# Circles in diffraction

Emily R. Korfanty  Department of Mathematical and Statistical Sciences,
University of Alberta, Edmonton, AB, T6G 2G1, Canada
<EMAIL_ADDRESS>and  Nicolae Strungaru  Department of Mathematical
Sciences, MacEwan University
10700 – 104 Avenue, Edmonton, AB, T5J 4S2, Canada
and
Institute of Mathematics “Simon Stoilow”
Bucharest, Romania<EMAIL_ADDRESS>https://sites.google.com/macewan.ca/nicolae-strungaru/home

###### Abstract.

Given a Fourier transformable measure in two dimensions, we find a formula for
the intensity of its Fourier transform along circles. In particular, we obtain
a formula for the diffraction measure along a circle in terms of the
autocorrelation measure. We look at some applications of this formula.

##  1\. Introduction

Since the early 1900s, crystallographers have used diffraction experiments to
determine the structure of crystals. Until the discovery of quasicrystals
[SBGC84] by Dan Shechtman in 1984, pure point diffraction was taken as
evidence of an underlying periodic structure in the material. Quasicrystals,
however, exhibit sharp diffraction patterns, and typically show symmetries
that are impossible in materials with a periodic structure. Shechtman was
later awarded the 2011 Nobel Prize in Chemistry for the discovery, which also
inspired a new interest in aperiodic structures and mathematical diffraction
theory. For an overview of this area, we recommend the monographs [BG13,
BG17].

The rigorous mathematical treatment of diffraction was introduced by Hof in
[Hof95]. He defined the diffraction measure as the Fourier transform of the
so-called _autocorrelation measure_ of the underlying structure. Diffraction
is now well understood for systems with pure point spectrum [LSS20, LSS24],
and systems coming from cut-and-project schemes [Sch00, BM04, RS17, Str05,
Str17, Str20, BHS17, KR16].

On the other hand, the diffraction for systems without finite local complexity
(FLC) is, in general, much less understood. One such example, the pinwheel
tiling [Rad94], is produced by a simple 2-dimensional substitution. The tiles
in the pinwheel tiling appear in infinitely many orientations, and therefore
the pinwheel does not have FLC. Its diffraction is rotationally invariant
[MPS06], and has a Bragg peak of unit intensity at the origin, but nothing
else is known about its diffraction. In particular, it is not known if the
diffraction is absolutely continuous, singular continuous or mixed.

There are many examples of rotational invariant diffraction patterns. For
example, any substitution with statistical circular symmetry, meaning that the
orientations of the tiles distribute uniformly around the unit circle ([Fre08]
[Sad98], [FHG24]), leading to a rotational invariant diffraction pattern. More
generally, the diffraction of any rotational invariant measure is also
rotational invariant.

Rotational invariance also appears in powder diffraction, which is a slightly
different diffraction technique, where the sample is ground into many smaller
samples prior to the usual diffraction process. During the grinding process,
the samples tend to rotate in random directions. This random distribution of
rotations on the unit circle is typically statistically circular, and hence
the powder diffraction is rotational invariant.

In [BFG07, BFG07a, GD11], numerical approximations of the pinwheel diffraction
were compared to a powder diffraction from ${\mathbb{Z}}^{2}$. These numerical
approximations suggest that the pinwheel diffraction may include bright
circles with the same radii as the circles in the powder diffraction from
${\mathbb{Z}}^{2}$, which would imply the existence of a singular continuous
component of the spectrum. Despite this observed similarity to the powder
diffraction, it is still unknown whether or not there exists any circle in the
pinwheel diffraction. Besides numerical simulations, which are not too precise
for this particular system, we are not aware of any progress made toward
understanding the diffraction of the pinwheel tiling in the last 20 years.

Motivated by the above, in this paper we develop a method for computing the
intensity of a circle in the diffraction. This may provide a new tool to study
the diffraction of substitution tilings with statistical circular symmetry,
such as the pinwheel and its related families [Sad98, Fre08, FHG24]. In
Section 5, we derive the following explicit limit formula for the diffraction
intensity
$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\gamma}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\gamma}{\tmpbox}(C_{r})$
along the circle $C_{r}=\\{x\in{\mathbb{R}}^{2}:\|x\|=r\\}$ in terms of the
autocorrelation measure $\gamma$:

######  Theorem 5.4.

Let $\gamma$ be an autocorrelation measure on ${\mathbb{R}}^{2}$ and let
$r>0$. Then

$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\gamma}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\gamma}{\tmpbox}(C_{r})=\lim_{n\rightarrow\infty}\frac{2r\pi\sqrt{\pi}}{n}\int_{{\mathbb{R}}^{2}}e^{-\pi^{2}\|y\|^{2}/n^{2}}J_{0}(2\pi
r\|y\|)\,\mbox{d}\gamma(y)\,.$

Here $J_{0}$ represents the Bessel function of the first kind (see Definition
3.1).

To obtain this result, our approach is as follows. Heuristically, we should
have

$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\gamma}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\gamma}{\tmpbox}(C_{r})=\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\gamma}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\gamma}{\tmpbox}(1_{C_{r}})\text{``=''}\gamma(\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{1_{C_{r}}}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{1_{C_{r}}}{\tmpbox})\,.$

Of course, the last equality above cannot hold, since $1_{C_{r}}=0$ almost
everywhere with respect to the Lebesgue measure on ${\mathbb{R}}^{2}$, and
hence
$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{1_{C_{r}}}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{1_{C_{r}}}{\tmpbox}=0$.
To get around this issue, we instead approximate $1_{C_{r}}$ by a sequence of
Schwartz functions of the form

$f_{n}:=g_{n}*\theta_{r}\,,$

where $\theta_{r}$ is the uniform probability measure on $C_{r}$ (see
Definition 3.1), and $g_{n}$ is an approximate identity sequence of Gaussian
functions, renormalized such that $f_{n}$ converges to $1$ on $C_{r}$. Using
the formula
$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\gamma}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\gamma}{\tmpbox}(f_{n})=\gamma(\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{f_{n}}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{f_{n}}{\tmpbox})$
and letting $n\to\infty$ we obtain Theorem 5.4. The Bessel function $J_{0}$
appears naturally in this approach through the Fourier transform of
$\theta_{r}$; see Proposition 3.4.

