###### Abstract

We say that a function $f\in L^{1}(\mathbb{R})$ _tiles_ at level $w$ by a
discrete translation set $\Lambda\subset\mathbb{R}$, if we have
$\sum_{\lambda\in\Lambda}f(x-\lambda)=w$ a.e. In this paper we survey the main
results, and prove several new ones, on the structure of tilings of
$\mathbb{R}$ by translates of a function. The phenomena discussed include
tilings of bounded and of unbounded density, uniform distribution of the
translates, periodic and non-periodic tilings, and tilings at level zero.
Fourier analysis plays an important role in the proofs. Some open problems are
also given.

title = Tiling by translates of a function: results and open problems, author
= Mihail N. Kolountzakis and Nir Lev, plaintextauthor = Mihail N. Kolountzakis
and Nir Lev, keywords = Tiling, translates, spectral gap, uncertainty
principle, quasicrystals., year=2021, number=12, received=22 September 2020,
published=13 September 2021, doi=10.19086/da.28122,

[classification=text]

##  1 Introduction

Let $f$ be a function in $L^{1}(\mathbb{R})$ and let
$\Lambda\subset\mathbb{R}$ be a discrete set. We say that _$f$ tiles
$\mathbb{R}$ at level $w$_ with the translation set $\Lambda$, or that
_$f+\Lambda$ is a tiling of $\mathbb{R}$ at level $w$_ (where $w$ is a
constant), if we have

$\sum_{\lambda\in\Lambda}f(x-\lambda)=w\quad\text{a.e.}$ (1.1)

and the series in (1.1) converges absolutely a.e.

In the same way one can define tiling by translates of an $L^{1}$ function on
$\mathbb{R}^{d}$, or more generally, on any locally compact abelian group. The
finite abelian groups, and in particular the cyclic ones, are an important
class being often considered.

If $f=\mathds{1}_{\Omega}$ is the indicator function of a set $\Omega$, and
$f+\Lambda$ is a tiling at level one, then this means that the translated
copies $\Omega+\lambda$, $\lambda\in\Lambda$, fill the whole space without
overlaps up to measure zero. To the contrary, for tilings by a general real or
complex-valued function $f$, the translated copies may have overlapping
supports and a wider variety of phenomena may (and do) occur.

In dimension one, translational tilings exhibit a stronger structure than in
higher dimensions, and there are interesting questions as to how rigid this
structure must be, e.g. how close the translation set $\Lambda$ is to being
periodic, or to being constructed out of periodic sets, or, at an even more
basic level, to being uniformly distributed in $\mathbb{R}$.

The subject has been studied by several authors, see, in particular, [LM91],
[KL96], [Kol04], [KL16], [Liu18], [Lev20]. The aim of this paper is to survey
the results obtained in earlier works, as well as to prove several new
results. At the end of the paper, some open problems are also given.

In this section we survey the main results on the structure of tilings by
translates of a function on $\mathbb{R}$, and we state the new results that
will be proved in this paper.

###  1.1 Tiling and density

We say that a set $\Lambda\subset\mathbb{R}$ has _bounded density_ if

$\sup_{x\in\mathbb{R}}\\#(\Lambda\cap[x,x+1))<+\infty.$ (1.2)

The set $\Lambda$ is said to be _uniformly distributed_ if there is a number
$D(\Lambda)$ satisfying

$\\#(\Lambda\cap[x,x+r))=D(\Lambda)\cdot r+o(r),\quad r\to+\infty$ (1.3)

uniformly with respect to $x\in\mathbb{R}$. In this case, $D(\Lambda)$ is
called the _uniform density_ of $\Lambda$.

The following result establishes a connection between tiling and density:

######  Theorem 1.1 ([KL96]).

Let $f+\Lambda$ be a tiling at some nonzero level $w$, where $f\in
L^{1}(\mathbb{R})$ and where $\Lambda\subset\mathbb{R}$ is a set of bounded
density. Then $f$ has nonzero integral, and $\Lambda$ has a uniform density
given by $D(\Lambda)=w\cdot(\int f)^{-1}$.

This was proved in [KL96, Lemma 2.3] for a weaker notion of density111In
[KL96] the density $d(\Lambda)$ of a set $\Lambda\subset\mathbb{R}$ is defined
by $d(\Lambda):=\lim_{r\to+\infty}\\#(\Lambda\cap(-r,r))/(2r)$., but a minor
adjustment to the proof in fact yields the stronger statement above for the
uniform density. A similar result is true also in $\mathbb{R}^{d}$.

###  1.2 Tiling at level zero

It is not known whether Theorem 1.1 has an analog for tilings of bounded
density _at level zero_. It was conjectured in [KL96, p. 660] that if
$f+\Lambda$ is such a tiling, then $f$ must have zero integral. In [KL96,
Lemma 2.4] this was proved under the extra assumption that $f$ has compact
support. Here we will prove that the conclusion is true in the general case:

######  Theorem 1.2.

Let $f+\Lambda$ be a tiling at level zero, where $f\in L^{1}(\mathbb{R})$ and
where $\Lambda\subset\mathbb{R}$ is a nonempty set of bounded density. Then
$f$ must have zero integral.

Do there exist tilings $f+\Lambda$ at level zero such that the set $\Lambda$
has density zero? Theorem 1.1 does not exclude such a possibility. In fact, in
dimensions two and higher it is easy to exhibit tilings of this kind. For
instance, in $\mathbb{R}^{2}$ one may take $f(x,y)=\varphi(x)\psi(y)$,
$\Lambda=\Gamma\times\\{0\\}$, where $\varphi,\psi\in L^{1}(\mathbb{R})$ and
$\varphi+\Gamma$ is a tiling of $\mathbb{R}$ at level zero.

We will show, however, that this is _not_ the case in dimension one. To state
the result, we recall that a set $\Lambda\subset\mathbb{R}$ is said to be
_relatively dense_ if there is $r>0$ such that any interval $[x,x+r)$ contains
at least one point from $\Lambda$. We will prove the following:

######  Theorem 1.3.

Let $f+\Lambda$ be a tiling at level zero, where $f\in L^{1}(\mathbb{R})$ is
nonzero and $\Lambda\subset\mathbb{R}$ is a nonempty set of bounded density.
Then $\Lambda$ must be a relatively dense set.

In particular this implies that $\Lambda$ cannot have density zero.

Does it follow from the assumptions in Theorem 1.3 that $\Lambda$ has a
uniform (positive) density $D(\Lambda)$ ? The answer to this question is not
known. The problem is nontrivial due to the existence of translation sets
$\Lambda$ that admit only tilings at level zero, so that Theorem 1.1 does not
apply to these sets:

######  Theorem 1.4 ([Lev20]).

There exists a nonempty set $\Lambda\subset\mathbb{R}$ of bounded density
which admits tilings $f+\Lambda$ with nonzero $f\in L^{1}(\mathbb{R})$, but
any such a tiling is necessarily a tiling at level zero.

###  1.3 Periodic tilings

We say that a set $\Lambda\subset\mathbb{R}$ has a _periodic structure_ if it
can be represented as a disjoint union of finitely many arithmetic
progressions, namely

$\Lambda=\biguplus_{j=1}^{N}(a_{j}\mathbb{Z}+b_{j})$ (1.4)

where $a_{j},b_{j}$ are real numbers and $a_{j}>0$. The sets $\Lambda$ with
this structure constitute the basic examples of translation sets for tilings
of $\mathbb{R}$. Indeed, one can check that if

$f=\mathds{1}_{[0,a_{1}]}\ast\mathds{1}_{[0,a_{2}]}\ast\dots\ast\mathds{1}_{[0,a_{N}]}$
(1.5)

and $\Lambda$ is given by (1.4), then $f+\Lambda$ is a tiling at some positive
level $w$.

The last example shows that any set $\Lambda$ of the form (1.4) admits a
tiling by a _compactly supported_ function $f\in L^{1}(\mathbb{R})$. A result
first proved in [LM91] and rediscovered in [KL96] asserts that there are no
other translation sets $\Lambda$ with this property:

######  Theorem 1.5 ([LM91], [KL96]).

Let $f\in L^{1}(\mathbb{R})$ be nonzero and have compact support. If $f$ tiles
at some level $w$ with a translation set $\Lambda$ of bounded density, then
$\Lambda$ has a periodic structure, namely, it must be of the form (1.4).

The proof of this result is based on Cohen’s characterization of idempotent
measures in locally compact abelian groups. The group on which Cohen’s theorem
is used in the proof is the _Bohr compactification_ of the real line (see also
[Mey70, p. 25]).

A discrete set $\Lambda\subset\mathbb{R}$ is said to have _finite local
complexity_ if $\Lambda$ can be enumerated as a sequence $\\{\lambda_{n}\\}$,
$n\in\mathbb{Z}$, such that $\lambda_{n}<\lambda_{n+1}$ and the successive
differences $\lambda_{n+1}-\lambda_{n}$ take only finitely many different
values. The following result establishes that tilings of finite local
complexity must be periodic, even if $f$ does not have compact support:

######  Theorem 1.6 ([IK13], [KL16]).

Let $\Lambda\subset\mathbb{R}$ have finite local complexity. If $f\in
L^{1}(\mathbb{R})$ is nonzero and $f+\Lambda$ is a tiling at some level $w$,
then $\Lambda$ must be a periodic set, namely, it has the form
$\Lambda=a\mathbb{Z}+\\{b_{1},\dots,b_{N}\\}$.

###  1.4 Non-periodic tilings

The papers [LM91], [KL96] leave the following question open: Does there exist
any set $\Lambda\subset\mathbb{R}$ _not_ of periodic structure, which can tile
with some function $f\in L^{1}(\mathbb{R})$ of unbounded support? Such a set
$\Lambda$ cannot have finite local complexity by Theorem 1.6. We settled this
question affirmatively in [KL16]:

######  Theorem 1.7 ([KL16]).

There exists a tiling $f+\Lambda$ at level one, where $f\in L^{1}(\mathbb{R})$
and $\Lambda\subset\mathbb{R}$ has bounded density, but such that $\Lambda$
has no periodic structure, i.e. the set $\Lambda$ is not of the form (1.4).

