Discrete Translates in Function Spaces
Alexander Olevskii\thanks{This author's research is supported in part by the Israel Science Foundation
We construct %prove that there is a Schwartz function $\varphi$ such that for every exponentially small perturbation of integers $\La$ % and a uniformly discrete sequence $\La$ (exponentially small perturbation of integers), such that the set of translates $\{\varphi(t-\lambda),\lambda\in\La\$ spans the space $L^p(\R)$ for every $p>1$. This result remains true for more general function spaces $X$, whose norm is "weaker" than $L^1$ (on bounded functions). %A set $\La\subset\R$ is called $ p-$spectral if there is a function $\varphi\in L^p(\R)$ whose $\La$-translates $\{\varphi(t-\lambda),\lambda\in\La\$ span $L^p(\R)$. We prove that exponentially small non-zero perturbations of the integers are $p-$spectral sets for all $p>1$.
1612.00811.tex

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\begin{document}
\begin{large}

\title{Discrete Translates in Function Spaces} %$L^p(\R)$}

\author{Alexander Olevskii\thanks{This author's research is supported in part  by the Israel Science  Foundation}\,  and  Alexander Ulanovskii}

\date{}\maketitle


\noindent A.O.: School of Mathematics, Tel Aviv University\\ Ramat Aviv,  69978 Israel. E-mail:
 olevskii@post.tau.ac.il

\medskip
%\medskip

\noindent A.U.: Stavanger University,  4036 Stavanger, Norway\\ E-mail: Alexander.Ulanovskii@uis.no


\begin{abstract}We construct %prove that there is
a Schwartz function  $\varphi$ such that for every exponentially small perturbation of integers $\La$ % and a uniformly discrete sequence $\La$ (exponentially small perturbation of integers),  such that
the set of translates $\{\varphi(t-\lambda),\lambda\in\La\}$                                    spans the space $L^p(\R)$ for every $p>1$.
                                  This result remains true  for more general function spaces $X$,  whose  norm is "weaker" than $L^1$  (on bounded functions).
 %A set $\La\subset\R$ is called $ p-$spectral  if there is a function             $\varphi\in L^p(\R)$           whose $\La$-translates  $\{\varphi(t-\lambda),\lambda\in\La\}$ span $L^p(\R)$.                             We prove that exponentially small non-zero  perturbations of the integers are   $p-$spectral sets for all $p>1$.
\end{abstract}




   \section{Introduction}

   \noindent{\bf 1. Completeness of translates}. In what follows we will use  the standard form of Fourier transform  \[\varphi(x)=\hat\Phi(x):=\int_\R e^{-2\pi i tx}\Phi(t)\,dt.\]

Classical Wiener's  theorems \cite{w}  provide  necessary and sufficient conditions on a function $\varphi$ whose  translates $\{\varphi(x-s), s\in\R\}$ span the space  $L^1(\R)$ or $L^2(\R)$:

\medskip

(i) The translates     of  $\g\in L^1(\R)$ span $L^1(\R)$  if and only if  $\hat\g$ does not vanish;


 (ii) The translates     of  $\g\in L^2(\R)$ span $L^2(\R)$       if and only if  $\hat\g$ is non-zero almost everywhere on $\R$.


\medskip

When $1<p<2$, Beurling \cite{B} proved that the translates of a function $\varphi\in L^p(\R)\cap L^1(\R)$ span $L^p(\R)$ provided
the  Hausdorff dimension of the zero set of  $\hat\varphi$ is  less than $ 2(p-1)/p$. However, this condition is not necessary. Moreover, the spanning property of the translates of  $\g\in L^p(\R)$ cannot be characterized in terms of the zero set of  $\hat\varphi$, see \cite{LO}.


\medskip\noindent{\bf 2. Discrete spectra}.
It is well known that sometimes even a {\it discrete} set of  translates may  span $L^p(\R).$



\begin{definition}  We say that a discrete set $\La\subset\R$ is a {\it $p-$spectrum} {\rm (}for translates{\rm )},   if there is a function $\g\in L^p(\R)$ such that the family of translates
 \[              \{\g(t-\lambda),   \lambda\in \La\}\]
        spans $L^p(\R)$.  Such a function $\g$ is called  {\it a $\La-$generator}.
\end{definition}


     For $p=2 $, using the Fourier transform one can re-formulate the definition above as follows:

      {\it A set $\La$ is a $2$-spectrum if and only if  there exists $\G\in L^2(\R)$ such that the  system
  \[\{\G(t) e^{2\pi i\lambda t}, \lambda\in \La\}\] spans the whole space $L^2(\R) $.}

  Using this, it is easy to see that the set of integers $\Z$ is not a $2$-spectrum.
        On the other hand, the following is true:

                 {\it  Every sequence $\La$ obtained by a "perturbation" of integers,
     \begin{equation}\label{ais}
\La=\{\lambda_n:=n+a_n, n\in\Z\},\quad a_n\ne0, n\in\Z,
\end{equation}where $a_n\to0$ as $|n|\to\infty,$ is a $2$-spectrum} (\cite{O}).