Our paper is organized as follows. We start by reviewing the basic definitions
and framework in Section 2.2. We use the Fourier transform of strongly
tempered measures (see Definition 2.5), which extends the classical theory of
Fourier transformable measures in ${\mathbb{R}}^{d}$. In Section 3 we briefly
review the Bessel functions $J_{0}$ and $I_{0}$ and some of their properties.
As emphasized above, $J_{0}$ appears as the Fourier transform of $\theta_{r}$,
while $I_{0}$ appears naturally in the convolution of $\theta_{r}$ with the
Gaussian function(s).

Section 4 provides the fundamental tool for proving Theorem 5.4. We show, via
a straightforward application of the dominated convergence theorem, that the
intensity
$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\gamma}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\gamma}{\tmpbox}(C_{r})$
can be calculated as the limit of integrals
$\int_{{\mathbb{R}}^{2}}f_{n}(x)\,\mbox{d}\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\gamma}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\gamma}{\tmpbox}(x)$,
whenever when the sequence $f_{n}$ of Schwartz functions satisfies certain
conditions (which we call approximate $r$-circles). Since
$\int_{{\mathbb{R}}^{2}}f_{n}(x)\,\mbox{d}\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\gamma}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\gamma}{\tmpbox}(x)=\int_{{\mathbb{R}}^{2}}\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{f_{n}}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{f_{n}}{\tmpbox}(y)\,\mbox{d}\gamma(y)$,
we obtain a more general version of Proposition 6.13 (see Corollary 4.4).
Theorem 5.4 is then obtained by constructing in Section 5 an explicit
approximate $r$-circle with easily calculable Fourier transform.

In Section 6, we complete the paper by looking at several examples and
applications of Theorem 5.4. In particular, the Poisson summation formula in
${\mathbb{R}}^{2}$ yields the following result for the sum of two squares
function:

######  Proposition 6.13.

The sum of two squares function $r_{2}(n)$ satisfies the following:

  * (a)

For all $k\in\mathbb{N}$ we have

$r_{2}(k)=\lim_{n\rightarrow\infty}\frac{2\pi\sqrt{\pi
k}}{n}\sum_{m=0}^{\infty}r_{2}(m)e^{-\frac{\pi^{2}m}{n^{2}}}J_{0}(2\pi\sqrt{km})\,.$

  * (b)

For all $r>0$ such that $r^{2}\notin\mathbb{N}$ we have

$\lim_{n\rightarrow\infty}\frac{2r\pi\sqrt{\pi}}{n}\sum_{m=0}^{\infty}r_{2}(m)e^{-\frac{\pi^{2}m\textbf{}}{n^{2}}}J_{0}(2\pi
r\sqrt{m})=0\,.$

In Appendix A, we construct a different family of approximate $r$-circles via
convolutions with annuli of small width.

##  2\. Preliminaries

###  2.1. Definitions and notations

Throughout this paper, ${\mathbb{R}}^{d}$ denotes the Euclidean space of
dimension $d\geq 1$. We will denote the standard Euclidean norm on
${\mathbb{R}}^{d}$ by $\|\cdot\|$. We will primarily be interested in the case
$d=2$.

Given $R>0$, we denote by $B_{R}$ the open ball

$B_{R}:=\\{x\in{\mathbb{R}}^{d}\,:\,\|x\|<R\\}\,,$

of radius $R$, and by $\overline{B_{R}}$ the closed ball

$\overline{B_{R}}:=\\{x\in{\mathbb{R}}^{d}\,:\,\|x\|\leq R\\}\,.$

We will use $1_{A}$ to denote the characteristic function of a set
$A\subseteq{\mathbb{R}}^{d}$, i.e.

$1_{A}(x):=\begin{cases}1&x\in A\\\ 0&x\notin A\,.\end{cases}$

As usual, $C_{\mathsf{c}}({\mathbb{R}}^{d})$ denotes the space of continuous
functions with compact support, and
$C_{\mathsf{c}}^{\infty}({\mathbb{R}}^{d})$ is the subspace of
$C_{\mathsf{c}}({\mathbb{R}}^{d})$ consisting of smooth (infinitely many times
differentiable) functions. The space of Schwartz functions is denoted by
${\mathcal{S}}({\mathbb{R}}^{d})$ and the space of tempered distributions is
denoted by ${\mathcal{S}}^{\prime}({\mathbb{R}}^{d})$.

If $f$ is any bounded function, we will use the notation

$\|f\|_{\infty}:=\sup_{x\in{\mathbb{R}}^{d}}|f(x)|\,,$

for the supremum norm of $f$.

