_m(\lambda_p)\right|\rightarrow 0,\quad m\rightarrow\infty, \quad
\forall p\in\mathds{N}.
$$
Therefore, $C_{ij}(m)\rightarrow 0$, $m\rightarrow \infty$, $i=\overline{1,n}$, $j=\overline{0,q_i-1}$, that is,
$\max\limits_{i=\overline{1,n},j= \overline{0,q_i-1}}\left|C_{ij}(m)\right|\rightarrow 0$, and this means that for each compact set $K_{\CC}$  we have
$$
\left|r_m(z)\right|\leq
Q\cdot\max\limits_{i=\overline{1,n},j=\overline{0,q_i-1}}\left|C_{ij}(m)\right|\max\limits_{z\in
K_{\CC}}|z|^{\max\limits_{i=\overline{1,n}}{q_i}}e^{C_{\mu}\cdot\max\limits_{z\in
K_{\CC}}|z|}\rightarrow 0,\quad m\rightarrow\infty, \quad z\in K_{\mathds{C}}.
$$
We also have the estimate for  $z\in\CC$
$$
\left|r_m(z)\right|\leq
Q\cdot|C_{ij}(m)\|z|^{\max\limits_{i=\overline{1,n}}{q_i}}e^{\RE(\mu_i\cdot
z)}\leq Q\cdot C \cdot
e^{\left(\max\limits_{i=\overline{1,n}}{q_i}\right)\ln|z|+C_{\mu}
|z|}\leq Q\cdot C\cdot
e^{2\max{\left(\max\limits_{i=\overline{1,n}}{q_i},C_{\mu}\right)}|z|}.
$$
We obtain that $r_m(z)\rightarrow 0$ in the $\tau_{\CC}$ topology, and
the set $N_{\varphi}$ is sufficient in
$\KER M_{\psi}$. \par 2. Let $N_{\varphi}$ be a sufficient set in $\KER M_{\psi}[\varphi\,\cdot\,]$. This means
that $N_{\varphi}$ is a uniqueness set in
$\KER M_{\psi}[\varphi\,\cdot\,]$, i.e., for each function $r(z)\in
\KER M_{\psi}[\varphi\,\cdot\,]$  vanishing on each element
of the set $N_{\varphi}$ we have $r(z)\equiv 0$.

 We substitute $Q$ elements of the set  $N_{\varphi}$
 into the function
$$
r(z)=\sum\limits_{i=1}^n\sum\limits_{j=0}^{q_i-1} C_{ij}z^j
e^{\mu_i\cdot z}
$$
and obtain the homogeneous system
$$
\sum\limits_{i=1}^n\sum\limits_{j=0}^{q_i-1} C_{ij}\lambda_p^j
e^{\mu_i\cdot \lambda_p}=0,\qquad p=1,\ldots,Q.
$$
Since this homogeneous system has only the trivial solution
($r(z)\equiv 0$), the determinant of this system is nonzero.
Therefore,
$$
D_{\mu\lambda}=\det\left(\lambda^j_p e^{\mu_i\lambda_p}\right)\neq
0, \quad i=\overline{1,n}, \quad j=\overline{0,q_i-1}, \quad p=1,\ldots,Q.$$
\end{proof}

\begin{corollary}
Let the set $N_{\psi}$ consist only of simple zeros of the function
$\psi$, where $\widetilde{N}_{\psi}=\{\mu_i\}_{i=1}^n$ is
the set of zeros of $\psi(z)$ located in the conjugate diagram
of function $r(z)\in \KER M_{\psi}[\varphi\,\cdot\,]$, that is,
$$r(z)=\sum\limits_{i=1}^n C_{ij} e^{\mu_i\cdot z}.$$
The set $N_{\varphi}$ is sufficient in $\KER M_{\psi}[\varphi\,\cdot\,]$ if and only if
there exist $n$ elements
of the set $N_{\varphi}$  obeying
$$
D_{\mu\lambda}=\det\left(e^{\mu_i\lambda_p}\right)\neq 0, \qquad i,p=\overline{1,n}.$$
\end{corollary}