          Assume  the "perturbations" $a_n$ are exponentially small:
\begin{equation}\label{an}
|a_n|<Cr^{|n|}, \quad n\in\Z;\ \mbox{ for some }  C>0,0<r<1.
\end{equation}
      Then    a stronger result is true  (\cite{OU2}):

      {\it If $\La$ satisfies (\ref{ais}) and (\ref{an}), then a $\La$-generator $\g$ can be chosen to be a Schwartz
           function. The decay condition (\ref{an}) is essentially sharp}.

         This result may seem surprising, since when (\ref{an}) holds, the set $\La$ is "very close" to the  "limiting case" $\La=\Z$ when
          no generator exists.

          Since every $p_0$-spectrum $\La$ is also a $p$-spectrum  for every
          $p> p_0$ (see \cite{Bl}), the result above holds for  $p>2.$
          However, in fact  no perturbations are "needed" in this case:

         {\it The set of integers $\Z$ is a $p-$spectrum for all $p > 2 $}, see \cite{ao}.

         On the other hand, for $p=1$ the spectrum is never uniformly discrete.
          The following is proved in \cite{BOU}:

         {\it A set $\La$ is a $1$-spectrum if and only if the Beurling--Malliavin density
          of $\La$ is infinite}.

          Notice, so far $p=1$ is the only case when an effective characterization
           of $p$-spectra is known.

          A  detailed survey on the topic, including the results above, may be
          found in \cite{ou}.

\medskip\noindent{\bf 3.  The case ${\bf 1<p<2}$}.          This case is much less investigated.  In particular, it has been
          an open problem  whether   a $p$-spectral set can be
          uniformly discrete, or at least have a finite density.




      \section{Results}                    In this  note we prove the following

   \begin{theorem}
                                  There is a function $\varphi \in  S(\R)  $, such that for every set   $\La$ satisfying
                               (2) and (3), the set of translates
$ \{\varphi(t-\lambda),   \lambda\in \La\}$
                            spans the whole space $L^p(\R)$ for each $p>1$.\end{theorem}

                             The result holds for more general function spaces
                              "arbitrarily closed" to the space $L^1(\R)$, see the last section.

         The function  $\varphi$  will be  effectively constructed as the Fourier
          transform of a Schwartz function $\Phi$ with  "exponentially deep"  zeros
         at the integers and at infinity.

        The proof is based on a certain uniqueness theorem for tempered distributions.


    %      In this note we give a positive answer to this question.           Our main result is the following \begin{theorem}  Every set $\La$ satisfying (\ref{ais}) and (\ref{an}) is a $p$-spectrum, for every $p>1$. \end{theorem}   We will construct  a function $\varphi$, which  is a $\Lambda$-generator for every $\La$ satisfying (\ref{ais}) and (\ref{an}) and all $p> 1$ simultaneously.   In addition, $\varphi$ will have the nice properties of analyticity and fast decay.           The construction  is based on a uniqueness theorem for a certain class           of tempered distributions.


%In fact, our approach  gives a  stronger result, see sec. 6,  showing   that  $L^1$   is  the    "critical"  norm  which  does not admit a uniformly discrete spectrum.

\section{Functions with Deep Zeros}


In this section we present a uniqueness result for a class of analytic functions whose inverse Fourier transforms  have "deep zeros". Then in the next section this result will be extended to a certain class of  temperate distributions.


\begin{definition} Denote by $K_0$ the class of continuous functions $\Phi$ satisfying  the condition
\begin{equation}\label{dz}|\G(t)|\leq C_1 e^{-C_2(|t|+\frac{1}{d(t,\Z)})}, \quad  t\in\R,
\end{equation}where $d(t,\Z)=\min_{n\in\Z}|t-n|$, and $C_1=C_1(\Phi),C_2=C_2(\Phi)$ are positive constants depending on $\G$.
\end{definition}


Condition (\ref{dz}) means that $\G$ has "deep" zeros on the set of integers and at infinity.


Next, let $\hat K_0$ denote the class of the Fourier transforms $\varphi=\hat\Phi$ of functions  $\Phi\in K_0$.