Let us recall that the Fourier transform of a function $f\in
L^{1}({\mathbb{R}}^{d})$ is the function
$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{f}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{f}{\tmpbox}:{\mathbb{R}}^{d}\to{\mathbb{C}}$
defined by

$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{f}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{f}{\tmpbox}(y):=\int_{{\mathbb{R}}^{d}}f(x)e^{-2\pi
ix\cdot y}\,\mbox{d}x\qquad\forall y\in{\mathbb{R}}^{d}\,.$

The map
$f\mapsto\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{f}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{f}{\tmpbox}$
is a bijection from ${\mathcal{S}}({\mathbb{R}}^{d})$ onto itself, and thus
induces a mapping
$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\cdot}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\cdot}{\tmpbox}:{\mathcal{S}}^{\prime}({\mathbb{R}}^{d})\to{\mathcal{S}}^{\prime}({\mathbb{R}}^{d})$,
via

(2.1)
$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{T}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{T}{\tmpbox}(f):=T(\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{f}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{f}{\tmpbox}\,)\qquad\forall
f\in{\mathcal{S}}({\mathbb{R}}^{d})\,.$

For a tempered distribution $T\in{\mathcal{S}}^{\prime}({\mathbb{R}}^{d})$ we
will refer to the tempered distribution
$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{T}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{T}{\tmpbox}$
as the _distributional Fourier transform of $T$_.

###  2.2. Strongly tempered measures

In this paper, we prove our results for a space of measures with a certain
property (see Definition 2.5 below) that is compatible with diffraction
theory. Let us first recall the definition of a measure.

######  Definition 2.1.

A Radon measure on ${\mathbb{R}}^{d}$ is a linear functional
$\mu:C_{\mathsf{c}}({\mathbb{R}}^{d})\rightarrow{\mathbb{C}}$ with the
property that for every compact set $K\subseteq{\mathbb{R}}^{d}$, there exists
a constant $a_{K}$ such that

(2.2) $|\mu(f)|\leq a_{K}\|f\|_{\infty}\,,$

holds for every $f\in C_{\mathsf{c}}({\mathbb{R}}^{d})$ with $\mbox{\rm
supp}(f)\subseteq K$. We write ${\mathcal{M}}({\mathbb{R}}^{d})$ for the space
of all Radon measures on ${\mathbb{R}}^{d}$. We will simply refer to Radon
measures as measures.

We will often use the so-called variation measure $|\mu|$, where $\mu$ is some
measure. The definition of $|\mu|$ is provided in the following theorem:

######  Theorem 2.2 ([RS17, Lemma C.2], [Ped95, 6.5.8]).

For every measure $\mu$, there exists a smallest positive measure $|\mu|$
satisfying

$|\mu(f)|\leq|\mu|(|f|)\qquad\forall f\in C_{\mathsf{c}}({\mathbb{R}}^{d})\,.$

Moreover, $|\mu|$ is the unique measure with the property that

$|\mu|(f)=\sup\\{|\mu(g)|:g\in C_{\mathsf{c}}({\mathbb{R}}^{d}),|g|\leq
f\\}\,,$

holds for all non-negative functions $f\in C_{\mathsf{c}}({\mathbb{R}}^{d})$.
∎

Below, we will be using the Fourier transform in the distributional sense for
measures. Let us first note that, in general,
$\int_{{\mathbb{R}}^{d}}f(x)\,\mbox{d}\mu(x)$ may not make sense for a measure
$\mu$ and Schwartz function $f$. Since we will work with integrals of this
type, we will need to restrict the class of measures we are studying. We start
by recalling the following definition from [BS23]:

######  Definition 2.3.

A measure $\mu$ is called111In [AG74] the authors call such a measure slowly
increasing. The term strongly tempered is used in [BS23]. _strongly tempered_
if there exists a polynomial $P\in{\mathbb{C}}[x_{1},x_{2},\ldots,x_{d}]$ such
that

$\int_{{\mathbb{R}}^{d}}\frac{1}{1+|P(x)|}\,\mbox{d}|\mu|(x)<\infty\,.$

The following fact about strongly tempered measures explains why this
definition is important for us.

######  Fact 2.4 ([BS23, Theorem 2.7]).

A measure $\mu$ is strongly tempered if and only if we have
$\int_{{\mathbb{R}}^{d}}|f(t)|\mbox{d}|\mu|(t)<\infty$ for all
$f\in{\mathcal{S}}({\mathbb{R}}^{d})$. Moreover, in this case, the map
$T_{\mu}:{\mathcal{S}}({\mathbb{R}}^{d})\rightarrow{\mathbb{C}}$ given by

$T_{\mu}(f):=\int_{{\mathbb{R}}^{d}}f(t)\mbox{d}\mu(t)\,,$

defines a tempered distribution. Furthermore, in this case, the measure $\mu$
and the tempered distribution $T_{\mu}$ uniquely determine each other.

Let us now discuss the Fourier transformability of strongly tempered measures
in the distributional sense. Every strongly tempered measure has a
distributional Fourier transform; however, this Fourier transform is a
tempered distribution that is not necessarily a measure. Even if it is a
measure, it may not be strongly tempered (see for example [AG74, Prop. 7.1] or
[BS23, Sect. 3]). Since our results rely on the distributional Fourier
transform being a strongly tempered measure, we introduce the following
definition.

######  Definition 2.5.

A measure $\mu$ is called Fourier transformable in the strongly tempered sense
if $\mu$ is a strongly tempered measure and there exists a strongly tempered
measure $\nu$ such that, as tempered distributions, we have

$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{T_{\mu}}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{T_{\mu}}{\tmpbox}=T_{\nu}\,.$

If this is the case, we write
$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\mu}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\mu}{\tmpbox}:=\nu$
and call it the Fourier transform of $\mu$ in the strongly tempered sense.

We write $\mathcal{S}\mathcal{M}({\mathbb{R}}^{d})$ for the space of all
strongly tempered measures $\mu$ which are Fourier transformable in the
strongly tempered sense.

In the rest of the paper, all measures are assumed to be strongly tempered.
Whenever we say that a measure $\mu$ is Fourier transformable, it is
understood to mean Fourier transformable in the strongly tempered sense. Let
us make some important remarks.