\begin{remark}\label{remark1}
The injectivity of a linear operator is equivalent to the triviality of its kernel, meaning that if the operator $M_{\psi}[\varphi\,\cdot\,]$ is injective, then
$$
D_{\mu\lambda}=\det\left(\lambda^j_p e^{\mu_i\lambda_p}\right)\neq 0, \quad i=\overline{1,n}, \quad j=\overline{0,q_i-1}, \quad p=1,\ldots,Q.$$
The proof of this fact is similar to the proof of Statement~2 of Theorem~\ref{mainTh}.
\end{remark}

In the case of simple  interpolation nodes, we can specify conditions, under which the determinant $D_{\mu\lambda}$ is nonzero. In order to do this, we   use the result in \cite{EvansIsaacs}.

\begin{theorem}
Let $p$ be a prime number, and $\varepsilon$ be the $p$ root of unity in a field of characteristics zero. Suppose that
$a_1,\ldots,a_n\in\mathds{Z}$ are pairwise incomparable modulo
$p$; similarly for $b_1,\ldots,b_n\in\mathds{Z}$. Then
$$
\det\left(\varepsilon^{a_i\cdot b_j}\right)\neq 0.
$$
\end{theorem}

Since the space of complex numbers is a field of characteristic zero, we can formulate the following statement.

\begin{corollary}\label{Cor1}
Let $p$ be a prime number, and  we fix a natural number
$k\in \{1,2,\ldots, p-1\}$. Suppose that
$a_1,\ldots,a_n\in\mathds{Z}$ are pairwise incomparable modulo
$p$; similarly for $b_1,\ldots,b_n\in\mathds{Z}$. Then
\begin{equation}\label{N2}
\det\left(e^{\frac{2\pi ik}{p}(a_i+i\cdot A)\cdot (b_j+i\cdot B)}\right)\neq 0,\quad \forall A,B\in\mathds{R}.
\end{equation}
\end{corollary}
\begin{proof}
 By the properties of the determinants and (\ref{N2}) for all  $ A,B\in\mathds{R}$ we have
$$
\det\left(e^{\frac{2\pi ik}{p}(a_i+i\cdot A)\cdot (b_j+i\cdot
B)}\right)=e^{-\frac{2\pi k B}{p}\sum\limits_{i=1}^n a_i}\cdot
e^{-\frac{2\pi k A}{p}\sum\limits_{j=1}^n b_j}\cdot e^{-\frac{2\pi
ik n}{p} A\cdot B}\cdot \det\left(e^{\frac{2\pi ik}{p} a_i\cdot
b_j}\right)\neq 0.
$$
The proof is complete.
\end{proof}


We are in position to formulate our main results.


\begin{theorem}
The following conditions are equivalent.

\vskip5pt

\begin{enumerate}
\item[(1)] The multiple interpolation problem   in the preimage of  convolution operator is solvable.

\vskip5pt

\item[(2)] The Fischer representation   $H(\CC)=\IM^{-1}M_{\varphi}+(\psi)$ holds.

\vskip5pt

\item[(3)] The operator $M_{\varphi}[\psi\,\cdot\,]+g_0(z)$ is surjective.
\vskip5pt

\item[(4)] The operator $M_{\psi}[\varphi\,\cdot\,]$ is injective and $\IM M_{\psi}[\varphi\,\cdot\,]$ is closed in $P_{\CC}$.


\vskip5pt

\item[(5)] The set $N_{\varphi}$ is sufficient in $\KER M_{\psi}$.