Condition (\ref{dz}) implies that $\Phi$ has an exponential decay at $\pm\infty$. Hence, every element $\varphi\in\hat K_0$ is analytic and bounded in the horizontal strip $|$Im$\,z|<C,$ $0<C<C_2/(2\pi)$, where $C_2$ is the constant in (\ref{dz}). Clearly,  every derivative of $\varphi$ also belongs to $\hat K_0$.


\begin{md}    Suppose  $\La$ satisfies  (\ref{ais}) and (\ref{an}) and $\varphi\in \hat K_0$. If $\g|_\La=0$, then $\g=0$.
\end{md}

In other words, the sets $\La$ satisfying (\ref{ais}) and (\ref{an}) are uniqueness sets for the class $\hat K_0$.


\begin{proof}[Sketch of            Proof]         Assume  $\varphi\in\hat K_0$ and  $\g|_\La=0$ for some $\La$ satisfying (\ref{ais}) and (\ref{an}). We have to show that $\g=0$.



      Since $\varphi'$ is bounded on $\R$, from (\ref{ais}) and (\ref{an}) we get
\begin{equation}\label{ra}
                      |\g(n)| = O(r^{|n|}),\quad |n|\to\infty,
\end{equation}where $0<r< 1.$

                 Consider the periodization  $P(\G)$ of  $\G $ defined by

\[
                       P(\G) (t):= \sum_{k\in\Z} \G(t+k).
\]
                   The Fourier decomposition of $P(\G)$ has the form
\[
                       P(\G) (t)= \sum_{n\in\Z} \g(n) e^{2\pi int}.
\]
                  By (\ref{ra}),  $P(\G)$ is  analytic in some horizontal strip containing $\R$. On the other hand, it easily follows from  (\ref{dz}) that it has zero of infinite order at the origin.
                    Hence, $P(\G)= 0$,  so that we have
\[
                    \g(n)=0, \quad  n\in\Z.
\]
                Iterating this argument, one may prove that every derivative of $\g $ vanishes
               at the integers, and the result follows.                   See  details in [OU16], Lecture~11.

\end{proof}



    \section{Distributions with Deep Zeros}

              Let $S(\R)$ denote the space of Schwartz test-functions,
                and  $S'(\R)$ the space of tempered distributions on $\R$.
We will denote by $\langle F, \Phi\rangle$ the action of the distribution $F\in S'(\R)$ on the test function $\Phi\in S(\R)$.

We would like to extend condition (\ref{dz}) to tempered distributions.

Recall  that every tempered distribution $F\in\mathcal{S}'(\R)$ is the distributional derivative of finite order, $F=\Phi^{(k)}$,  of a continuous function   $\Phi$ having at most polynomial growth on the real axis (see \cite{FJ}, Theorem 8.3.1).


 \begin{definition} Denote by  $K$ the class of all tempered distributions  $F\in S'(\R)$ which  admit a representation \begin{equation}\label{dd}F=\Phi_1^{(k_1)}+\dots+\Phi_l^{(k_l)},\end{equation} where  $l\in\N,k_1,\dots,k_l\in\N\cup\{0\}$ and  $\Phi_1,\dots,\Phi_l$ are continuous functions satisfying (\ref{dz}).
\end{definition}



Let $\hat K$ denote the class the distributional Fourier transforms of the elements of $K$.

Assume $\Phi$ satisfies (\ref{dz}). Then the Fourier transform $\varphi=\hat\Phi$ is analytic and bounded in some horizontal strip. The distributional Fourier transform of $\Phi^{(k)}$ is the function
\[
\widehat{\Phi^{(k)}}= (2\pi ix )^k\varphi(x).
\]
It is therefore an analytic function of at most polynomial growth in the strip.

By the definition of class $\hat K$, we have
\begin{equation}\label{0}
x^k\varphi(x)\in \hat K,\quad \mbox{for every } k\geq0.
\end{equation}
From (\ref{dd}) it follows that   $\hat K$ consists of the functions $f$  admitting a representation
\[
f(x) = \sum_{j=1}^l(2\pi ix )^{k_j}\varphi_j(x), \quad \varphi_j=\hat\Phi_j, \Phi_j\in K_0,\quad j=1,\dots,l.
\]


 We will now list several simple properties of the class $\hat K$.



\begin{lemma} (a) $\hat K$ is a linear space over $\C$.

(b) Every element $f\in \hat K$ is analytic  and of at most polynomial growth in some strip $|$Im$\,z|<C, C=C(f)>0.$

(c) If $f\in \hat K$, then $xf(x)\in \hat K$ and $f'\in\hat K$.

(d) If $f\in \hat K$ then  Re$\, f(x)\in \hat K$ and Im$\, f(x)\in\hat K$.

(e) Assume $\varphi=\hat\Phi$, where  $\Phi\in S(\R)$ is such that  $\Phi,\Phi'\in K_0$.
Then for every $1\leq q\leq\infty$ and every $h\in L^q(\R)$ we have $\varphi\ast h\in \hat K.$

\end{lemma}

\begin{proof}
Statements (a) and (b) are obvious.