######  Remark 2.6.

  * (a)

Let $\mu$ be a strongly tempered measure. Then $\mu$ is Fourier transformable
in the measure sense (see [AG74, MS17] for a definition) if and only if
$\mu\in\mathcal{S}\mathcal{M}({\mathbb{R}}^{d})$ and
$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\mu}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\mu}{\tmpbox}$
is translation bounded [SS21]. Moreover, in this case the two Fourier
transforms coincide.

In particular, if $\mu$ is a strongly tempered measure that is also Fourier
transformable in the measure sense, then
$\mu\in\mathcal{S}\mathcal{M}({\mathbb{R}}^{d})$ and the Fourier transforms
coincide. This allows us to use the same notation for both Fourier transforms.

  * (b)

Typically, the diffraction measure
$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\gamma}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\gamma}{\tmpbox}$
is defined as the Fourier transform in the measure sense of the
autocorrelation measure $\gamma$. In general, $\gamma$ is a strongly tempered
measure. Therefore, by (a), the diffraction measure
$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\gamma}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\gamma}{\tmpbox}$
is the Fourier transform in the strongly tempered sense of $\gamma$.

We now list some easy observations about
$\mathcal{S}\mathcal{M}({\mathbb{R}}^{d})$:

######  Fact 2.7.

Let $\mu\in\mathcal{S}\mathcal{M}({\mathbb{R}}^{d})$. Then the following
statements hold:

  * (a)

The measures $\mu$ and
$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\mu}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\mu}{\tmpbox}$
uniquely determine each other.

  * (b)

For all $f\in{\mathcal{S}}({\mathbb{R}}^{d})$ we have
$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{f}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{f}{\tmpbox}\in
L^{1}(|\mu|)$, $f\in
L^{1}(|\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\mu}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\mu}{\tmpbox}|)$,
and

$\int_{{\mathbb{R}}^{d}}\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{f}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{f}{\tmpbox}(y)\mbox{d}\mu(y)=\int_{{\mathbb{R}}^{d}}f(x)\mbox{d}\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\mu}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\mu}{\tmpbox}(x)\,.$

  * (c)

$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\mu}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\mu}{\tmpbox}\in\mathcal{S}\mathcal{M}({\mathbb{R}}^{d})$
and

$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\mu}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\mu}{\tmpbox}}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\mu}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\mu}{\tmpbox}}{\tmpbox}=\mu^{\dagger}\,,$

where $\mu^{\dagger}$ is the measure defined by
$\mu^{\dagger}(f):=\int_{{\mathbb{R}}^{d}}f(-x)\,\mbox{d}\mu(x)$.

In particular, the Fourier transform is a bijection from
$\mathcal{S}\mathcal{M}({\mathbb{R}}^{d})$ onto
$\mathcal{S}\mathcal{M}({\mathbb{R}}^{d})$.

The results in this paper rely on these observations.

Let us conclude this section by recalling a classical result about
convolutions that we will need in Section 5.

######  Theorem 2.8 ([Gra08, Theorem 2.3.20]).

Let $T\in{\mathcal{S}}^{\prime}({\mathbb{R}}^{d})$ be a tempered distribution,
and let $f\in{\mathcal{S}}({\mathbb{R}}^{d})$ be a Schwartz function. Then
$f*T$ is a $C^{\infty}$ function, where

$f*T(\varphi):=T(f^{\dagger}*\varphi)\qquad\forall\varphi\in{\mathcal{S}}({\mathbb{R}}^{d})\,,$

and $f^{\dagger}$ denotes the function given by $f^{\dagger}(x)=g(-x)$ for all
$x\in{\mathbb{R}}^{d}$. Furthermore, if $T$ has compact support, then $f*T$ is
a Schwartz function.∎

######  Remark 2.9.

Any measure of compact support is a strongly tempered measure. Moreover, a
strongly tempered measure $\mu$ has compact support if and only if the
tempered distribution $T_{\mu}$ is compactly supported as a distribution.

##  3\. Bessel functions

In Section 5 we are going to run into certain integrals. It turns out that
these are exactly the integrals defining the so called Bessel function $J_{0}$
of first kind and order zero, and the modified Bessel function $I_{0}$ of the
first kind and order zero. The asymptotic behaviour of $I_{0}$ also plays an
important role in our computations. In this section we collect the results
about these functions that will be important for us.

###  3.1. $J_{0}$ and the Fourier transform of circles

######  Definition 3.1.

For each $t\geq 0$, define

$J_{0}(t):=\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{-it\cos\theta}\,\mbox{d}\theta\,.$

$J_{0}$ is called the _Bessel function of the first kind and order 0_.

Since $\cos\theta$ is $2\pi$-periodic, it is immediate that for all
$a\in{\mathbb{R}}$ we have

$\frac{1}{2\pi}\int_{a}^{a+\pi}e^{-it\cos\theta}\,\mbox{d}\theta=J_{0}(t)\,.$

The Bessel function $J_{0}$ will be important for the following reason: the
Fourier transform of the uniform probability measure of a circle is given by a
$J_{0}$ function. Let us make this statement precise.

######  Definition 3.2.

For $r>0$, define
$\theta_{r}:C_{\mathsf{c}}({\mathbb{R}}^{2})\rightarrow{\mathbb{C}}$ by

$\theta_{r}(f)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(r\cos\theta,r\sin\theta)\,\mbox{d}\theta\qquad\forall
f\in C_{\mathsf{c}}({\mathbb{R}}^{2})\,.$

Then, $\theta_{r}$ is a probability measure with compact support
$C_{r}:=\\{x\in{\mathbb{R}}^{2}:\|x\|=r\\}$.