The proof was based on the implicit function method due to Kargaev [Kar82],
and it yields a set $\Lambda$ which is a small perturbation of the integers.
The proof moreover allows to choose the function $f$ in the Schwartz class.
However it yields a function $f$ satisfying
$\operatorname{supp}(f)=\mathbb{R}$, where $\operatorname{supp}(f)$ is the
closed support of $f$ (the smallest closed set such that $f$ vanishes a.e. on
its complement).

In this paper we will prove a stronger version of Theorem 1.7, which
establishes the existence of non-periodic tilings $f+\Lambda$ of bounded
density, such that $f$ has “sparse” support. Precisely, we will show that the
support (which must be unbounded, due to Theorem 1.5) can be localized inside
any given set $\Omega\subset\mathbb{R}$ which contains arbitrarily long
intervals:

######  Theorem 1.8.

There is a discrete set $\Lambda\subset\mathbb{R}$ of bounded density, but
which is not of the form (1.4), with the following property: given any scalar
$w$, and any set $\Omega\subset\mathbb{R}$ which contains arbitrarily long
intervals, one can find a nonzero $f\in L^{1}(\mathbb{R})$,
$\operatorname{supp}(f)\subset\Omega$, such that $f+\Lambda$ is a tiling at
level $w$.

We note that the set $\Omega$ in this theorem may be chosen very sparse, for
example, one may take $\Omega=\bigcup_{j=1}^{\infty}[\tau_{j},\tau_{j}+j]$,
where the numbers $\tau_{j}$ grow arbitrarily fast.

Theorem 1.8 implies in particular the existence of non-periodic tilings by a
function $f$ such that $\operatorname{supp}(f)\subset[0,+\infty)$, i.e. the
support is _bounded from below_ (while it cannot be bounded from both above
and below, due to Theorem 1.5).

The proof of Theorem 1.8 relies on a recent result due to Kurasov and Sarnak
[KS20], who constructed a new type of _crystalline measures_ on $\mathbb{R}$.
These are pure point measures with discrete closed support, whose Fourier
transform is a measure of the same type.

###  1.5 Tilings of unbounded density

It seems that very little attention has been paid to tilings which are _not_
of bounded density. In fact, we are not aware of any example of such a tiling
in the literature. In [KL96, Example 7.1] some examples are given in a more
general context, where the points of the translation set $\Lambda$ are endowed
with nonnegative integer multiplicities (so that $\Lambda$ is actually not a
set, but a multi-set).

We will prove that there exist tilings of unbounded density in the proper
sense. Let us say that a set $\Lambda\subset\mathbb{R}$ has _tempered growth_
if there is $N$ such that

$\\#(\Lambda\cap(-r,r))=O(r^{N}),\quad r\to+\infty.$ (1.6)

######  Theorem 1.9.

There is a set $\Lambda\subset\mathbb{R}$ which is not of bounded density but
has tempered growth, with the following property: given any scalar $w$ one can
find a nonzero function $f$ in the Schwartz class, such that $f+\Lambda$ is a
tiling at level $w$.

Moreover, in our example the set $\Lambda$ is contained in a small
neighborhood of the integers. The construction is done using a variant of
Kargaev’s implicit function method.

It follows from [KL96, Lemma 2.1] that the function $f$ in Theorem 1.9 must
change sign, i.e. $f$ cannot be chosen nonnegative. This indicates that
cancellations are playing a decisive role in the tiling. We can even prove a
stronger claim:

######  Theorem 1.10.

Let $f+\Lambda$ be a tiling at some level $w$, where the set
$\Lambda\subset\mathbb{R}$ is not of bounded density but has tempered growth,
and where $f$ is a function in the Schwartz class. Then $f$ must have zero
integral.

It may seem counter-intuitive at first glance that one can tile $\mathbb{R}$
at a nonzero level $w$ by translates of a Schwartz function $f$ whose integral
is zero.

###  1.6 Organization of the paper

The rest of the paper is organized as follows.

In Section 2 some preliminary background is given, and Fourier analytic
conditions for tiling are discussed.

In Section 3 we prove that if $f+\Lambda$ is a tiling at level zero, where
$f\in L^{1}(\mathbb{R})$ is a nonzero function and $\Lambda\subset\mathbb{R}$
is a nonempty set of bounded density, then $f$ has zero integral (Theorem 1.2)
and $\Lambda$ is relatively dense (Theorem 1.3).

In Section 4 we establish the existence of non-periodic tilings $f+\Lambda$ of
bounded density, such that the function $f$ has “sparse” support (Theorem
1.8).

In Section 5 we construct tilings $f+\Lambda$ of unbounded density, such that
$\Lambda$ has tempered growth and $f$ is in the Schwartz class (Theorem 1.9).
We also show that in any such a tiling, the function $f$ must have zero
integral (Theorem 1.10).

In the last section, Section 6, we pose some open problems.

##  2 Preliminaries. Fourier analytic conditions for tiling.

It is well-known that in the study of translational tilings, Fourier analysis
plays an important role, see e.g. [Kol04]. If $f\in L^{1}(\mathbb{R})$ then
its Fourier transform is defined by

$\widehat{f}(t)=\int_{\mathbb{R}}f(x)\,e^{-2\pi itx}\,dx.$ (2.1)

If $\alpha$ is a tempered distribution on $\mathbb{R}$, then $\alpha(\varphi)$
denotes the action of $\alpha$ on a Schwartz function $\varphi$. The Fourier
transform $\widehat{\alpha}$ is defined by
$\widehat{\alpha}(\varphi)=\alpha(\widehat{\varphi})$.

If $\alpha$ is a tempered distribution on $\mathbb{R}$, and if $\varphi$ is a
Schwartz function, then the convolution $\alpha\ast\varphi$ is a tempered
distribution whose Fourier transform is
$\widehat{\alpha}\cdot\widehat{\varphi}$.

We use $\operatorname{supp}(\alpha)$ to denote the closed support of a
tempered distribution $\alpha$.

If $f\in L^{1}(\mathbb{R})$ then $\operatorname{supp}(f)$ is the smallest
closed set such that $f$ vanishes a.e. on its complement. This set coincides
with the support of $f$ in the distributional sense.

For more details on distribution theory we refer the reader to [Rud91].

###  2.1 Necessary conditions for tiling

For a discrete set $\Lambda\subset\mathbb{R}$ we define the measure

$\delta_{\Lambda}:=\sum_{\lambda\in\Lambda}\delta_{\lambda}.$ (2.2)

The tiling condition (1.1) can then be restated as

$f\ast\delta_{\Lambda}=w\quad\text{a.e.}$ (2.3)

If $\Lambda$ has bounded density then the measure $\delta_{\Lambda}$ is a
tempered distribution on $\mathbb{R}$. So, at least formally, taking the
Fourier transform of both sides of (2.3) yields

$\widehat{f}\cdot\widehat{\delta}_{\Lambda}=w\cdot\delta_{0}.$ (2.4)

If $f$ is a Schwartz function, then condition (2.4) makes sense and it is
equivalent to (2.3). To the contrary, for an arbitrary function $f\in
L^{1}(\mathbb{R})$ (not assumed to be in the Schwartz class) the product
$\widehat{f}\cdot\widehat{\delta}_{\Lambda}$ is not well-defined, and the
condition (2.4) can only be interpreted as a heuristic principle.

The following result, inspired by the heuristic condition (2.4), provides a
necessary condition for tiling which holds for any $f\in L^{1}(\mathbb{R})$.

######  Theorem 2.1 ([KL16]).

Let $f\in L^{1}(\mathbb{R})$, and $\Lambda\subset\mathbb{R}$ be a discrete set
of bounded density. If $f+\Lambda$ is a tiling at some level $w$, then

$\operatorname{supp}(\widehat{\delta}_{\Lambda})\setminus\\{0\\}\subset\\{\widehat{f}=0\\}.$
(2.5)

The proof of this result is based on Wiener’s tauberian theorem. In the
earlier works [KL96], [Kol00a], [Kol00b] the result was proved under various
extra assumptions.

###  2.2 Sufficient conditions for tiling

One may ask whether Theorem 2.1 admits a converse, i.e. if the condition (2.5)
implies that $f+\Lambda$ is a tiling at some level $w$. If the distribution
$\widehat{\delta}_{\Lambda}$ happens to be a _measure_ on $\mathbb{R}$, then
the product $\widehat{f}\cdot\widehat{\delta}_{\Lambda}$ is a well-defined
measure and the condition (2.4) makes sense. In this case, the two conditions
(2.4) and (2.5) are equivalent, and can be shown to imply that $f+\Lambda$ is
a tiling:

######  Theorem 2.2.

Let $\Lambda\subset\mathbb{R}$ have bounded density and suppose that
$\widehat{\delta}_{\Lambda}$ is a locally finite measure. If $f\in
L^{1}(\mathbb{R})$ satisfies (2.5) then $f+\Lambda$ is a tiling at level
$w=\widehat{\delta}_{\Lambda}(\\{0\\})\cdot\int f$.

A proof of this result is given below. In [KL96] the result was proved under
the extra assumption that $\widehat{f}$ is a smooth function.

As an example, Theorem 2.2 applies if the set $\Lambda$ has a periodic
structure, i.e. if it has the form (1.4). In this case
$\widehat{\delta}_{\Lambda}$ is a (pure point) measure by Poisson’s summation
formula, and the condition (2.5) is both necessary and sufficient for a
function $f\in L^{1}(\mathbb{R})$ to tile at some level $w$ with the
translation set $\Lambda$.

Theorem 2.2 leaves open, though, the question as to whether the condition
(2.5) is sufficient for tiling in the general case, i.e. for an arbitrary set
$\Lambda\subset\mathbb{R}$ of bounded density. This question was addressed
recently in [Lev20] where it was answered in the negative:

######  Theorem 2.3 ([Lev20]).

There is a set $\Lambda\subset\mathbb{R}$ of bounded density and a function
$f\in L^{1}(\mathbb{R})$, such that (2.5) is satisfied however $f+\Lambda$ is
not a tiling at any level.