Clearly, it suffices to  prove statements  (c) and (d) for the case $l=1$ in (\ref{dd}), i.e. when $f(x)=\widehat{\Phi^{(k)}}=(2\pi i x)^k\varphi(x)$.

By  (\ref{0}), we see that $xf(x)\in K.$

Since the function $t\Phi(t)$ also satisfies (\ref{dz}), then
\[\varphi'= \widehat{(-2\pi i t)\Phi(t)}\in \hat K.\] From (\ref{0}) we get $x^k\varphi'(x)\in\hat K$ and $x^{k-1}\varphi(x)\in \hat K$ which yields
$f'\in\hat K$.  This proves (c).

Now,
write
\[\Phi(t):=\Phi_r(t)+i\Phi_i(t):=\frac{\Phi(t)+\Phi(-t)}{2}+i\frac{\Phi(t)-\Phi(-t)}{2i}.\]
Clearly, $\Phi_r$ and $\Phi_i$ satisfy (\ref{dz}). So, we have $\varphi=\varphi_r+i\varphi_i$, where the functions $\varphi_r=\hat \Phi_r\in\hat K$ and $\varphi_i=\hat\Phi_i\in\hat K$ are real on the real line. Hence, if $k$ is even, we see that
\[\mbox{Re}\,f(x)=(2\pi i x)^k\varphi_r(x)\in\hat K,\quad \mbox{Im}\,f(x)=(2\pi i x)^k\varphi_i(x)\in\hat K.\] Similarly,  one proves (d) for the odd $k$.

To prove (e), write $h(x)=h_1(x)+ h_2(x)$, where   \[ h_1(x):=\frac{h(x)}{1-2\pi i x}, \quad h_2(x):= -2\pi i x h_1(x).\] Hence,
$\varphi\ast h=\varphi\ast h_1+\varphi\ast  h_2.$

Clearly,   $h_1\in L^1(\R)$, and so  its  inverse Fourier transform  $H_1$ is continuous and bounded. This gives
 $\varphi\ast h_1= \widehat{\Phi H_1}\in\hat K$.

 Observe that $h_2=\widehat{H_1'}$, and so the inverse Fourier transform of
 $\varphi\ast h_2$ is the distribution $\Phi H_1'= (\Phi H_1)'- \Phi' H_1$. Since, by assumption, $\Phi'$ satisfies (\ref{dz}), we conclude that $\varphi\ast h_2\in\hat K$, which completes the proof of Lemma 1.
\end{proof}





\section{Uniqueness Theorem for the Class $\hat K$}


   \begin{theorem}   Assume $\La$  satisfies (\ref{ais}) and (\ref{an})
                and  $f \in\hat  K$.
                 If  $f|_\La =0$  then  $f=0$.
\end{theorem}

Hence, the sets $\La$ satisfying (\ref{ais}) and (\ref{an}) are uniqueness sets for the class $\hat K.$


\begin{ml} Assume $\La$ satisfies (\ref{ais}) and (\ref{an}). Then
\[
                    f\in\hat K, f |_\La= 0\Rightarrow f |_\Z=0 .
\]
\end{ml}


\begin{proof}  For convenience, in what follows we denote by $C$ different positive constants.

 Take any function $f\in\hat K$ which vanishes on $\La$.
 We have to show that $f|_\Z=0$.

 By Lemma 1, $f'\in K$ and  has at most polynomial growth on $\R$. So,   it follows from (\ref{ais}) and (\ref{an}) that \[ |f(n)|<C|n|^Cr^{|n|},\quad n\in\Z,\]
 where $0<r<1.$
 This shows that  the function
\[
R(t):=\sum_{n\in\Z}f(n)e^{2\pi i nt}, \quad t\in\R.
\]
is analytic in some strip $|$Im$\,z|<C$.

Clearly, to prove the lemma it suffices to  show that
$R(t)\equiv0$.
For this purpose we introduce the function $f_\epsilon:=fh_\epsilon$, where $\epsilon>0$ and
\[
h_\epsilon(x):=\left(\frac{\sin(2\pi \epsilon x)}{2\pi\epsilon x}\right)^N.
\]
 We will assume that $N$ is an even integer so large that  $f_\epsilon\in L^1(\R)$.

It is easy to see that $h_\epsilon=\hat H_\epsilon$, where
\[ H_\epsilon(t)= \left(\frac{1}{2\epsilon}1_{\epsilon}(t)\right)^{N*}.\] Here $1_\epsilon$ is the indicator function of $[-\epsilon,\epsilon]$. Hence, $H_\epsilon(t)\geq 0, t\in\R,$  and so
\[
\|H_\epsilon\|_1=h_\epsilon(0)=1.
\]

Given  $d>0$,  we denote by $S(d)$ the subspace of $S(\R)$ of functions $\Psi$ vanishing outside $(-d,d).$
In what follows we  assume that $0<d<1/2$ and $\epsilon>0$ is so small  that $d+N\epsilon<1/2$.