######  Remark 3.3.

Note that $C_{r}$ is a compact Abelian group under the group operation

$(re^{i\theta})\diamond(re^{i\varphi}):=re^{i(\theta+\varphi)}\,.$

Then, $\theta_{r}$ can be seen as the unique probability Haar measure on
$(C_{r},\diamond)$, embedded into ${\mathbb{R}}^{2}$.

Since $\theta_{r}$ has compact support, it defines a strongly tempered
measure, and we can easily compute its Fourier transform using (2.1).

######  Proposition 3.4.

Let $r>0$. The Fourier transform of $\theta_{r}$ is an absolutely continuous
measure with density function

$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\theta_{r}}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\theta_{r}}{\tmpbox}(y)=J_{0}(2\pi
r\|y\|)\qquad\forall y\in{\mathbb{R}}^{2}\,.$

###### Proof.

Since $\theta_{r}$ is a distribution with compact support, its Fourier
transform is an absolutely continuous measure with density function given by

$\displaystyle\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\theta_{r}}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\theta_{r}}{\tmpbox}(y$
$\displaystyle=\int_{{\mathbb{R}}^{d}}e^{-2\pi ix\cdot
y}\,\mbox{d}\theta_{r}(x)=\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{-2\pi
ir(y_{1}\cos\theta+y_{2}\sin\theta)}\mbox{d}\theta\qquad\forall
y=(y_{1},y_{2})\in{\mathbb{R}}^{2}\,.$

Consider $y=(R\cos\phi,R\sin\phi)$ for some $R\geq 0$ and $\phi\in[-\pi,\pi)$.
We have

$\displaystyle\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\theta_{r}}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\theta_{r}}{\tmpbox}(R\cos(\phi),R\sin(\phi)$
$\displaystyle=\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{-2\pi
irR\cos(\phi+\theta)}\mbox{d}\theta$
$\displaystyle\stackrel{{\scriptstyle\psi=\phi+\theta}}{{=\joinrel=\joinrel=\joinrel=\joinrel=}}\frac{1}{2\pi}\int_{\phi-\pi}^{\phi+\pi}e^{-2\pi
irR\cos\psi}\mbox{d}\psi=J_{0}(2\pi rR)\,.$

This shows that
$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\theta_{r}}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\theta_{r}}{\tmpbox}(y)=J_{0}(2\pi
r\|y\|)$ for all $y\in{\mathbb{R}}^{2}$, as desired. ∎

###  3.2. $I_{0}$ and its asymptotic behaviour

######  Definition 3.5.

For each $t\geq 0$, define

$I_{0}(t):=\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{t\cos\theta}\,\mbox{d}\theta\,.$

$I_{0}$ is called the _modified Bessel function of the first kind of order 0_.

The function $I_{0}$ will appear naturally in Section 5 as the convolution of
a Gaussian function and $\theta_{r}$. This type of convolution will be
essential to our approach. The following asymptotic property of $I_{0}$ will
be crucial in our computations:

######  Theorem 3.6 (Asymptotic behavior of $I_{0}$).

$\lim_{t\rightarrow\infty}\frac{\sqrt{2\pi t}I_{0}(t)}{e^{t}}=1\,.$

This result is well-known and follows from a standard application of Laplace’s
method for asymptotic expansions of functions of the form

$I(x)=\int_{a}^{b}e^{xg(t)}f(t)\,\mbox{d}t\,,$

(refer to [BO99, Ch. 6.4] for the general theory of Laplace’s method). The
limit can easily be deduced from this general theory; anyhow, for
completeness, we include a short direct proof of Theorem 3.6 below.

First, we need the following lemma.

######  Lemma 3.7.

For each $t\geq 0$, we have

$I_{0}(t)=\frac{e^{t}}{\pi}\int_{0}^{2}\frac{e^{-t\xi}}{\sqrt{\xi(2-\xi)}}\,\mbox{d}\xi\,.$

###### Proof.

Using symmetry and applying the substitution $\xi=1-\cos\theta$, we have

$\displaystyle I_{0}(t)$
$\displaystyle=\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{t\cos\theta}\,\mbox{d}\theta=\frac{1}{\pi}\int_{0}^{\pi}e^{t\cos\theta}\,\mbox{d}\theta\stackrel{{\scriptstyle\xi=1-\cos\theta}}{{=\joinrel=\joinrel=\joinrel=\joinrel=\joinrel=\joinrel=}}\frac{1}{\pi}\int_{0}^{2}\frac{e^{t(1-\xi)}}{\sqrt{\xi(2-\xi)}}\,\mbox{d}\xi$
$\displaystyle=\frac{e^{t}}{\pi}\int_{0}^{2}\frac{e^{-t\xi}}{\sqrt{\xi(2-\xi)}}\,\mbox{d}\xi\,.$

∎

We can now prove the asymptotic result.