The proof of this result is based on the relation of the problem to
Malliavin’s _non-spectral synthesis_ example.

Theorem 2.3 implies that a converse to Theorem 2.1 can only be valid under
certain extra assumptions. One example of such a converse is given by Theorem
2.2. Another example is the following result:

######  Theorem 2.4.

Let $f\in L^{1}(\mathbb{R})$, and let $\Lambda\subset\mathbb{R}$ be a discrete
set of bounded density. If the set
$\operatorname{supp}(\widehat{\delta}_{\Lambda})\setminus\\{0\\}$ is closed
and is disjoint from $\operatorname{supp}(\widehat{f})$, then $f+\Lambda$ is a
tiling at some level $w$.

We will not use this result in the paper and we do not include its proof. For
other results of similar type see [Kol00a, Theorem 3], [Kol00b, Theorem 5].

###  2.3 Translation-bounded measures

A measure $\mu$ on $\mathbb{R}$ satisfying the condition

$\sup_{x\in\mathbb{R}}|\mu|([x,x+1))<\infty$ (2.6)

is said to be _translation-bounded_. If a measure $\mu$ is translation-
bounded, then it is a tempered distribution. If $\mu$ is a translation-bounded
measure on $\mathbb{R}$, and if $\nu$ is a finite measure on $\mathbb{R}$,
then the convolution $\mu\ast\nu$ is a translation-bounded measure.

######  Theorem 2.5.

Let $\mu$ be a translation-bounded measure on $\mathbb{R}$, and suppose that
$\widehat{\mu}$ is a locally finite measure. If $\nu$ is a finite measure on
$\mathbb{R}$, then the Fourier transform of the convolution $\mu\ast\nu$ is
the measure $\widehat{\mu}\cdot\widehat{\nu}$.

In particular, let $\Lambda\subset\mathbb{R}$ be a discrete set of bounded
density, and let $f\in L^{1}(\mathbb{R})$. Then the measure $\delta_{\Lambda}$
is translation-bounded, and the convolution $f\ast\delta_{\Lambda}$ is the sum
of the series $\sum_{\lambda\in\Lambda}f(x-\lambda)$ which converges
absolutely a.e., see [KL96, Lemma 2.2]. If the distribution
$\widehat{\delta}_{\Lambda}$ is assumed to be a locally finite measure, then
using Theorem 2.5 we obtain that the three conditions (2.3), (2.4) and (2.5)
are equivalent. We thus see that Theorem 2.2 is a consequence of Theorem 2.5.

Theorem 2.5 should be known to experts but we could not find a proof in the
literature, so we include one below for completeness.

###### Proof of Theorem 2.5.

We will use $\sigma$ to denote the measure $\widehat{\mu}$. The assertion of
the theorem means that if $\psi$ is a Schwartz function with compact support
then

$\int_{\mathbb{R}}\widehat{\psi}(x)\,d(\mu\ast\nu)(x)=\int_{\mathbb{R}}\psi(t)\,\widehat{\nu}(t)\,d\sigma(t).$
(2.7)

Let $\chi$ be a Schwartz function whose Fourier transform $\widehat{\chi}$ is
non-negative, has compact support, $\int\widehat{\chi}(t)dt=1$, and for each
$\varepsilon>0$ let $\chi_{\varepsilon}(x):=\chi(\varepsilon x)$. Let
$h_{\varepsilon}:=(\psi\cdot\widehat{\nu})\ast\widehat{\chi}_{\varepsilon}$,
then $h_{\varepsilon}$ is an infinitely smooth function with compact support.
As $\varepsilon\to 0$, the function $h_{\varepsilon}$ remains supported on a
certain interval $I=[a,b]$ that does not depend on $\varepsilon$, and
$h_{\varepsilon}$ converges to $\psi\cdot\widehat{\nu}$ uniformly on $I$.
Hence

$\lim_{\varepsilon\to 0}\widehat{\mu}(h_{\varepsilon})=\lim_{\varepsilon\to
0}\int_{\mathbb{R}}h_{\varepsilon}(t)d\sigma(t)=\int_{\mathbb{R}}\psi(t)\,\widehat{\nu}(t)\,d\sigma(t).$
(2.8)

The function $\psi$ is the Fourier transform of some function $\varphi$ in the
Schwartz class. Let
$g_{\varepsilon}:=(\varphi\ast\nu)\cdot\chi_{\varepsilon}$, then
$g_{\varepsilon}$ is a smooth function in $L^{1}(\mathbb{R})$ and we have
$\widehat{g}_{\varepsilon}=h_{\varepsilon}$. Since $h_{\varepsilon}$ belongs
to the Schwartz space, the same is true for $g_{\varepsilon}$, and it follows
that

$\widehat{\mu}(h_{\varepsilon})=\mu(\widehat{h}_{\varepsilon})=\int_{\mathbb{R}}{g_{\varepsilon}}(-x)\,d\mu(x)=\int_{\mathbb{R}}(\varphi\ast\nu)(-x)\,\chi_{\varepsilon}(-x)\,d\mu(x).$
(2.9)

We observe that $|\chi_{\varepsilon}(-x)|\leqslant 1$ and
$\chi_{\varepsilon}(-x)\to 1$ pointwise as $\varepsilon\to 0$. We now wish to
apply the dominated convergence theorem to (2.9). Using Fubini’s theorem we
obtain

$\int_{\mathbb{R}}(|\varphi|\ast|\nu|)(-x)\,|d\mu|(x)=\int_{\mathbb{R}}|\varphi(-x)|\,d(|\mu|\ast|\nu|)(x).$
(2.10)

The function $\varphi$ has fast decay being in the Schwartz class, while
$|\mu|\ast|\nu|$ is a translation-bounded measure, hence the integral in
(2.10) converges. We may therefore apply to (2.9) the dominated convergence
theorem, which yields

$\lim_{\varepsilon\to
0}\widehat{\mu}(h_{\varepsilon})=\int_{\mathbb{R}}(\varphi\ast\nu)(-x)\,d\mu(x)=\int_{\mathbb{R}}\varphi(-x)\,d(\mu\ast\nu)(x).$
(2.11)

But $\varphi(-x)=\widehat{\psi}(x)$, so we see that (2.7) follows from (2.8)
and (2.11) as needed. ∎

##  3 Tiling at level zero

In this section we prove that if $f+\Lambda$ is a tiling at level zero, where
$f\in L^{1}(\mathbb{R})$ is nonzero and $\Lambda\subset\mathbb{R}$ is nonempty
and has bounded density, then $f$ has zero integral (Theorem 1.2) and
$\Lambda$ is a relatively dense set (Theorem 1.3).

###  3.1 The function $f$ has zero integral

We begin with Theorem 1.2, for which we give two proofs. The first proof is
Fourier analytic, and is based on the following result:

######  Theorem 3.1 ([KL16]).

Let $f\in L^{1}(\mathbb{R})$ have nonzero integral. If
$\Lambda\subset\mathbb{R}$ is a set of bounded density and $f+\Lambda$ is a
tiling at some level $w$, then there is $a>0$ such that
$\widehat{\delta}_{\Lambda}=c\cdot\delta_{0}$ in $(-a,a)$, where
$c=w\cdot(\int f)^{-1}$.

This result follows from the proof of [KL16, Theorem 4.1] and it is stated in
that paper as a remark on p. 4598. The proof of Theorem 3.1 is based on
Wiener’s tauberian theorem.

###### First proof of Theorem 1.2.

Let $f+\Lambda$ be a tiling at level $w=0$, and suppose to the contrary that
$f$ has nonzero integral. Then by Theorem 3.1 there is $a>0$ such that
$\widehat{\delta}_{\Lambda}=c\cdot\delta_{0}$ in $(-a,a)$, where
$c=w\cdot(\int f)^{-1}$. Since $w=0$ this means that
$\widehat{\delta}_{\Lambda}$ vanishes in a neighborhood $(-a,a)$ of the
origin. We will show that this cannot happen.

Indeed, choose a Schwartz function $\varphi$ such that
$\operatorname{supp}(\varphi)\subset(-a,a)$ and $\widehat{\varphi}>0$. Then we
have $\widehat{\delta}_{\Lambda}(\varphi)=0$ since
$\widehat{\delta}_{\Lambda}$ vanishes in a neighborhood of
$\operatorname{supp}(\varphi)$. On the other hand,

$\widehat{\delta}_{\Lambda}(\varphi)=\delta_{\Lambda}(\widehat{\varphi})=\sum_{\lambda\in\Lambda}\widehat{\varphi}(\lambda)>0,$

since $\Lambda$ is nonempty and $\widehat{\varphi}$ is everywhere positive. We
thus arrive at a contradiction. ∎

As a remark, we record here the following observation:

######  Lemma 3.2.

Let $\Lambda\subset\mathbb{R}$ be a nonempty set of bounded density, and
suppose that there is $a>0$ such that
$\widehat{\delta}_{\Lambda}=c\cdot\delta_{0}$ in $(-a,a)$. Then $c>0$, and
$\Lambda$ has a uniform density given by $D(\Lambda)=c$.

###### Proof.

Let $\varphi$ be a Schwartz function such that $\varphi>0$, $\int\varphi=1$
and $\operatorname{supp}(\widehat{\varphi})\subset(-a,a)$. Then we have
$\widehat{\varphi}\cdot\widehat{\delta}_{\Lambda}=c\cdot\delta_{0}$ and hence
$\varphi+\Lambda$ is a tiling at level $c$. Since $\varphi$ is a positive
function and $\Lambda$ is nonempty, the tiling level $c$ must also be
positive. Finally, we obtain from Theorem 1.1 that the set $\Lambda$ has a
uniform density given by $D(\Lambda)=c$. ∎

Next we give our second proof of Theorem 1.2. This proof, which does not
involve Fourier analysis, is based on the following result due to Ruzsa and
Székely [RS83].