Fix any function  $\Psi\in S(-d,d)$. Then  $$\Psi\ast H_\epsilon\in S(d+N\epsilon),$$ and we have
\begin{equation}\label{hh}
\|(\Psi\ast H_\epsilon)^{(k)}\|_\infty\leq \|\Psi^{(k)}\|_\infty\|H_\epsilon\|_1=\|\Psi^{(k)}\|_\infty.
\end{equation}


Denote by $F_\epsilon$ the inverse Fourier transform of $f_\epsilon$, and by $\psi$ the Fourier transform of $\Psi$. From (\ref{dd}), for every $n\in\Z$ we obtain:
\[
|\langle F_\epsilon,\Psi(t-n)\rangle|=|\langle f_\epsilon,e^{2\pi inx}\psi\rangle|=|\langle f,e^{2\pi inx}\psi h_\epsilon\rangle|=\]\[
|\langle \sum_{j=1}^l\Phi_j^{(k_j)},(\Psi\ast H_\epsilon)(t-n)\rangle|
\leq\|(\Psi\ast H_\epsilon)^{(k)}\|_\infty\sum_{j=1}^l\int_{n-d-N\epsilon}^{n+d+N\epsilon}|\Phi_j|,\] where $k:=\max_j{k_j}$.
Recalling that the functions $\Phi_j$  satisfy (\ref{dz}), by  (\ref{hh}) we obtain
\begin{equation}\label{4}
|\langle F_\epsilon,\Psi(t-n)\rangle|\leq Ce^{-C(|n|+\frac{1}{N\epsilon+d})}\| \Psi^{(k)}\|_{\infty}, \quad n\in\Z.
\end{equation}


Set
\[
R_\epsilon(t):=\sum_{n\in\Z}f_\epsilon(n)e^{2\pi i n t},
\]
and let us calculate the product $\langle R_\epsilon,\Psi\rangle.$

By the Poisson formula,
\[
R_\epsilon(t)=\sum_{n\in\Z}f_\epsilon(n)e^{2\pi i n t}= \sum_{n\in\Z}F_\epsilon(t+n).
\]
Therefore, by (\ref{4}),
\[
|\langle R_\epsilon, \Psi\rangle|=|\langle\sum_{n\in\Z}F_\epsilon(t+n), \Psi(t)\rangle|=|\langle F_\epsilon(t),\sum_{n\in\Z} \Psi(t-n)\rangle|
\]
\[
\leq C\| \Psi^{(k)}\|_{\infty}e^{-\frac{C}{N\epsilon+d}}.
\]
Letting $\epsilon\to 0$  we arrive at the inequality:
\begin{equation}\label{5}
|\langle R, \psi\rangle| \leq C\|\Psi^{(k)}\|_{\infty}e^{-\frac{C}{d}},
\end{equation}  which holds for some  $k$ and every $0< d<1/2$ and  $ \Psi\in S(d)$.

Now, the Main Lemma follows from

\begin{lemma} Assume a function $R$ is  analytic  in a strip $|${\rm Im}$\,z|<C$  and  satisfies (\ref{5}). Then $R=0$.\end{lemma}

The proof is simple: One may use the standard calculus to check that $R$ must have zero of infinite order at the origin, and so $R=0$. We omit the details.
\end{proof}

\section{Proofs of Theorem 1 and 2}

\begin{proof}[Proof of Theorem 2] Assume $\La$ satisfies (1) and (2), $f\in\hat K$  and $f|_\La=0$. Write $f=f_r+if_i,$ where $f_r(x):=$Re$\,f(x)$ and $f_i(x):=$Im$\, f(x)$.

 Let us show that $f_r=0$. This function vanishes on $\La$ and by Lemma 1,  $f_r\in K$.  It follows from the Main Lemma that $f_r|_\Z=0$.