\vskip5pt

\item[(6)] We have $D_{\mu\lambda}=\det\left(\lambda^j_p e^{\mu_i\lambda_p}\right)\neq
0,$ $i=\overline{1,n},$ $j=\overline{0,q_i-1},$ $p=1,\ldots,$ $Q=\sum\limits_{i=1}^n q_i.$
\end{enumerate}
\end{theorem}

\begin{proof}

\begin{enumerate}
\item[1.] (1) $\Leftrightarrow$ (2) was proved in Lemma~\ref{lemma1}.
\item[2.] (2) $\Leftrightarrow$ (3) was proved in Theorem~\ref{T1}.
\item[3.] (3) $\Leftrightarrow$ (4) is implied by Theorem~\ref{T5}.
\item[4.] (4) $\Rightarrow$ (6) is implied by Remark~\ref{remark1}.
\item[5.] (5) $\Leftrightarrow$ (6) was proved in Theorem~\ref{mainTh}.
\item[6.] (5) $\Rightarrow$ (2) is implied by Theorem~\ref{T7}.
\end{enumerate}
The proof is complete.
\end{proof}
%--------------------------------------------------------------------------

\vskip20pt

\section{Particular cases}
%--------------------------------------------------------------------------
\subsection{Linear differential equations with constant coefficients}
If the characteristic function of a convolution operator is a polynomial, then
the convolution operator becomes a linear differential operator with constant coefficients of finite order. We take two polynomials
$P(z)$ and $Q(z)$ over the field of complex numbers and consider the equation
$$
P\left(\frac{d}{dz}\right)f_0(z)=g_0(z).
$$
Here $g_0(z)\in H(\CC)$ is chosen arbitrarily and is fixed. For
the introduced polynomials $P(z)$, $Q(z)$ and the function $g_0(z)$, we define
the operator
$$
T[f]=P\left(\frac{d}{dz}\right)(Q\cdot f)+g_0(z),
$$
which all acts in the space of entire functions.


All theorems of Sections 2 and 3 of this article for the functions $\psi(z)\in H(\CC)$, $\varphi(z)\in P_{\CC}$ hold for the case of polynomials
$P(z)$ and $Q(z)$ over the field of complex numbers. By Corollary \ref{C3} we obtain
 the next theorem.

\begin{theorem}\label{P1}
$$
H(\CC)=(Q)\oplus \KER P\left(\frac{d}{dz}\right)\Leftrightarrow
P_{\CC}=(P)\oplus \KER Q\left(\frac{d}{dz}\right).
$$
\end{theorem}

Theorem \ref{P1} was also obtained in   \cite{MerilStruppa}. In \cite{MerilYger} the following theorem was proved.

\begin{theorem}
A pair of polynomials $(P,Q)$ in $\CC$ satisfies
$$H(\CC)=(Q)\oplus \KER P\left(\frac{d}{dz}\right)$$ if and only if   the degrees of the polynomials $P(z)$ and $Q(z)$ are equal  and the operator $\displaystyle P\left(\frac{d}{dz}\right)(Q\,\cdot\,)$ is injective.
\end{theorem}


In particular, when the zeros of the polynomials $P(z)$ and $Q(z)$ are simple, we denote them by $\lambda_k$ and $\mu_k$,  $k=\deg P(z)=\deg Q(z)$, and the
injectivity means
\begin{equation}\label{addN6}
\det\left|e^{\lambda_k\mu_k}\right|\neq 0.
\end{equation}
Moreover, since there is no Fisher decomposition in $H(\CC)$ for the functions $(\varphi,\psi)$, there is no Fisher decomposition for the pair of polynomials $(P,Q)$
$$
H(\CC)=(Q)\oplus \IM^{-1}\,P\left(\frac{d}{dz}\right),
$$
but the Fisher representation
$$
H(\CC)=(Q)+\IM^{-1}\,P\left(\frac{d}{dz}\right)
$$
holds.