######  Theorem 3.3 ([RS83]).

Let $f\in L^{1}(\mathbb{R})$ be real-valued with $\int f>0$. Then there is a
nonnegative (nonzero) $g\in L^{1}(\mathbb{R})$ such that the convolution
$f\ast g$ is nonnegative a.e.

In fact this is proved in [RS83] not only for functions on $\mathbb{R}$, but
on a wider class of locally compact abelian groups, which in particular
includes also $\mathbb{R}^{d}$ for every $d\geqslant 1$.

###### Second proof of Theorem 1.2.

Let $f+\Lambda$ be a tiling at level zero, and suppose to the contrary that
$\int f$ is nonzero. By replacing $f(x)$ with the function
$\operatorname{Re}\big{\\{}(\int f)^{-1}f(x)\big{\\}}$, we may assume that $f$
is real-valued and that $\int f>0$. So we can use Theorem 3.3 to find a
nonnegative, nonzero $g\in L^{1}(\mathbb{R})$ such that $f\ast g$ is
nonnegative a.e. Notice that also the function $f\ast g$ is nonzero, since we
have $\int(f\ast g)=(\int f)(\int g)>0$.

On the other hand we have $f\ast\delta_{\Lambda}=0$ a.e., and the measure
$\delta_{\Lambda}$ is translation-bounded since $\Lambda$ has bounded density.
Using Fubini’s theorem this implies that

$(f\ast g)\ast\delta_{\Lambda}=g\ast(f\ast\delta_{\Lambda})=g\ast
0=0\quad\text{a.e.,}$

hence $(f\ast g)+\Lambda$ is a tiling at level zero. But this is a
contradiction, since $f\ast g$ is a nonnegative, nonzero function in
$L^{1}(\mathbb{R})$, and $\Lambda$ is a nonempty set. ∎

###  3.2 The set $\Lambda$ is relatively dense

We now move on to the proof of Theorem 1.3. First we note that this theorem
_cannot_ be deduced based on Lemma 3.2 above, since there exist tilings
$f+\Lambda$ at level zero such that $\widehat{\delta}_{\Lambda}$ is _not_ a
scalar multiple of $\delta_{0}$ in any neighborhood of the origin; see [Lev20,
Section 5] for a construction of such tilings.

We fix the following terminology: given a sequence of measures $\\{\mu_{n}\\}$
on $\mathbb{R}$, we say that the sequence is _uniformly translation-bounded_
if there is a constant $M>0$ not depending on $n$, such that
$\sup_{x\in\mathbb{R}}|\mu_{n}|([x,x+1))\leqslant M$ for every $n$.

If $\\{\mu_{n}\\}$ is a uniformly translation-bounded sequence of measures on
$\mathbb{R}$, then we say that $\mu_{n}$ _converges vaguely_ to a measure
$\mu$ if for every continuous, compactly supported function $\varphi$ we have
$\int\varphi\,d\mu_{n}\to\int\varphi\,d\mu$ (see, for instance, [Fol99,
Section 7.3]). In this case, the vague limit $\mu$ must also be a translation-
bounded measure. For a uniformly translation-bounded sequence of measures
$\\{\mu_{n}\\}$ to converge vaguely, it is necessary and sufficient that
$\\{\mu_{n}\\}$ converge in the space of tempered distributions. From any
uniformly translation-bounded sequence of measures $\\{\mu_{n}\\}$ one can
extract a vaguely convergent subsequence $\\{\mu_{n_{j}}\\}$.

###### Proof of Theorem 1.3.

Let $f+\Lambda$ be a tiling at level zero, and suppose to the contrary that
$\Lambda$ is not relatively dense. Then for each $r>0$ one can find an open
interval $I(r)$ of length $r$ which is disjoint from $\Lambda$. By translating
the interval $I(r)$ we may assume that one of its endpoints lies in $\Lambda$
(since the set $\Lambda$ is nonempty). Then for arbitrarily large values of
$r$ the right endpoint of $I(r)$ is in $\Lambda$, or for arbitrarily large
values of $r$ the left endpoint is in $\Lambda$. We will assume that the
former is the case (the latter case can be treated similarly). It follows that
there exist two sequences $r_{j}\to+\infty$ and $\lambda_{j}\in\Lambda$, such
that for each $j$ the interval $I_{j}:=(\lambda_{j}-r_{j},\lambda_{j})$ does
not intersect $\Lambda$.

Define $\Lambda_{j}:=\Lambda-\lambda_{j}$. Since $\Lambda$ has bounded
density, the measures $\delta_{\Lambda_{j}}$ are uniformly translation-
bounded. By passing to a subsequence if necessary, we may assume that
$\delta_{\Lambda_{j}}$ converges vaguely to some (also translation-bounded)
measure $\mu$ on $\mathbb{R}$. The vague limit $\mu$ is supported on
$[0,+\infty)$, since $\delta_{\Lambda_{j}}$ vanishes on $(-r_{j},0)$.
Moreover, each measure $\delta_{\Lambda_{j}}$ is positive and has a unit mass
at the origin, which implies that $\mu$ is also positive and has an atom at
the origin, of mass at least one. In particular the measure $\mu$ is nonzero.

On the other hand, since $f+\Lambda$ is a tiling (at level zero), then by
Theorem 2.1 the support of $\widehat{\delta}_{\Lambda}$ is contained in
$\\{\widehat{f}=0\\}\cup\\{0\\}$. Since $\widehat{f}$ is a nonzero continuous
function, it follows that there is an open interval $(a,b)$ on which
$\widehat{\delta}_{\Lambda}$ vanishes. In turn this implies that
$\widehat{\delta}_{\Lambda_{j}}$ also vanishes on $(a,b)$ for each $j$. But as
the sequence $\widehat{\delta}_{\Lambda_{j}}$ converges to $\widehat{\mu}$ in
the distributional sense, we obtain that $\widehat{\mu}$ vanishes on $(a,b)$
as well.

We conclude that $\mu$ is a nonzero, translation-bounded, positive measure on
$\mathbb{R}$, supported on $[0,+\infty)$ and whose Fourier transform
$\widehat{\mu}$ vanishes on some open interval $(a,b)$. But this contradicts
classical results on boundary values of functions analytic in the upper half-
plane. To be more concrete: choose two Schwartz functions $\varphi,\psi$ such
that $\varphi>0$,
$\operatorname{supp}(\widehat{\varphi})\subset(-\delta,\delta)$, $\psi$ is
nonnegative, $\psi$ is supported on $[0,+\infty)$ and $\int\psi=1$. Then
$h:=(\mu\cdot\varphi)\ast\psi$ is a nonzero function belonging to
$L^{1}(\mathbb{R})$, $\operatorname{supp}(h)\subset[0,+\infty)$, and the
Fourier transform
$\widehat{h}=(\widehat{\mu}\ast\widehat{\varphi})\cdot\widehat{\psi}$ is also
in $L^{1}(\mathbb{R})$ and $\widehat{h}$ vanishes on some nonempty open
interval provided that $\delta$ is small enough. This contradicts the
uniqueness theorem for functions in the Hardy space $H^{1}$, see e.g. [Gar07,
Chapter II]. ∎

##  4 Non-periodic tilings by functions with sparse support

In this section we establish the existence of non-periodic tilings $f+\Lambda$
of bounded density, such that the function $f$ has “sparse” support (Theorem
1.8).

Recall that the first example of a tiling $f+\Lambda$ such that $f\in
L^{1}(\mathbb{R})$, $\Lambda\subset\mathbb{R}$ has bounded density, but the
set $\Lambda$ does _not_ have a periodic structure, was given in [KL16]. The
proof was based on the implicit function method due to Kargaev [Kar82], and it
yields a set $\Lambda$ which is a small perturbation of the integers. However,
in this example the function $f$ is _analytic_ , which implies that
$\operatorname{supp}(f)=\mathbb{R}$.

In order to prove Theorem 1.8 we will use a different approach, which is based
on two main ingredients. The first one is a recent construction from [KS20] of
a new type of _crystalline measures_ on $\mathbb{R}$. The second main
ingredient is a result concerning interpolation of discrete functions by
continuous ones with a sparse spectrum.

###  4.1 Crystalline measures

A tempered distribution $\mu$ on $\mathbb{R}$ satisfying

$\mu=\sum_{\lambda\in\Lambda}a_{\lambda}\delta_{\lambda},\quad\widehat{\mu}=\sum_{s\in
S}b_{s}\delta_{s}$ (4.1)

(i.e. both $\mu$ and $\widehat{\mu}$ are pure point measures), where $\Lambda$
and $S$ are discrete, closed sets in $\mathbb{R}$, is called a _crystalline
measure_ [Mey16]. This notion plays a role in the mathematical theory of
quasicrystals, i.e. atomic arrangements having a discrete diffraction pattern.

The classical example of a crystalline measure is $\mu=\delta_{\mathbb{Z}}$,
which satisfies $\widehat{\mu}=\mu$ by Poisson’s summation formula. More
generally, the measure $\delta_{\Lambda}$ is crystalline for any set $\Lambda$
of the form (1.4), i.e. any set $\Lambda\subset\mathbb{R}$ having a periodic
structure.

There exist also examples of crystalline measures on $\mathbb{R}$, whose
support is _not_ contained in any set $\Lambda$ with a periodic structure.
Constructions of such examples, using different approaches, were given in the
papers [LO16], [Kol16], [Mey16], [Mey17], [RV19].

Recently, new progress was achieved by Kurasov and Sarnak [KS20] who
constructed examples of crystalline measures on $\mathbb{R}$ enjoying some
remarkable properties, answering several questions left open by the above
mentioned papers. To state the result, we recall the terminology introduced in
Section 1.5 above:

######  Definition 4.1.

We say that a set $S\subset\mathbb{R}$ has _tempered growth_ if there is $N$
such that

$\\#(S\cap(-r,r))=O(r^{N}),\quad r\to+\infty.$ (4.2)

We note that if a set $S\subset\mathbb{R}$ has tempered growth then the
measure $\delta_{S}$ is a tempered distribution, and also the converse is
true.