 Since $f_r$  is  real  for real $x$ and vanishes both on $\Z$  and on $\La=\{n+a_n, n\in\Z\}, a_n\ne0$,  its derivative $f'_r$  vanishes on some set $\La_1:=\{n+a_1^{(1)}\}$, where each point $a^{(1)}_n$ lies inside the open interval between $0$ and $a_n$. Since $\La$ satisfies (\ref{ais}) and (\ref{an}),  so does   $\La_1$.
  By Lemma 1, $f_r'\in  K.$ We now we apply the Main Lemma again to get $f_r'|_\Z=0$. A straightforward iteration of this argument proves that $f_r^{(j)}|_\Z=0$ for every $j\in\N$.  Since $f_r$ is analytic in some strip $|$Im$\,z|<C$, we conclude that $f_r=0.$ Similarly, one proves that $f_i=0$, and so $f=0.$
\end{proof}


\begin{proof}[Proof of Theorem 1]
Choose any real function $\Phi\in S(\R)$ such that $\Phi$ and $\Phi '$ satisfy (\ref{dz}) and such that $\Phi(-t)=\Phi(t)>0$ for $t\not\in\Z$. Then its Fourier   transform $\varphi$ is real and $\varphi(-x)=\varphi(x), x\in\R$.


Let $\La$ satisfy (\ref{ais}) and (\ref{an}).
 Assume the set of  translates  $\{\varphi(x-\lambda),\lambda\in\La\}$ does not span $L^p(\R)$. Then there is a non-trivial function $h\in L^{q}(\R), 1/p+1/q=1,$ such that $$\int_\R \varphi(x-\lambda)h(x)\,dx= (\varphi\ast h)(\lambda)=0,\quad\lambda\in\La.$$ By Lemma 1, $\varphi\ast h\in\hat  K$.  Theorem 2 yields  $\varphi \ast h=0.$ This means that all translates  $\{\varphi(x-s), s\in\R\}$ do not span the space $L^p(\R)$. However, this  contradicts Beurling's theorem cited in the introduction.
 Theorem 1 is proved.
\end{proof}
%\section{Acknowledgments}  This research was supported through the program "Research in Pairs" by the Mathematisches Forschungsinstitut Oberwolfach in April 2014. The authors would like to thank the MFO   for providing a stimulating and pleasant atmosphere.


\section{Extension}
Below we assume that $X$ is a Banach function space on $\R$ with the property that  the Schwartz  space $S(\R)$ is continuously embedded  and dense in $X$.
%Observe that this property holds for every space $L^p(\R), p\geq 1.$

The following extension of Theorem 1 hods true:

\begin{theorem}    Assume that no non-trivial distribution  $h\in X^*$ has spectrum on $\Z$.
Then there exists $\varphi\in S(\R)$ such that for every set   $\La$ satisfying
                               (2) and (3), the set of translates
$$ \{\varphi(t-\lambda),   \lambda\in \La\}$$
                            spans $X$.
\end{theorem}

\begin{proof}  Choose a Schwartz function $\varphi$ as in the proof of Theorem 1, where we additionally require that  {\it every derivative}  $\Phi^{(j)}$  satisfies (3).  If the $\La$-translates of $\varphi$ do not span $X$, then  there is a distribution $h\in X^\ast$   such that the convolution $\varphi * h$ vanishes on $\La$. The distributional inverse Fourier transform of  $\varphi * h$ is the the distribution $\Phi H,$ where $h=\hat H.$ Since  $H$ is the distributional derivative of finite order of some continuous function, similarly to Lemma 1 (e) one may check that $\varphi * h\in \hat K$. Hence, by Theorem 2,
$$
                           \Phi H  = 0.
$$
It follows that the support of  $H$ lies in $\Z$   (the zero set of $\Phi$). Contradiction.\end{proof}


  \begin{exa}   Let $w>0$ be a weight  vanishing at infinity.
                      Then Theorem 3 holds for the space
$$
                          X=   L^1(w, \R) := \{f:    \|f\|_w= \int_\R | f(x)|w(x)\,dx < \infty\}.
$$\end{exa}
                 Indeed, the elements of the dual space  $X^\ast=L^\infty(1/w, \R)$ are functions vanishing at infinity. One may prove that no non-trivial function $h\in X^\ast$   may have spectrum on $\Z$.


   However, as mentioned in Introduction,  Theorem 3 ceases to be true for $w=1 $, that is for the space $L^1(\R)$.
  %  A set of translates of a single function, which spans this space must    have infinite Beurling-Malliavin density.

       Theorem 3 is also applicable to the separable Lorentz spaces and also to more general symmetric spaces, see definition of these spaces in \cite{kps}, Ch. 2.

%*** Here we should be more accurate :  it may be non separable.       I'll look at  the book  ....












\begin{thebibliography}{99}


\bibitem[AO96]{ao}
A. Atzmon and A. Olevskii,   `Completeness of integer translates in function spaces on $\R$', {\it J. Approx. Theory} 87, (1996), no. 3, 291--327.


\bibitem[B51]{B} A. Beurling, `On a closure problem', {\it Ark. Mat. } 1, (1951). 301--303. See also: {\it The Collected Works of Arne Beurling}, in: Harmonic Analysis, vol. 2, Harmonic Analysis. Birkhäuser, Boston, 1989.