###### Proof of Theorem 3.6.

By Lemma 3.7, we have

$\displaystyle\frac{\pi I_{0}(t)}{e^{t}}$
$\displaystyle=\int_{0}^{2}\frac{e^{-t\xi}}{\sqrt{\xi(2-\xi)}}\,\mbox{d}\xi$
$\displaystyle=\underbrace{\int_{0}^{2}\frac{e^{-t\xi}}{\sqrt{2\xi}}\,\mbox{d}\xi}_{I_{1}}+\underbrace{\int_{0}^{1}\frac{e^{-t\xi}}{\sqrt{\xi}}\left(\frac{1}{\sqrt{2-\xi}}-\frac{1}{\sqrt{2}}\right)\,\mbox{d}\xi}_{I_{2}}+\underbrace{\int_{1}^{2}\frac{e^{-t\xi}}{\sqrt{\xi}}\left(\frac{1}{\sqrt{2-\xi}}-\frac{1}{\sqrt{2}}\right)\,\mbox{d}\xi}_{I_{3}}\,.$

We now look at each of the three integrals one by one. First, with a simple
substitution we get

$\displaystyle I_{1}$ $\displaystyle\stackrel{{\scriptstyle
u=\sqrt{t\xi}}}{{=\joinrel=\joinrel=\joinrel=\joinrel=}}\sqrt{\frac{2}{t}}\int_{0}^{\sqrt{2t}}e^{-u^{2}}\mbox{d}u\,.$

Then, using $\int_{-\infty}^{\infty}e^{-u^{2}}\,\mbox{d}u=\sqrt{\pi}$, we get

$\displaystyle\left|\sqrt{\frac{2t}{\pi}}I_{1}-1\right|=\left|\frac{2}{\sqrt{\pi}}\int_{0}^{\sqrt{2t}}e^{-u^{2}}\mbox{d}u-1\right|$
$\displaystyle=\frac{2}{\sqrt{\pi}}\int_{\sqrt{2t}}^{\infty}e^{-u^{2}}\mbox{d}u$
$\displaystyle\leq\frac{2}{\sqrt{\pi}}\int_{\sqrt{2t}}^{\infty}e^{-u\sqrt{2t}}\mbox{d}u=\frac{2}{\sqrt{\pi}}\frac{e^{-2t}}{\sqrt{2t}}\,.$

Next, observe that the function $f:[0,1]\to{\mathbb{R}}$

$f(\xi):=\left\\{\begin{array}[]{cc}\frac{1}{\sqrt{\xi}}\left(\frac{1}{\sqrt{2-\xi}}-\frac{1}{\sqrt{2}}\right)&\mbox{
if }\xi\neq 0\\\ 0&\mbox{ if }\xi=0\,,\end{array}\right.$

is continuous and hence bounded on $[0,1]$. Therefore, there exists some $M>0$
such that

$\displaystyle|I_{2}|$
$\displaystyle\leq\int_{0}^{1}Me^{-t\xi}\,\mbox{d}\xi=\frac{M}{t}(1-e^{-t})\,.$

For the third integral, we use that $e^{-t\xi}\leq e^{-t}$ for $t\geq 1$ to
obtain

$\displaystyle|I_{3}|$ $\displaystyle\leq
e^{-t}\int_{1}^{2}\frac{1}{\sqrt{\xi}}\left|\frac{1}{\sqrt{2-\xi}}-\frac{1}{\sqrt{2}}\right|\,\mbox{d}\xi=Ce^{-t}\,,$

where

$C:=\int_{1}^{2}\frac{1}{\sqrt{\xi}}\left|\frac{1}{\sqrt{2-\xi}}-\frac{1}{\sqrt{2}}\right|\,\mbox{d}\xi<\infty\,.$

Putting everything together, we get

$\displaystyle\left|\frac{\sqrt{2\pi t}I_{0}(t)}{e^{t}}-1\right|$
$\displaystyle=\left|\sqrt{\frac{2t}{\pi}}(I_{1}+I_{2}+I_{3})-1\right|$
$\displaystyle\leq\left|\sqrt{\frac{2t}{\pi}}I_{1}-1\right|+\sqrt{\frac{2t}{\pi}}\left(|I_{2}|+|I_{3}|\right)$
$\displaystyle\leq\frac{2}{\sqrt{\pi}}\frac{e^{-2t}}{\sqrt{2t}}+\sqrt{\frac{2t}{\pi}}\left(\frac{M}{t}(1-e^{-t})+Ce^{-t}\right)\,.$

The claim follows. ∎

##  4\. The measure of circles

We now restrict our attention to ${\mathbb{R}}^{2}$. In this section, we show
that the measure of a circle can computed as a limit via a sequence of
Schwartz functions. For a measure $\mu$ which is Fourier transformable, this
will allow us to express the measure
$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\mu}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\mu}{\tmpbox}(C_{r})$
as a formula in terms of the original measure $\mu$.

We will more generally show that we can use any sequence of Schwartz functions
that satisfy a condition, which we will call and an approximate $r$-circle.
While approximate $r$-circles are easy to construct, for applications we will
need such a sequence with nice, easily computable Fourier transforms. We will
construct one such example in Section 5.

We now introduce the notion of an approximate $r$-circle.

######  Definition 4.1.

Let $r\geq 0$. We say that a sequence
$\left\\{f_{n}\right\\}_{n\in\mathbb{N}}$ in ${\mathcal{S}}({\mathbb{R}}^{2})$
is an approximate $r$-circle if the following conditions are satisfied:

  1. (C1)

$\lim\limits_{n\rightarrow\infty}f_{n}(x)=1_{C_{r}}(x)$ for all
$x\in{\mathbb{R}}^{2}$;

  2. (C2)

there exists some $F\in{\mathcal{S}}({\mathbb{R}}^{2})$ such that

$|f_{n}(x)|\leq F(x)\qquad\forall x\in{\mathbb{R}}^{2}\,,\forall
n\in\mathbb{N}\,.$

In the following lemma, we see that for any measure in
$\mathcal{S}\mathcal{M}({\mathbb{R}}^{2})$, the measure of $C_{r}$ can be
computed as a limit via an approximate $r$-circle.