######  Theorem 4.2 ([KS20]).

There exist crystalline measures $\mu$ of the form (4.1) with the following
properties:

  1. (i)

$\Lambda$ is a set of bounded density;

  2. (ii)

$a_{\lambda}=1$ for all $\lambda\in\Lambda$, i.e. we have
$\mu=\delta_{\Lambda}$;

  3. (iii)

$\Lambda$ does not have a periodic structure, i.e. it is not of the form
(1.4);

  4. (iv)

$S$ is a set of tempered growth.

Actually, the crystalline measures constructed in [KS20] have even stronger
properties than stated above – we only mentioned the properties that will be
used in this paper. In the more recent works [Mey20], [OU20], alternative
approaches to the construction of crystalline measures with these properties
can be found.

###  4.2 Interpolation by functions with a sparse spectrum

We now turn to the second main ingredient in our proof of Theorem 1.8. It can
be described in the context of the classical uncertainty principle, which
states that a nonzero function $f$ and its Fourier transform $\widehat{f}$
cannot be “too small” at the same time. This general principle has many
concrete versions, see [HJ94].

In particular, using complex analysis one can show that if $f\in
L^{1}(\mathbb{R})$ has compact support, and if $\widehat{f}$ vanishes on a set
$S\subset\mathbb{R}$ satisfying

$\limsup_{r\to+\infty}\frac{\\#(S\cap(-r,r))}{r}=+\infty,$ (4.3)

then $f=0$ a.e. This fact was used in [LM91], [KL96] in order to prove that if
a nonzero function $f\in L^{1}(\mathbb{R})$ has compact support, and if $f$
tiles at some level $w$ by a translation set $\Lambda$ of bounded density,
then $\Lambda$ must have a periodic structure (Theorem 1.5).

On the other hand, suppose that $\Omega\subset\mathbb{R}$ is a set which
contains arbitrarily long intervals (and so $\Omega$ must be unbounded, but
can be very sparse). Then given any discrete set $S\subset\mathbb{R}$ of
tempered growth, there exists a nonzero function $f\in L^{1}(\mathbb{R})$,
$\operatorname{supp}(f)\subset\Omega$, such that $\widehat{f}$ vanishes on
$S$. This is a consequence of the following result:

######  Theorem 4.3.

Let $\Omega\subset\mathbb{R}$ be a set which contains arbitrarily long
intervals, and let $S\subset\mathbb{R}$ be a set of tempered growth. Then
given any values $\\{c(s)\\}\in\ell^{1}(S)$, one can find a smooth function
$f\in L^{1}(\mathbb{R})$, $\operatorname{supp}(f)\subset\Omega$, such that
$\widehat{f}(s)=c(s)$ for all $s\in S$.

If we call $\operatorname{supp}(f)$ the “spectrum” of the function
$\widehat{f}$, then the result means that every discrete function in
$\ell^{1}(S)$ can be interpolated by a continuous function (the Fourier
transform of an $L^{1}$ function) with spectrum in $\Omega$. We refer the
reader to [OU16] where the problem of interpolation by functions with a given
spectrum is discussed in detail.

The approach that we use to prove Theorem 4.3 is inspired by [OU09, Section
3].

###  4.3 Proof of Theorem 4.3

We begin with a simple lemma.

######  Lemma 4.4.

Let $S\subset\mathbb{R}$ be a discrete set of tempered growth, and let $N$ be
a sufficiently large number such that (4.2) holds. Then for any
$t\in\mathbb{R}\setminus S$ and any $p>N$ we have

$\sum_{s\in S}|s-t|^{-p}<+\infty.$ (4.4)

###### Proof.

The condition (4.2) remains valid if we replace $S$ with $S-t$, so we may
assume that $t=0$. The function $n(r):=\\#(S\cap(-r,r))$ is then a step
function vanishing near the origin. For each $R>0$ which is not a jump
discontinuity point of $n(r)$, we have

$\sum_{|s|\leqslant
R}|s|^{-p}=\int_{0}^{R}\frac{dn(r)}{r^{p}}=\frac{n(R)}{R^{p}}+p\int_{0}^{R}\frac{n(r)}{r^{p+1}}dr$
(4.5)

which follows from integration by parts. But as $n(r)=O(r^{N})$ and $p>N$, the
right-hand side of (4.5) remains bounded as $R\to\infty$. The condition (4.4)
is thus established. ∎

###### Proof of Theorem 4.3.

We suppose that $\Omega\subset\mathbb{R}$ is a set which contains arbitrarily
long intervals, and that $S\subset\mathbb{R}$ is a discrete set of tempered
growth. We also choose and fix some enumeration $\\{s_{j}\\}$ of the set $S$.

Let $\Phi$ be an infinitely smooth function on $\mathbb{R}$ supported on the
interval $[-1,1]$, and such that $\int\Phi=1$. For each $r>0$ we denote
$\Phi_{r}(x):=r^{-1}\Phi(r^{-1}x)$. Define

$f_{j}(x):=\Phi_{r_{j}}(x-\tau_{j})\,e^{2\pi is_{j}x}$

where the numbers $r_{j}>0$ and $\tau_{j}\in\mathbb{R}$ will be determined
later on. Then we have

$\widehat{f}_{j}(t)=\widehat{\Phi}(r_{j}(t-s_{j}))\,e^{-2\pi
i\tau_{j}(t-s_{j})}.$

In particular, $\widehat{f}_{j}(s_{j})=\widehat{\Phi}(0)=1$.

Let $N$ be a sufficiently large number such that (4.2) holds, and let $p>N$.
Since $\widehat{\Phi}$ is a Schwartz function, there is a constant
$C=C(\Phi,p)>0$ such that $|\widehat{\Phi}(t)|\leqslant C|t|^{-p}$ for every
nonzero $t$. For $j$ fixed, we have

$\sum_{k\neq j}|\widehat{f}_{j}(s_{k})|=\sum_{k\neq
j}|\widehat{\Phi}(r_{j}(s_{k}-s_{j}))|\leqslant Cr_{j}^{-p}\sum_{k\neq
j}|s_{k}-s_{j}|^{-p}.$ (4.6)

If we use Lemma 4.4 with the set $\\{s_{k}:k\neq j\\}$ and $t=s_{j}$, the
lemma yields that the sum on the right-hand side of (4.6) converges. We thus
see that given any $\varepsilon>0$, we can choose
$r_{j}=r_{j}(S,\Phi,p,\varepsilon)>0$ large enough such that the right-hand
side of (4.6) does not exceed $\varepsilon$. It follows that we can choose the
numbers $r_{j}$ in such a way that, no matter how the numbers $\tau_{j}$ are
chosen, we have

$M:=\sup_{j}\sum_{k\neq j}|\widehat{f}_{j}(s_{k})|\leqslant\varepsilon.$ (4.7)

To any sequence $\mathbf{b}=\\{b_{j}\\}$ belonging to $\ell^{1}$, we associate
a corresponding sequence $\mathbf{c}=\\{c_{k}\\}$ defined by

$c_{k}=\sum_{j}\widehat{f}_{j}(s_{k})b_{j}=b_{k}+\sum_{j\neq
k}\widehat{f}_{j}(s_{k})b_{j}.$

It follows from (4.7) that the mapping $\mathbf{b}\mapsto\mathbf{c}$ defines a
bounded linear operator $A:\ell^{1}\to\ell^{1}$ such that
$\|A-I\|=M\leqslant\varepsilon$, where $I$ is the identity operator. If we
choose $\varepsilon<1$ then this implies that $A$ is invertible in $\ell^{1}$.

Now suppose that we are given a sequence $\mathbf{c}=\\{c_{k}\\}\in\ell^{1}$.
Let $\mathbf{b}=\\{b_{j}\\}\in\ell^{1}$ be the solution of the equation
$A\mathbf{b}=\mathbf{c}$, and define

$f(x):=\sum_{j}b_{j}f_{j}(x).$ (4.8)

We observe that

$\|f_{j}\|_{L^{1}(\mathbb{R})}=\|\Phi\|_{L^{1}(\mathbb{R})},\quad\sum|b_{j}|<\infty,$

which implies that the series (4.8) converges in $L^{1}(\mathbb{R})$. It
follows that

$\widehat{f}(t)=\sum_{j}b_{j}\widehat{f}_{j}(t),$ (4.9)

where the series (4.9) converges uniformly on $\mathbb{R}$. In particular we
have

$\widehat{f}(s_{k})=\sum_{j}b_{j}\widehat{f}_{j}(s_{k})=c_{k}$ (4.10)

for each $k$.

Finally we observe that $f_{j}$ is supported on the interval
$I_{j}:=[\tau_{j}-r_{j},\tau_{j}+r_{j}]$. Since the $r_{j}$’s were chosen in a
way that does not depend on the values of the $\tau_{j}$’s, we can use the
assumption that $\Omega$ contains arbitrarily long intervals in order to
choose each $\tau_{j}$ such that the interval $I_{j}$ lies in $\Omega$.
Moreover, by choosing these intervals e.g. such that
$\operatorname{dist}(I_{j},I_{k})\geqslant 1$ $(j\neq k)$ this implies that
$\operatorname{supp}(f)\subset\Omega$ and ensures that $f$ is infinitely
smooth. Thus $f$ has all the required properties, and Theorem 4.3 is proved. ∎

###  4.4 Proof of Theorem 1.8

Now we can finish the proof of Theorem 1.8 based on the two results above,
namely, Theorem 4.2 and Theorem 4.3.

Let $\mu=\delta_{\Lambda}$ be the crystalline measure given by Theorem 4.2,
that is, $\Lambda\subset\mathbb{R}$ is a set of bounded density but not of a
periodic structure, such that $\widehat{\delta}_{\Lambda}$ is a pure point
measure, $\widehat{\delta}_{\Lambda}=\sum_{s\in S}b_{s}\delta_{s}$, where
$S\subset\mathbb{R}$ is a set of tempered growth.