%\bibitem[BM62]{BM1} Beurling, A. and  Malliavin, P. On Fourier transforms of measures with compact support.  Acta. Math., 107,  (1962) 291--309.

%\bibitem[BM67]{BM} Beurling, A., Malliavin, P. On the closure of characters and the zeros of entire functions.     Acta Math., 118, (1967), 79--93.


\bibitem[Bl06]{Bl} N. Blank, `Generating sets for Beurling algebras', {\it J. Approx. Theory} 140, (2006), no. 1, 61--70.

\bibitem[BOU06]{BOU} J. Bruna, A. Olevskii and A. Ulanovskii, `Completeness in $L^1(\R)$ of discrete translates', {\it Rev. Mat. Iberoam.} 22, (2006), no. 1, 1--16.

\bibitem[FJ98]{FJ}
 F.J. Friedlander and M.S. Joshi, `Introduction to the Theory of Distributions.'
Cambridge University Press, 1998.

\bibitem[KPS82]{kps}
S. G. Kre\v{i}n, Ju. I. Petunin and E. M. Semenov,  `Interpolation of linear operators'. Translations of Mathematical Monographs, vol. 54, AMS, Providence, R.I., 1982.


\bibitem[LO11]{LO} N. Lev and A.  Olevskii, `Wiener's "closure of translates" problem and Piatetski--Shapiro's uniqueness phenomenon', {\it Ann. of Math.} (2) 174, (2011), no. 1, 519--541.



%\bibitem[NO07]{NO} Nitzan, S., Olevskii, A. Sparse exponential systems: completeness with estimates. Israel J. Math. 158, (2007), 205--215.

 \bibitem[Ol97]{O} A. Olevskii, `Completeness in $L^2(\R)$ of almost integer translates', {\it
 C. R. Acad. Sci. Paris S$\acute{e}$r. I Math.} 324, (1997), no. 9, 987--991.

 \bibitem[OU04]{OU2}  A. Olevskii and A. Ulanovskii, `Almost integer translates. Do nice generators exist?', {\it J. Fourier
Anal. Appl. } 10, (2004), no. 1, 93--104.


\bibitem[OU16]{ou} A. Olevskii and A. Ulanovskii,  `Functions with Disconnected Spectrum: Sampling, Interpolation, Translates.
AMS,  University Lecture Series,  65, 2016.

%\bibitem[Ul01]{u} Ulanovskii, A. Sparse systems of functions closed on large sets in $\R^n$. J. London Math. Soc. (2) 63,   (2001), no. 2, 428--440.




%Ulanovskii, A. On Landau's phenomenon in $\R^n$. Math. Scand. 88 (2001), no. 1, 72-78.
\bibitem[W32]{w}
N. Wiener, `Tauberian Theorems', {\it Annals of Math.} 33 (1): 1--100, 1932.
\end{thebibliography}


\end{large}
\end{document}

{\sl
\noindent A. Olevskii \\ School of Mathematics, Tel Aviv University\\ Ramat Aviv,  69978 Israel\\
 olevskii@post.tau.ac.il

\medskip

\noindent A. Ulanovskii\\ Stavanger University,  4036 Stavanger, Norway\\  Alexander.Ulanovskii@uis.no
}

     It is shown in sec 6 ,that  Th.1 remains true for any function space X
                  weaker (for bdd functions) then L1-norm.

                    THEOREM 3.  LET X  be a function space on R,in which S(R) is embedded continuously
                                and densely.
                                  If no distributions  h \in X* has the spectrum in Z ,then
                                the conclusion of Th.1 remains true.

                   Indeed:  .............

                      TH 3 is applicable for any Orlitz zpace.

                      On the other it is not for L1 ,since the dual space contains periodic distr.
                      And indeed ,it is well known ,see ... ,that no u.d. set of transl f a single
                      f may spend L1.







  In what follows we assume that  $X$ is a Banach function space on $\R$ such that the
                   Schwartz space $S(\R)$ is continuously embedded in $X$.
                   Then every element of the dual space, $f\in X^\ast$,
                   is a tempered distribution.
                    Clearly, if $\varphi \in S(\R)$    then  the actions  $\langle f,\varphi\rangle$  of  $f$
                  as the  functional over $X$ and as the distribution
                   are equal.

                   %Let $\La$ be a discrete set and $\varphi$ be a Shwartz function. Then the $\La$- translates

                   \begin{theorem} Let $\varphi=\hat\Phi\in S(\R)$, where the function  $\Phi$ and its every derivative $\Phi^{(k)}$ satisfies (3). Assume  $\La$ satisfies (1) and (2).
                     If  the space $X^\ast$ does not  contain a  (non-trivial) distribution with spectrum on $\Z$
                                          then the $\La$-translates $\{\varphi(x-\lambda),\lambda\in\La\}$ span  $X$.\end{theorem}

                    \begin{proof}[Sketch of proof]   % Recall that $\varphi \in S(\R)$.
                                      Suppose the family of translates $\{\varphi(x-\lambda),\lambda\in\La\}$
                                   does not span $X$.   Then there is  a non-zero functional
                                   $h\in X^\ast$ such that
$\langle h, \varphi(\cdot-\lambda)\rangle  = 0,  \lambda\in \La$.