\subsection{Fisher representation with conjugate zeros.} For a function $f(z)\in H(\CC)$, the function $f^*(z):=\overline{f(\overline{z})}\in H(\CC)$  has complex conjugate zeros.
For a pair of polynomials $(P(z),P^*(z))$ in one variable, it is known   \cite{MerilYger} ) that the decomposition
$$
H(\CC)=(P^*)\oplus \KER  P\left(\frac{d}{dz}\right)
$$
holds. Let us give another way of proving
$$
H(\CC)=(P^*)+\IM^{-1} P\left(\frac{d}{dz}\right)
$$
for the case when all zeros of the polynomial $P(z)$ are simple, namely,  by using condition
(\ref{addN6}). Let $\lambda_1$, $\lambda_2$, $\dots$, $\lambda_m$ be a set
of points from ${\mathds C}$. The matrix $A$ reads
\begin{equation}
\label{vidA}
 A=
\begin{pmatrix}
e^{\lambda_1\overline{\lambda_1}}&e^{\lambda_1\overline{\lambda_2}}&\dots&
e^{\lambda_1\overline{\lambda_m}}\\
e^{\lambda_2\overline{\lambda_1}}&e^{\lambda_2\overline{\lambda_2}}&\dots&
e^{\lambda_2\overline{\lambda_m}}\\
\hdotsfor{4}\\
e^{\lambda_m\overline{\lambda_1}}&e^{\lambda_m\overline{\lambda_2}}&\dots&
e^{\lambda_m\overline{\lambda_m}}
\end{pmatrix}.
\end{equation}

\begin{proposition}
\label{prop1} If the numbers $\lambda_1$, $\lambda_2$, $\dots$, $\lambda_m$ are pairwise
distinct, then the determinant of  matrix $A$ is nonzero. If the set
of numbers $\lambda_1$, $\lambda_2$, $\dots$, $\lambda_m$ contains coincinding
numbers, then  $$\det A=0.$$
\end{proposition}

\begin{proof} Let us construct the following finite--dimensional unitary
space $R$ (see definition in \cite{Gant}).
We take a set of functions of the variable $z\in {\mathds C}$ generated
by the numbers $\lambda_1$, $\lambda_2$, $\dots$, $\lambda_m$
\[
%{\cal E}
 \mathcal{E}=\{e^{\lambda_1 z},e^{\lambda_2 z},\dots,e^{\lambda_m z} \}.
\]
All functions from the set $\mathcal{E}$ are entire and, moreover,
$\mathcal{E}\subset F$, where $F$ is a Fock space, i.e.
\[
F=\{f\in H({\mathds C})\colon \|f\|_{F}^2=\frac1\pi\int\limits_{\mathds C}|f(z)|^2 e^{-|z|^2}\, dv(z)<\infty\}.
\]
Let $R\stackrel{def}{=}\SPAN \mathcal{E}$, then
the elements of the space $R$ are entire functions, which
are represented as finite linear combinations of functions from
$\mathcal{E}$. For functions $p(z), q(z)\in R$, $z\in {\mathds C}$, we introduce the scalar product
$$
(p,q)_{R}\stackrel{def}{=}(p,q)_{F}=\frac1\pi\int\limits_{\mathds C}
p(z)\cdot \overline{q(z)}\, e^{-|z|^2}\, dv(z).
$$
We obtain that $R$ is a finite--dimensional closed subspace of the Fock Hilbert space.
It is also easy to verify that for each $z_0\in {\mathds C}$
the  estimate holds
\[
|p(z_0)|\leq C_{z_0}\|p\|_{R},\quad \forall p\in R.
\]
where $C_{z_0}$ is a constant depending only on $z_0$.
This means that the delta functional
$\delta_{z_0}\colon p\rightarrow p(z_0)$ is a linear
continuous functional on $R$. This means that $R$ is a space with
a reproducing kernel~\cite{Aron}.