Since the set $S$ is discrete and closed, it follows that there is $a>0$ such
that we have $\widehat{\delta}_{\Lambda}=c\cdot\delta_{0}$ in $(-a,a)$.
Moreover, we must have $c>0$ due to Lemma 3.2. By rescaling the set $\Lambda$
if needed, we may assume that $\widehat{\delta}_{\Lambda}=\delta_{0}$ in
$(-a,a)$. In particular, this means that the set $S$ contains the origin.

Now suppose that we are given a scalar $w$, and a set
$\Omega\subset\mathbb{R}$ which contains arbitrarily long intervals. Let $\xi$
be any real number, $\xi\notin S$, and define $S^{\prime}:=S\cup\\{\xi\\}$.
Then also the set $S^{\prime}$ has tempered growth. We prescribe the values
$c(\xi)=1$, $c(0)=w$, and $c(s)=0$ for all $s\in S\setminus\\{0\\}$, then
these values belong to $\ell^{1}(S^{\prime})$. Hence using Theorem 4.3 we can
find a smooth function $f\in L^{1}(\mathbb{R})$,
$\operatorname{supp}(f)\subset\Omega$, such that $\widehat{f}(s)=c(s)$ for all
$s\in S^{\prime}$. This means that we have $\widehat{f}(0)=w$ and
$\widehat{f}(s)=0$ for all $s\in S\setminus\\{0\\}$. It also implies that
$\widehat{f}(\xi)\neq 0$, which ensures that the function $f$ is nonzero.

We conclude that $\widehat{f}$ vanishes on the set
$\operatorname{supp}(\widehat{\delta}_{\Lambda})\setminus\\{0\\}$, i.e. the
condition (2.5) is satisfied, and moreover we have
$\widehat{\delta}_{\Lambda}(\\{0\\})\cdot\int f=w$. Due to Theorem 2.2 this
implies that $f+\Lambda$ is a tiling at level $w$, and the proof is thus
complete. ∎

##  5 Tilings of unbounded density

In this section we construct examples of tilings $f+\Lambda$ such that the set
$\Lambda$ does _not_ have bounded density (Theorem 1.9). We are not aware of
any previous example of such a tiling in the literature. Moreover, in our
example the set $\Lambda$ has tempered growth (see Definition 4.1) and the
function $f$ is in the Schwartz class. We will also show that in any such a
tiling, the function $f$ must have zero integral, no matter what the level $w$
of the tiling is (Theorem 1.10).

###  5.1 Translation sets with unbounded density

We will construct tilings $f+\Lambda$ such that the translation set
$\Lambda\subset\mathbb{R}$ has the form

$\Lambda=\bigcup_{n\in\mathbb{Z}}\Lambda_{n},\quad\Lambda_{n}\subset(n-\varepsilon,n+\varepsilon),\quad\\#\Lambda_{n}=n^{2}+1,$
(5.1)

where $\varepsilon>0$ is an arbitrarily small number. Then the condition (1.2)
is not satisfied and $\Lambda$ does not have bounded density. On the other
hand it follows from (5.1) that $\\#(\Lambda\cap(-r,r))=O(r^{3})$ as
$r\to+\infty$, so the set $\Lambda$ has tempered growth and $\delta_{\Lambda}$
is a tempered distribution on $\mathbb{R}$. We will prove the following
theorem:

######  Theorem 5.1.

Given any $\varepsilon>0$ there is a set $\Lambda\subset\mathbb{R}$ of the
form (5.1), such that

$\widehat{\delta}_{\Lambda}=\delta_{0}-\frac{\delta^{\prime\prime}_{0}}{4\pi^{2}}\quad\text{in
$(-\tfrac{1}{2},\tfrac{1}{2})$.}$ (5.2)

In order to better understand what is behind condition (5.2), consider the
measure $\mu$ on $\mathbb{R}$ which assigns the mass $n^{2}+1$ to each point
$n\in\mathbb{Z}$. Using Poisson’s summation formula one can check that the
Fourier transform of $\mu$ is given by
$\widehat{\mu}=\sum_{k\in\mathbb{Z}}(\delta_{k}-(4\pi^{2})^{-1}\delta^{\prime\prime}_{k})$.
Theorem 5.1 now says that one can “redistribute” the mass $n^{2}+1$ assigned
at each point $n\in\mathbb{Z}$ into equal unit masses at $n^{2}+1$ distinct
points of a set $\Lambda_{n}$ contained in a small neighborhood of $n$, in
such a way that the Fourier transform of the new measure $\delta_{\Lambda}$
thus obtained remains unchanged on the interval
$(-\tfrac{1}{2},\tfrac{1}{2})$.

The role of condition (5.2) in the tiling problem is clarified by the
following lemma:

######  Lemma 5.2.

Let $\Lambda\subset\mathbb{R}$ be a set of tempered growth, and suppose that
there is $a>0$ such that

$\operatorname{supp}(\widehat{\delta}_{\Lambda})\cap(-a,a)=\\{0\\}.$ (5.3)

Then given any scalar $w$ one can find a nonzero Schwartz function $f$, such
that $f+\Lambda$ is a tiling at level $w$.

###### Proof.

It is well-known that a distribution supported by the origin is a finite
linear combination of derivatives of $\delta_{0}$ (see [Rud91, Theorem 6.25]).
Hence (5.3) implies that

$\widehat{\delta}_{\Lambda}=\sum_{j=0}^{n}c_{j}\delta_{0}^{(j)}\quad\text{in
$(-a,a),$}$ (5.4)

and we may assume that the highest order coefficient $c_{n}$ is nonzero. It
follows that

$t^{n}\,\widehat{\delta}_{\Lambda}(t)=c_{n}n!(-1)^{n}\cdot\delta_{0}(t)\quad\text{in
$(-a,a)$}$ (5.5)

(see e.g. [Fol99, Section 9.1, Exercise 3]).

Let $\varphi$ be a nonzero Schwartz function,
$\operatorname{supp}(\widehat{\varphi})\subset(-a,a)$,
$\widehat{\varphi}(0)=w$. Then

$f(x):=\frac{\varphi^{(n)}(x)}{c_{n}n!(-2\pi i)^{n}}$ (5.6)

is also a nonzero Schwartz function. We claim that $f+\Lambda$ is a tiling at
level $w$. Indeed,

$\widehat{f}(t)\,\widehat{\delta}_{\Lambda}(t)=\frac{\widehat{\varphi}(t)t^{n}}{c_{n}n!(-1)^{n}}\cdot\widehat{\delta}_{\Lambda}(t)=\widehat{\varphi}(t)\,\delta_{0}(t)=w\cdot\delta_{0}(t),$
(5.7)

where in the second equality we used (5.5). This implies that
$f\ast\delta_{\Lambda}=w$ as needed. ∎

Since condition (5.2) implies (5.3), we see that Theorem 1.9 is a consequence
of Theorem 5.1 and Lemma 5.2. It therefore remains to prove Theorem 5.1.

###  5.2 Kargaev’s implicit function method

In order to prove Theorem 5.1 we will use a variant of Kargaev’s implicit
function method [Kar82]. See also [KL16], [Lev20] for applications of the
method in the construction of translational tiling examples.

The proof will be done in several steps.

#### 5.2.1

For each $k=1,2,3,\dots$ let $\chi_{k}$ be the function on $\mathbb{R}$
defined by

$\chi_{k}(x)=k-j+1,\quad
x\in\big{[}\tfrac{2(j-1)}{k(k+1)},\,\tfrac{2j}{k(k+1)}\big{)},\quad 1\leqslant
j\leqslant k,$ (5.8)

and $\chi_{k}(x)=0$ outside the interval $[0,\frac{2}{k+1})$.

Let $\\{\alpha_{n}\\}$, $n\in\mathbb{Z}$, be a bounded sequence of real
numbers. To such a sequence we associate a function $F$ on the real line,
defined by

$F(x):=\sum_{n\in\mathbb{Z}}F_{n}(x),\quad x\in\mathbb{R},$ (5.9)

where we let

$F_{n}(x):=\operatorname{sign}(\alpha_{n})\cdot\chi_{n^{2}+1}(\tfrac{x-n}{\alpha_{n}})$
(5.10)

for each $n\in\mathbb{Z}$ such that $\alpha_{n}\neq 0$, while if
$\alpha_{n}=0$ then we let $F_{n}(x):=0$.

(The notation $\operatorname{sign}(\alpha_{n})$ means $+1$ or $-1$ depending
on whether $\alpha_{n}>0$ or $\alpha_{n}<0$.)

One can verify, using the assumption that the sequence $\\{\alpha_{n}\\}$ is
bounded, that the series (5.9) converges in the space of tempered
distributions. For instance, this follows from the fact that
$(1+x^{2})^{-1}\sum|F_{n}(x)|$ is a bounded function.

######  Theorem 5.3.

Let $0<\varepsilon<1$. Then for any sufficiently small $r>0$ one can find a
real sequence $\alpha=\\{\alpha_{n}\\}$, $n\in\mathbb{Z}$, satisfying

$(1-\varepsilon)r\leqslant|\alpha_{n}|\leqslant(1+\varepsilon)r,\quad
n\in\mathbb{Z},$ (5.11)

and such that $\widehat{F}=0$ in $(-\tfrac{1}{2},\tfrac{1}{2})$, where $F$ is
the function defined by (5.9).

We will first prove Theorem 5.3 and then use it to deduce Theorem 5.1.

#### 5.2.2

We will need the following lemma.

######  Lemma 5.4.

The function $\chi_{k}$ has the following properties:

  1. (i)

$\int\chi_{k}(x)dx=1$;

  2. (ii)

$\int x\chi_{k}(x)dx\leqslant Ck^{-1}$;

  3. (iii)

$|\widehat{\chi}_{k}(-s)-1|\leqslant C|s|k^{-1}$;

  4. (iv)

$\big{|}v\big{(}\widehat{\chi}_{k}(-v)-1\big{)}-u\big{(}\widehat{\chi}_{k}(-u)-1\big{)}\big{|}\leqslant
Ck^{-1}|v-u|\cdot\max\\{|u|,|v|\\}$

for every $s,u,v\in\mathbb{R}$, where $C>0$ is an absolute constant.