######  Lemma 4.2.

Let $\mu$ be a strongly tempered measure. Then for any approximate $r$-circle
$\left\\{f_{n}\right\\}_{n\in\mathbb{N}}$, we have

(4.1)
$\mu(C_{r})=\lim_{n\rightarrow\infty}\int_{{\mathbb{R}}^{2}}f_{n}(x)\,\mbox{d}\mu(x)\,.$

###### Proof.

By Fact 2.4, if $\mu$ is strongly tempered then $f\in L^{1}(|\mu|)$ for every
$f\in{\mathcal{S}}({\mathbb{R}}^{2})$. In particular, if
$\left\\{f_{n}\right\\}_{n\in\mathbb{N}}$ is an approximate $r$-circle and $F$
is as in condition (C2), then $F\in L^{1}(|\mu|)$. Then by condition (C1) and
the dominated convergence theorem, we have

$\mu(C_{r})=\int_{{\mathbb{R}}^{2}}1_{C_{r}}(x)\,\mbox{d}\mu(x)=\int_{{\mathbb{R}}^{2}}\lim_{n\rightarrow\infty}f_{n}(x)\,\mbox{d}\mu(x)=\lim_{n\rightarrow\infty}\int_{{\mathbb{R}}^{2}}f_{n}(x)\,\mbox{d}\mu(x)\,.$

∎

######  Remark 4.3.

Using the monotone convergence theorem, we also get that (4.1) holds for any
decreasing sequence $\left\\{f_{n}\right\\}_{n\in\mathbb{N}}$ of functions in
$L^{1}(|\mu|)$ satisfying (C1).

As an immediate consequence of Lemma 4.2 we get the following result, which
allows us calculate the intensity of a circle in the diffraction in terms of
the autocorrelation measure.

######  Corollary 4.4.

If $\mu\in\mathcal{S}\mathcal{M}({\mathbb{R}}^{d})$ and
$\left\\{f_{n}\right\\}_{n\in\mathbb{N}}$ is an approximate $r$-circle, then

$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\mu}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\mu}{\tmpbox}(C_{r})=\lim_{n\rightarrow\infty}\int_{{\mathbb{R}}^{2}}f_{n}(x)\,\mbox{d}\,\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\mu}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\mu}{\tmpbox}(x)=\lim_{n\rightarrow\infty}\int_{{\mathbb{R}}^{2}}\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{f_{n}}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{f_{n}}{\tmpbox}(x)\,\mbox{d}\mu(x)\,.\qed$

This formula will become useful once we construct examples of approximate
$r$-circles with functions that have nice Fourier transforms, which we do in
the next section.

##  5\. Circles in diffraction

In this section, we construct an example of an approximate $r$-circle via the
convolution of $\theta_{r}$ with a family of Gaussian functions. We then use
this approximate $r$-circle and Corollary 4.4 to derive an explicit formula
for the diffraction measure of a circle in terms of the autocorrelation. A
different example can be found in Section A.

Let $G_{n}$ be the following family of two dimensional Gaussian functions

(5.1) $G_{n}(x)=e^{-n^{2}\|x\|^{2}}\qquad\forall x\in{\mathbb{R}}^{2}\,,\
\forall n\in\mathbb{N}\,,$

and for each $r>0$, let $\theta_{r}$ be the uniform probability measure on the
circle $C_{r}$. Consider the convolution $G_{n}*\theta_{r}$, defined by

$G_{n}*\theta_{r}(x)=\frac{1}{2\pi}\int_{-\pi}^{\pi}G_{n}(x_{1}-r\cos\theta,x_{2}-r\sin\theta)\,\mbox{d}\theta\qquad\forall
x=(x_{1},x_{2})\in{\mathbb{R}}^{2}\,.$

Since both $\theta_{r}$ and $G_{n}$ are rotationally invariant, their
convolution $G_{n}*\theta_{r}$ is also rotationally invariant. With this
observation, it is easy to show the following.

######  Lemma 5.1.

For each $r>0$ and $n\in\mathbb{N}$, we have

(5.2)
$G_{n}*\theta_{r}(x)=I_{0}(2n^{2}r\|x\|)e^{-n^{2}(\|x\|^{2}+r^{2})}\qquad\forall
x\in{\mathbb{R}}^{2}\,.$

###### Proof.

Using rotational invariance of $G_{n}*\theta_{r}$, for each
$x\in{\mathbb{R}}^{2}$ we have:

$\displaystyle
G_{n}*\theta_{r}(x)=G_{n}*\theta_{r}(\|x\|,0)=\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{-n^{2}((\|x\|-r\cos\theta)^{2}+r^{2}\sin^{2}\theta)}\,\mbox{d}\theta$
$\displaystyle=\frac{1}{2\pi}e^{-n^{2}(\|x\|^{2}+r^{2})}\int_{-\pi}^{\pi}e^{2n^{2}r\|x\|\cos\theta}\,\mbox{d}\theta=I_{0}(2n^{2}r\|x\|)e^{-n^{2}(\|x\|^{2}+r^{2})}\,.$

∎

Throughout the rest of this section, we fix $r>0$ and define the functions

$g_{n}:=2nr\sqrt{\pi}G_{n}\,,\qquad
f_{n}=g_{n}*\theta_{r}=2nr\sqrt{\pi}G_{n}*\theta_{r}\,.$

We will prove that $f_{n}$ is an approximate $r$-circle. Note that by (5.2) we
have

$f_{n}(x)=2nr\sqrt{\pi}I_{0}(2n^{2}r\|x\|)e^{-n^{2}(\|x\|^{2}+r^{2})}\qquad\forall
x\in{\mathbb{R}}^{2}\,.$

First, we show that $f_{n}$ converges pointwise to $1_{C_{r}}$.

######  Lemma 5.2.

For each $r>0$, the sequence $\left\\{f_{n}\right\\}_{n\in\mathbb{N}}$ is an
approximate $r$-circle.