%\begin{proposition}\label{nul}
 %  Функция $p(z)$, $z\in \CC$
%является нулевым элементом пространства $R$ тогда и только тогда,
%когда $p(z)\equiv 0$, $z\in \mathds{C}$.
%\end{proposition}

It is easy to calculate
\begin{equation}
\label{skal_e} (e^{\lambda_j z},e^{\lambda_k
z})_{R}\stackrel{def}{=} \frac1\pi\int\limits_{\mathds C}e^{\lambda_j z}
\cdot \overline{e^{\lambda_k z}}\, e^{-|z|^2}\,
dv(z)=e^{\lambda_j\overline{\lambda_k}}, \quad j,k=1,\dots,m.
\end{equation}
In  \cite{Gant},  there is a criterion  for the linear
independence of vectors $x_1$, $x_2$, $\dots$, $x_m$ in an abstract
finite--dimensional unitary space $R$. It is  proved there that
a system of vectors $x_1$, $x_2$, $\dots$, $x_m$ is linearly independent if and only if the determinant of the Gram matrix of this system
$$
 \det \Gamma(x_1,x_2,\dots, x_m)= \det
\begin{pmatrix}
(x_1,x_1)_R&(x_1,x_2)_R&\dots&
(x_1,x_m)_R\\
(x_2,x_1)_R&(x_2,x_2)_R&\dots&
(x_2,x_m)_R\\
\hdotsfor{4}\\
(x_m,x_1)_R&(x_m,x_2)_R&\dots& (x_m,x_m)_R
\end{pmatrix}
$$
is nonzero. The same reasoning applies to $R$.
As vectors $x_1$, $x_2$, $\dots$, $x_m$, we take the functions
\[
e^{\lambda_1 z},\quad e^{\lambda_2 z},\quad \dots, \quad e^{\lambda_m z}.
\]
Linear independence of this system means that   the condition
\begin{equation}
\label{lin_komb} c_1e^{\lambda_1 z}+c_2e^{\lambda_2 z}+\dots+c_m e^{\lambda_m z}=\text{\O},
\end{equation}
where $\text{\O}$ is the zero element in the space $R$, and
$c_1$, $c_2$, $\dots$, $c_m$ are complex numbers, implies  $c_j=0$, $j=1,2,\dots, m$.


This is why, by
 \cite{Aron} the  linear independence of a system of functions
\[
e^{\lambda_1 z},\quad e^{\lambda_2 z},\quad \dots, \quad e^{\lambda_m z}
\]
means that   the condition
$$ c_1e^{\lambda_1 z}+c_2e^{\lambda_2 z}+\dots+c_m e^{\lambda_m z}\equiv 0,\quad z\in {\mathds C}$$
implies   $c_j=0$, $j=1,2,\dots, m$. In our case the Gram matrix is the matrix $A$ (see~\eqref{vidA}, \eqref{skal_e}).

We are going to prove that if the set  $\lambda_1$, $\lambda_2$, $\dots$, $\lambda_m$
contains coincinding numbers, then the system of functions
\[
e^{\lambda_1 z},\quad e^{\lambda_2 z},\quad \dots, \quad e^{\lambda_m z}
\]
is linearly dependent in  $R$.


If all the numbers $\lambda_1$, $\lambda_2$, $\dots$, $\lambda_m$ are pairwise distinct, then the system of functions
\[
e^{\lambda_1 z},\quad e^{\lambda_2 z},\quad \dots, \quad e^{\lambda_m z}
\]
is linearly independent in  $R$.

Indeed, let us suppose the contrary and there exist distinct complex numbers $\lambda_1$, $\lambda_2$, $\dots$, $\lambda_m$ such that the system of functions
\[
e^{\lambda_1 z},\quad e^{\lambda_2 z},\quad \dots, \quad e^{\lambda_m z}
\]
is linearly  dependent in $R$. This means that
there exists a set of complex numbers
$c_1$, $c_2$, $\dots$, $c_l$, $l\leq m$, $c_j\neq0$ such that
$$
\sum_{j=1}^lc_je^{\lambda_j z}=\text{\O},
$$
see~\eqref{lin_komb}.