###### Proof.

The property (i) can be checked directly. In order to establish (ii) we can
estimate the left hand side by $k\int_{0}^{2/(k+1)}xdx=2k/(k+1)^{2}$. Together
with the estimate

$\textstyle|\widehat{\chi}_{k}(-s)-1|=\Big{|}\int\chi_{k}(x)(e^{2\pi
isx}-1)dx\Big{|}\leqslant 2\pi|s|\int x\chi_{k}(x)dx$ (5.12)

this yields (iii). Finally, to establish (iv) consider the function
$\varphi(x):=x(e^{2\pi ix}-1)$. We may suppose that $u<v$, then the left hand
side of (iv) is equal to

$\displaystyle\Big{|}\int\chi_{k}(x)\frac{\varphi(vx)-\varphi(ux)}{x}dx\Big{|}=\Big{|}\int\chi_{k}(x)\int_{u}^{v}\varphi^{\prime}(sx)dsdx\Big{|}$
(5.13)
$\displaystyle\qquad\qquad\leqslant|v-u|\cdot\max_{s\in[u,v]}\int\chi_{k}(x)|\varphi^{\prime}(sx)|dx.$
(5.14)

If we use the estimate $|\varphi^{\prime}(sx)|\leqslant C|sx|$, then (5.14)
and (ii) imply (iv). ∎

#### 5.2.3

Let $X$ be the space of all bounded sequences of real numbers
$\alpha=\\{\alpha_{n}\\}$, $n\in\mathbb{Z}$, endowed with the norm

$\|\alpha\|_{X}:=\sup_{n\in\mathbb{Z}}|\alpha_{n}|$

that makes $X$ into a real Banach space.

Denote $I:=[-\tfrac{1}{2},\tfrac{1}{2}]$. Let $Y$ be the space of all
continuous functions $\psi:I\to\mathbb{C}$ satisfying
$\psi(-t)=\overline{\psi(t)}$ for all $t\in I$. If we endow $Y$ with the norm
$\|\psi\|_{Y}=\sup|\psi(t)|$, $t\in I$, then also $Y$ is a real Banach space.

Let $\alpha=\\{\alpha_{n}\\}$, $n\in\mathbb{Z}$, be a sequence in $X$. Define

$(R\alpha)(t):=\sum_{n\in\mathbb{Z}}e^{2\pi
int}\cdot\alpha_{n}\big{(}\widehat{\chi}_{n^{2}+1}(-\alpha_{n}t)-1\big{)}.$
(5.15)

We notice that the terms of the series (5.15) are elements of $Y$. By Lemma
5.4(iii), the $n$’th term of the series is bounded by
$C|\alpha_{n}|^{2}(n^{2}+1)^{-1}$ uniformly on $I$, where $C>0$ does not
depend on $\alpha$ or $n$. Hence the series (5.15) converges uniformly on $I$
to an element of $Y$, and we have

$\|R\alpha\|_{Y}\leqslant C\|\alpha\|^{2}_{X},$ (5.16)

where the constant $C$ does not depend on $\alpha$.

We note that the mapping $R:X\to Y$ defined by (5.15) is _nonlinear_.

#### 5.2.4

For each $r>0$ let $U_{r}$ denote the closed ball of radius $r$ around the
origin in $X$:

$U_{r}:=\\{\alpha\in X:\|\alpha\|_{X}\leqslant r\\}.$ (5.17)

######  Lemma 5.5.

Given any $\rho>0$ there is $r>0$ such that

$\|R\beta-R\alpha\|_{Y}\leqslant\rho\|\beta-\alpha\|_{X},\quad\alpha,\beta\in
U_{r}.$ (5.18)

In particular, if $r$ is small enough then $R$ is a contractive (nonlinear)
mapping on $U_{r}$.

###### Proof.

Let $\alpha,\beta\in U_{r}$. Then using (5.15) we have

$(R\beta-R\alpha)(t)=\sum_{n\in\mathbb{Z}}e^{2\pi
int}\cdot\Big{[}\beta_{n}\big{(}\widehat{\chi}_{n^{2}+1}(-\beta_{n}t)-1\big{)}-\alpha_{n}\big{(}\widehat{\chi}_{n^{2}+1}(-\alpha_{n}t)-1\big{)}\Big{]}.$
(5.19)

By Lemma 5.4(iv), the $n$’th term of the series is bounded by
$Cr(n^{2}+1)^{-1}|\beta_{n}-\alpha_{n}|$ uniformly on $I$, where $C>0$ does
not depend on $r$, $\alpha$, $\beta$ or $n$. Hence the series converges
uniformly on $I$ and $\|R\beta-R\alpha\|_{Y}\leqslant Cr\|\beta-\alpha\|_{X}$,
where $C$ is a constant not depending on $r$, $\alpha$ or $\beta$. It thus
suffices to choose $r$ small enough so that $Cr\leqslant\rho$. ∎

#### 5.2.5

For each element $\psi\in Y$ we denote by $\mathcal{F}(\psi)$ the sequence

$\widehat{\psi}(n)=\int_{I}\psi(t)e^{-2\pi int}dt,\quad n\in\mathbb{Z},$
(5.20)

of Fourier coefficients of $\psi$. Since the Fourier coefficients
$\widehat{\psi}(n)$ are real and bounded, we have a linear mapping
$\mathcal{F}:Y\to X$ satisfying
$\|\mathcal{F}(\psi)\|_{X}\leqslant\|\psi\|_{Y}$.

######  Lemma 5.6.

Given any $\varepsilon>0$ there is $\delta>0$ with the following property: Let
$\beta\in X$, $\|\beta\|_{X}\leqslant\delta$. Then one can find an element
$\alpha\in X$, $\|\alpha-\beta\|_{X}\leqslant\varepsilon\|\beta\|_{X}$, which
solves the equation $\alpha+\mathcal{F}(R\alpha)=\beta$.

###### Proof.

Fix $\beta\in X$ such that $\|\beta\|_{X}\leqslant\delta$, and let

$B=B(\beta,\varepsilon):=\\{\alpha\in
X:\|\alpha-\beta\|_{X}\leqslant\varepsilon\|\beta\|_{X}\\}.$

We observe that if $\alpha\in B$ then
$\|\alpha\|_{X}\leqslant(1+\varepsilon)\|\beta\|_{X}$. Define a map $H:B\to X$
by

$H(\alpha):=\beta-\mathcal{F}(R\alpha),\quad\alpha\in B,$

and notice that an element $\alpha\in B$ is a solution to the equation
$\alpha+\mathcal{F}(R\alpha)=\beta$ if and only if $\alpha$ is a fixed point
of the map $H$.

Let us show that if $\delta$ is small enough then $H(B)\subset B$. Indeed, if
$\alpha\in B$ then using (5.16) we have

$\|H(\alpha)-\beta\|_{X}=\|\mathcal{F}(R\alpha)\|_{X}\leqslant\|R\alpha\|_{Y}\leqslant
C\|\alpha\|^{2}_{X}\leqslant C(1+\varepsilon)^{2}\|\beta\|^{2}_{X}.$

Hence if we choose $\delta$ such that
$C(1+\varepsilon)^{2}\delta\leqslant\varepsilon$ then we obtain

$\|H(\alpha)-\beta\|_{X}\leqslant\varepsilon\|\beta\|_{X},$

and it follows that $H(B)\subset B$.

It also follows from Lemma 5.5 that if $\delta$ is small enough, then $H$ is a
contractive mapping from the closed set $B$ into itself. Indeed, let
$\alpha^{\prime},\alpha^{\prime\prime}\in B$, then we have

$\|H(\alpha^{\prime\prime})-H(\alpha^{\prime})\|_{X}=\|\mathcal{F}(R\alpha^{\prime\prime}-R\alpha^{\prime})\|_{X}\leqslant\|R\alpha^{\prime\prime}-R\alpha^{\prime}\|_{Y}\leqslant\rho\|\alpha^{\prime\prime}-\alpha^{\prime}\|_{X},$

where $0<\rho<1$. Then the Banach fixed point theorem implies that $H$ has a
(unique) fixed point $\alpha\in B$, which yields the desired solution. ∎

####  5.2.6 Proof of Theorem 5.3

Let $r>0$, and let $\beta=\\{\beta_{n}\\}$, $n\in\mathbb{Z}$, be a real
sequence defined by $\beta_{n}:=(-1)^{n}r$. Then $\|\beta\|_{X}=r$. If
$r=r(\varepsilon)>0$ is small enough, then by Lemma 5.6 there is an element
$\alpha\in X$, $\|\alpha-\beta\|_{X}\leqslant\varepsilon\|\beta\|_{X}$, which
solves the equation $\alpha+\mathcal{F}(R\alpha)=\beta$.

We observe that the estimate
$\|\alpha-\beta\|_{X}\leqslant\varepsilon\|\beta\|_{X}$ implies (5.11).

The relation $\alpha+\mathcal{F}(R\alpha)=\beta$ means that $\beta-\alpha$ is
the sequence of Fourier coefficients of the function $R\alpha$. This implies
that the series

$\sum_{n\in\mathbb{Z}}(\beta_{n}-\alpha_{n})e^{2\pi int}$

converges in $L^{2}(I)$ to $R\alpha$.

Since we have $\beta_{n}=(-1)^{n}r$, $n\in\mathbb{Z}$, the series

$\sum_{n\in\mathbb{Z}}\beta_{n}e^{2\pi int}$

converges in the distributional sense to the measure
$r\cdot\delta_{\mathbb{Z}+\frac{1}{2}}$ on $\mathbb{R}$. In particular, this
series converges to zero in the open interval $(-\frac{1}{2},\frac{1}{2})$.