                                  As in the proof of Theorem 1, consider the distribution    $\varphi\ast h$.
                                  It is an analytic function on $\R$ which vanishes  on $\La$. Similarly to  Lemma 1 (e), one may check that it belongs to $\hat K$.
                                 Theorem 2 yields    $\varphi \ast  h = 0$.
                                 So, $\Phi \hat h  = 0$.
                                It follows that  the distribution   $\hat h$   is supported by $\Z$  (the zeros of $\Phi$).
                                Contradiction.

\end{proof}

\begin{corollary}                       Let $X$ be a translation-invariant function space on $\R$ such that  $X^\ast$  contains no (non-trivial) algebraic polynomial.
                       Then the conclusion of Theorem 3 holds.
\end{corollary}
                       In  particular,    this can be applied  to Orlicz spaces.



1612.00811.extracted_bib
[AO96]{ao A. Atzmon and A. Olevskii, `Completeness of integer translates in function spaces on $\R$', {J. Approx. Theory 87, (1996), no. 3, 291--327.
1612.00811.extracted_bib
[B51]{B A. Beurling, `On a closure problem', {Ark. Mat. 1, (1951). 301--303. See also: {The Collected Works of Arne Beurling, in: Harmonic Analysis, vol. 2, Harmonic Analysis. Birkhäuser, Boston, 1989. %
1612.00811.extracted_bib
[BM62]{BM1 Beurling, A. and Malliavin, P. On Fourier transforms of measures with compact support. Acta. Math., 107, (1962) 291--309. %
1612.00811.extracted_bib
[BM67]{BM Beurling, A., Malliavin, P. On the closure of characters and the zeros of entire functions. Acta Math., 118, (1967), 79--93.
1612.00811.extracted_bib
[Bl06]{Bl N. Blank, `Generating sets for Beurling algebras', {J. Approx. Theory 140, (2006), no. 1, 61--70.
1612.00811.extracted_bib
[BOU06]{BOU J. Bruna, A. Olevskii and A. Ulanovskii, `Completeness in $L^1(\R)$ of discrete translates', {Rev. Mat. Iberoam. 22, (2006), no. 1, 1--16.
1612.00811.extracted_bib
[FJ98]{FJ F.J. Friedlander and M.S. Joshi, `Introduction to the Theory of Distributions.' Cambridge University Press, 1998.
1612.00811.extracted_bib
[KPS82]{kps S. G. Kre\v{in, Ju. I. Petunin and E. M. Semenov, `Interpolation of linear operators'. Translations of Mathematical Monographs, vol. 54, AMS, Providence, R.I., 1982.
1612.00811.extracted_bib
[LO11]{LO N. Lev and A. Olevskii, `Wiener's "closure of translates" problem and Piatetski--Shapiro's uniqueness phenomenon', {Ann. of Math. (2) 174, (2011), no. 1, 519--541. %
1612.00811.extracted_bib
[NO07]{NO Nitzan, S., Olevskii, A. Sparse exponential systems: completeness with estimates. Israel J. Math. 158, (2007), 205--215.
1612.00811.extracted_bib
[Ol97]{O A. Olevskii, `Completeness in $L^2(\R)$ of almost integer translates', {C. R. Acad. Sci. Paris S$\acute{e$r. I Math. 324, (1997), no. 9, 987--991.
1612.00811.extracted_bib
[OU04]{OU2 A. Olevskii and A. Ulanovskii, `Almost integer translates. Do nice generators exist?', {J. Fourier Anal. Appl. 10, (2004), no. 1, 93--104.
1612.00811.extracted_bib
[OU16]{ou A. Olevskii and A. Ulanovskii, `Functions with Disconnected Spectrum: Sampling, Interpolation, Translates. AMS, University Lecture Series, 65, 2016. %
1612.00811.extracted_bib
[Ul01]{u Ulanovskii, A. Sparse systems of functions closed on large sets in $\R^n$. J. London Math. Soc. (2) 63, (2001), no. 2, 428--440. %Ulanovskii, A. On Landau's phenomenon in $\R^n$. Math. Scand. 88 (2001), no. 1, 72-78.
1612.00811.extracted_bib
[W32]{w N. Wiener, `Tauberian Theorems', {Annals of Math. 33 (1): 1--100, 1932.