We consider the space $\widehat F$
\[
\widehat F=\{\widehat f\colon \widehat f(\lambda)=(e^{\lambda z},f)_{F},\,f\in F\}
\]
Laplace transforms of functionals on $F$, $R\subset F$. For each
function $f\in F$ we have
\begin{equation}
\label{sopr_sps} 0=(\text{\O},
f)_F=\left(\sum_{j=1}^lc_je^{\lambda_j z},
f(z)\right)_F=\sum_{k=1}^l c_k \widehat f(\lambda_k ).
\end{equation}
We have found that there are nonzero constants $c_1$, $\dots$, $c_l$ such that for each function $\widehat f\in \widehat F$ identity~\eqref{sopr_sps} holds.
It is well--known that the space $\widehat F$ coincides with the
space $F$  \cite{Barg}.

Therefore, the space $\widehat F$ contains the same functions as the
space $F$. This implies that for each function $g\in F$ the identity
\begin{equation}
\label{sopr_sps1} \sum_{k=1}^l c_k g(\lambda_k )=0
\end{equation}
holds and  $c_1,\dots,c_l\neq 0$. Relation~\eqref{sopr_sps1}  is obviously impossible for each function $g\in F$.

Hence, the system of functions
\[
e^{\lambda_1 z},\quad e^{\lambda_2 z},\quad \dots, \quad e^{\lambda_m z}
\]
is linearly independent in $R$. By  \cite{Gant}
the Gram determinant of this system is nonzero, and therefore
the determinant of the matrix $A$ is nonzero. The proof is complete.
\end{proof}

It is obvious that Proposition \ref{prop1} is also valid for an arbitrary finite
set of points from ${\mathds C}^n$, $n\in\mathds{N}$.

\begin{proposition}
Let $\lambda_1$, $\lambda_2$, $\dots$, $\lambda_m$ be a set of points from
${\mathds C}^n$, $n\in\mathds{N}$. The matrix $A$ reads
$$
 A=
\begin{pmatrix}
e^{\langle\lambda_1,\overline{\lambda_1}\rangle}&e^{\langle\lambda_1,\overline{\lambda_2}\rangle}&\dots&
e^{\langle\lambda_1,\overline{\lambda_m}\rangle}\\
e^{\langle\lambda_2,\overline{\lambda_1}\rangle}&e^{\langle\lambda_2,\overline{\lambda_2}\rangle}&\dots&
e^{\langle\lambda_2,\overline{\lambda_m}\rangle}\\
\hdotsfor{4}\\
e^{\langle\lambda_m,\overline{\lambda_1}\rangle}&e^{\langle\lambda_m,\overline{\lambda_2}\rangle}&\dots&
e^{\langle\lambda_m,\overline{\lambda_m}\rangle}
\end{pmatrix}.
$$
If the points $\lambda_1$, $\lambda_2$, $\dots$, $\lambda_m$ from ${\mathds
C}^n$ are  pairwise distinct, then the determinant of  matrix $A$ is nonzero.
If the set of points $\lambda_1$, $\lambda_2$, $\dots$, $\lambda_m$ from
${\mathds C}^n$ contains identical points, then $\det A=0$.
\end{proposition}
The proof of this proposition is almost literally reproduces  the proof of Proposition~\ref{prop1}.

In conclusion of this section, we note that, according to   \cite[Thm. 37]{NapalkovMullabaeva} and Theorem~\ref{T7} of this paper, the next theorem is true.



\begin{theorem}
If  $\varphi(z)\in P_{\CC}$,  $\psi(z)\equiv\varphi^*(z)$,  then the Fisher representation
$$
H(\CC)=(\varphi^*)+\IM^{-1} M_{\varphi}
$$
holds.
\end{theorem}

The authors express their deep gratitude to Yulmukhametov R.S. for valuable comments that were taken into account in this paper.


\bigskip
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\end{document}