Let $F$ be the function given by (5.9) associated to the sequence
$\alpha=\\{\alpha_{n}\\}$. Then

$\widehat{F}(-t)=\lim_{N\to\infty}\sum_{|n|\leqslant N}\widehat{F}_{n}(-t)$

in the sense of distributions, and by (5.10) we have

$\widehat{F}_{n}(-t)=e^{2\pi
int}\cdot\alpha_{n}\,\widehat{\chi}_{n^{2}+1}(-\alpha_{n}t).$ (5.21)

Hence

$\widehat{F}(-t)=\lim_{N\to\infty}\Big{[}\sum_{|n|\leqslant N}\beta_{n}e^{2\pi
int}-\sum_{|n|\leqslant N}(\beta_{n}-\alpha_{n})e^{2\pi
int}+\sum_{|n|\leqslant N}e^{2\pi
int}\cdot\alpha_{n}\big{(}\widehat{\chi}_{n^{2}+1}(-\alpha_{n}t)-1\big{)}\,\Big{]}.$

The first sum converges in the distributional sense to zero in
$(-\tfrac{1}{2},\tfrac{1}{2})$. The second sum converges in $L^{2}(I)$ to
$R\alpha$. The third sum converges to $R\alpha$ uniformly on $I$. It follows
that the distribution $\widehat{F}$ vanishes in the open interval
$(-\tfrac{1}{2},\tfrac{1}{2})$. ∎

###  5.3 Proof of Theorem 5.1

Let $0<\varepsilon<1$ be given, and for $r=r(\varepsilon)>0$ sufficiently
small let $\\{\alpha(n)\\}$, $n\in\mathbb{Z}$, be the sequence given by
Theorem 5.3. Define

$\Lambda_{n}=\Big{\\{}n+\frac{2j\alpha_{n}}{(n^{2}+1)(n^{2}+2)},\quad
1\leqslant j\leqslant n^{2}+1\Big{\\}}$ (5.22)

and

$\Lambda=\bigcup_{n\in\mathbb{Z}}\Lambda_{n}.$ (5.23)

We have $\alpha_{n}\neq 0$ for every $n\in\mathbb{Z}$, due to (5.11). Hence
$\Lambda_{n}$ is a set with exactly $n^{2}+1$ elements. The set $\Lambda_{n}$
is contained in the interval $[n-|\alpha_{n}|,n+|\alpha_{n}|]$. This yields
(5.1) provided that $r>0$ is small enough, again due to (5.11). In particular
we may assume that the sets $\Lambda_{n}$ are pairwise disjoint.

Observe that the distributional derivative of the function $F$ in (5.9) is

$F^{\prime}=\sum_{n\in\mathbb{Z}}F^{\prime}_{n}=\sum_{n\in\mathbb{Z}}((n^{2}+1)\delta_{n}-\delta_{\Lambda_{n}}),$

and hence

$\delta_{\Lambda}=\sum_{n\in\mathbb{Z}}(n^{2}+1)\delta_{n}-F^{\prime}.$

The Fourier transform of the measure
$\sum_{n\in\mathbb{Z}}(n^{2}+1)\delta_{n}$ is
$\sum_{k\in\mathbb{Z}}\big{(}\delta_{k}-(4\pi^{2})^{-1}\delta^{\prime\prime}_{k}\big{)}$,
which follows from Poisson’s summation formula. This implies that

$\widehat{\delta}_{\Lambda}=\delta_{0}-\frac{\delta^{\prime\prime}_{0}}{4\pi^{2}}-\widehat{F^{\prime}}\quad\text{in
$(-\tfrac{1}{2},\tfrac{1}{2})$.}$

But since $\widehat{F}$ vanishes in $(-\tfrac{1}{2},\tfrac{1}{2})$, then the
same is true for $\widehat{F^{\prime}}$, so (5.2) is established. ∎

###  5.4 Proof of Theorem 1.10

Finally we show that if $f+\Lambda$ is a tiling at some level $w$, where
$\Lambda\subset\mathbb{R}$ is any set of tempered growth but not of bounded
density, and $f$ is any function in the Schwartz class, then $f$ must have
zero integral.

Suppose to the contrary that $\int f=\widehat{f}(0)$ is nonzero. Then, due to
the continuity and smoothness of $\widehat{f}$, there is a Schwartz function
$g$ such that $\widehat{f}\cdot\widehat{g}=1$ in some neighborhood $(-a,a)$ of
the origin. Let $h>0$ be a Schwartz function with
$\operatorname{supp}(\widehat{h})\subset(-a,a)$, then

$\widehat{h}\cdot\widehat{g}\cdot\widehat{f}=\widehat{h}$

and hence

$h\ast g\ast f=h.$

It follows that

$h\ast\delta_{\Lambda}=(h\ast g\ast f)\ast\delta_{\Lambda}=(h\ast
g)\ast(f\ast\delta_{\Lambda})=w\cdot\textstyle\int(h\ast g),$

where in the last equality we used the tiling assumption
$f\ast\delta_{\Lambda}=w$ (the associativity of the convolution is justified
since $\delta_{\Lambda}$ is a tempered distribution and $f,g,h$ are Schwartz
functions, see [Rud91, Theorem 7.19]).

We conclude that $h+\Lambda$ is a tiling (at a certain level). But it is
known, see [KL96, Lemma 2.1], that if $h$ is a _nonnegative_ , nonzero
function and if $h+\Lambda$ is a tiling, then the set $\Lambda$ must have
bounded density. We thus arrive at a contradiction. ∎

##  6 Open problems

We conclude the paper by posing some open problems.

###  6.1 Tiling at level zero

The following problem was already mentioned in Section 1.2 above: Let
$f+\Lambda$ be a tiling at level zero, where the function $f\in
L^{1}(\mathbb{R})$ is nonzero and the set $\Lambda\subset\mathbb{R}$ is
nonempty and has bounded density. Does it follow that $\Lambda$ has a _uniform
density_ $D(\Lambda)$ ?

In Fourier analytic terms, the problem can be equivalently stated as follows:
Let $\Lambda\subset\mathbb{R}$ be nonempty and have bounded density, and
suppose that $\widehat{\delta}_{\Lambda}$ vanishes on some open interval
$(a,b)$. Does $\Lambda$ necessarily have a uniform density?

What makes the problem nontrivial is the existence of tilings $f+\Lambda$ at
level zero such that $\widehat{\delta}_{\Lambda}$ is not a scalar multiple of
$\delta_{0}$ in any neighborhood of the origin, see [Lev20, Section 5]. In
particular, Lemma 3.2 does not apply.

We note that by Theorem 1.3 the set $\Lambda$ must be relatively dense, so if
the density $D(\Lambda)$ exists then it is a strictly positive number.

###  6.2 Non-periodic tilings

Let $f$ be a nonzero function in $L^{1}(\mathbb{R})$, and suppose that the set
$\\{x:f(x)\neq 0\\}$ has _finite measure_. If $f$ tiles at some level $w$ by a
translation set $\Lambda\subset\mathbb{R}$ of bounded density, does it follows
that $\Lambda$ has a periodic structure?

Theorem 1.5 does not apply here, since $f$ is _not_ assumed to have compact
support.

Does there exist a measurable set $\Omega\subset\mathbb{R}$,
$0<\operatorname{mes}(\Omega)<+\infty$, whose indicator function
$\mathds{1}_{\Omega}$ can tile at level one, or, a weaker requirement, at some
other integer level $w$, with a translation set $\Lambda\subset\mathbb{R}$
that _does not_ have a periodic structure?

Notice that such a set $\Omega$ (if it exists) must be unbounded, again due to
Theorem 1.5.

###  6.3 Tilings of unbounded density

Let $f\in L^{1}(\mathbb{R})$ be nonzero and have _compact support_ , and
suppose that $f+\Lambda$ is a tiling at some level $w$, where
$\Lambda\subset\mathbb{R}$ is a discrete set (not a multi-set) of tempered
growth. Does it follow that $\Lambda$ is of the form (1.4), i.e. $\Lambda$ is
a set of bounded density having a periodic structure?

In other words, the question is whether Theorem 1.5 remains valid if the set
$\Lambda$ is not assumed to have bounded density, but only tempered growth.

We note that Theorem 1.9 does _not_ provide a negative answer to this
question, since the function $f$ constructed in the proof of this theorem has
_unbounded_ support.

###  6.4 Lattice tilings

Let $f\in L^{1}(\mathbb{R}^{d})$, $d\geqslant 1$, and suppose that $f$ tiles
at some level $w$ with a translation set $\Lambda\subset\mathbb{R}^{d}$ of
bounded density.222A set $\Lambda\subset\mathbb{R}^{d}$ is said to have
_bounded density_ if there exists $M>0$ such that
$\\#(\Lambda\cap(x+B))\leqslant M$ for all $x\in\mathbb{R}^{d}$, where $B$ is
the open unit ball in $\mathbb{R}^{d}$. Does there necessarily exist a
_lattice_ $L\subset\mathbb{R}^{d}$ such that $f+L$ is also a tiling, possibly
at a different level $w^{\prime}$ ?

The answer is known to be affirmative in the special case where $\Lambda$ is
assumed to be a disjoint union of finitely many translated lattices, namely,
$\Lambda=\biguplus_{j=1}^{N}(L_{j}+\tau_{j})$ where each $L_{j}$ is a lattice
in $\mathbb{R}^{d}$ and the $\tau_{j}$ are translation vectors. This was
proved in dimension one in [KL96, p. 673], while in several dimensions the
result was proved more recently in [Liu18, Theorem 1.6]. In both proofs,
number theory plays an essential role: the proof in $\mathbb{R}$ uses the
classical Skolem–Mahler–Lech theorem, while in $\mathbb{R}^{d}$ the proof
relies on a result due to Evertse, Schlickewei and Schmidt [ESS02].

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[kolount] M. N. Kolountzakis
Department of Mathematics and Applied Mathematics
University of Crete
Voutes Campus, GR-700 13
Heraklion, Crete
Greece
<EMAIL_ADDRESS>
http://mk.eigen-space.org/ [lev] Nir Lev
Department of Mathematics
Bar-Ilan University
Ramat-Gan 5290002
Israel
<EMAIL_ADDRESS>
https://u.math.biu.ac.il/~levnir/