###### Proof.

Let us first note that $f_{n}\in{\mathcal{S}}({\mathbb{R}}^{2})$ for each
$n\in\mathbb{N}$ by Theorem 2.8, since $G_{n}$ is a Schwartz function and
$\theta_{r}$ is compactly supported.

Next, observe that

(5.3)
$\begin{split}f_{n}(x)&=\frac{2nr\sqrt{\pi}I_{0}(2n^{2}r\|x\|)}{e^{n^{2}(\|x\|^{2}+r^{2})}}=\frac{2n\sqrt{r\pi\|x\|}I_{0}(2n^{2}r\|x\|)}{e^{2n^{2}r\|x\|}}\cdot\frac{\sqrt{r}e^{2n^{2}r\|x\|}}{\sqrt{\|x\|}e^{n^{2}(\|x\|^{2}+r^{2})}}\\\
&=u(2n^{2}r\|x\|)\sqrt{\frac{r}{\|x\|}}e^{-n^{2}(\|x\|-r)^{2}}\end{split}$

where

$u(t):=\frac{\sqrt{2\pi t}I_{0}(t)}{e^{t}}\,.$

We now can show that the two conditions (C1) and (C2) of Definition 4.1 hold.

(C1): Since $I_{0}(0)=1$, when $x=0$ we have

$\lim_{n\rightarrow\infty}f_{n}(0)=2nr\sqrt{\pi}e^{-n^{2}r^{2}}=0\,.$

Next, by Theorem 3.6 and (5.3), for all $x\neq 0$ we have

$\lim_{n\rightarrow\infty}f_{n}(x)=\lim_{n\rightarrow\infty}\sqrt{\frac{r}{\|x\|}}e^{-n^{2}(\|x\|-r)^{2}}=\begin{cases}1&\|x\|=r\\\
0&\|x\|\neq 0\ \text{and}\ \|x\|\neq r\,.\end{cases}\,.$

This proves (C1).

(C2): The function $u$ is continuous on $[0,\infty)$ and satisfies

$\lim_{t\rightarrow\infty}u(t)=1\,.$

This implies that there exists some constant $A$ such that $|u(t)|\leq A$ for
all $t\in[0,\infty)$. In particular,

$|u(2n^{2}r\|x\|)|\leq A\qquad\forall n\in\mathbb{N},\forall
x\in{\mathbb{R}}^{2}\,.$

By (5.3) we then have

$|f_{n}(x)|\leq A\sqrt{\frac{r}{\|x\|}}e^{-n^{2}(\|x\|-r)^{2}}\leq
A\sqrt{\frac{r}{\|x\|}}e^{-(\|x\|-r)^{2}}=:\psi_{1}(x)\,.$

In particular, we get

(5.4) $|f_{n}(x)|\leq
A\sqrt{2}e^{-(\|x\|-r)^{2}}\qquad\forall\|x\|\geq\frac{r}{2}\,.$

Next, let $x\in B_{\frac{r}{2}}$. Then, for any $y\in C_{r}$ we have

$\|x-y\|\geq r-\|x\|\geq\frac{r}{2}\,.$

From this, we see that

$\displaystyle 0\leq g_{n}*\theta_{r}(x)$
$\displaystyle=\int_{{\mathbb{R}}^{2}}g_{n}(x-y)\,\mbox{d}\theta_{r}(y)=\int_{C_{r}}g_{n}(x-y)\,\mbox{d}\theta_{r}(y)$
$\displaystyle=2rn\sqrt{\pi}\int_{C_{r}}e^{-n^{2}\|x-y\|^{2}}\,\mbox{d}\theta_{r}(y)$
$\displaystyle\leq
2rn\sqrt{\pi}\int_{C_{r}}e^{-n^{2}r^{2}/4}\,\mbox{d}\theta_{r}(y)$
$\displaystyle=2rn\sqrt{\pi}e^{-n^{2}r^{2}/4}\leq 2r\sqrt{\pi}\cdot
c_{1}=:c_{2}\,,$

where

$c_{1}:=\sup\\{ne^{-n^{2}r^{2}/4}:n\in\mathbb{N}\\}<\infty\,.$

Now, pick some $\psi_{2}\in C_{\mathsf{c}}^{\infty}({\mathbb{R}}^{2})$ such
that

$\psi_{2}\geq c_{2}1_{B_{\frac{r}{2}}}\,.$

Then, we have

(5.5) $|f_{n}(x)|\leq\psi_{2}(x)\qquad\forall\|x\|\leq\frac{r}{2}\,.$

Therefore, combining (5.4) and (5.5) we get

$|f_{n}(x)|\leq\psi_{1}(x)+\psi_{2}(x)\qquad\forall\in{\mathbb{R}}^{2}\,.$

Setting $F:=\psi_{1}+\psi_{2}\in{\mathcal{S}}({\mathbb{R}}^{2})$, we showed
condition (C2). ∎

We have thus constructed an explicit approximate $r$-circle. Since we can
easily compute
$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{f_{n}}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{f_{n}}{\tmpbox}$,
we can now construct an explicit formula for the diffraction measure
$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\gamma}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\gamma}{\tmpbox}(C_{r})$
for a circle $C_{r}$ of a given radius $r>0$, in terms of the autocorrelation
$\gamma$. To derive the formula, we simply need to apply Corollary 4.4 for the
approximate $r$-circle $\left\\{f_{n}\right\\}_{n\in\mathbb{N}}$, and compute
$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{f_{n}}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{f_{n}}{\tmpbox}$.

We proved in Lemma 5.2 that $\left\\{f_{n}\right\\}_{n\in\mathbb{N}}$ is an
approximate $r$-circle, so by Corollary 4.4, we have the formula

$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\mu}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\mu}{\tmpbox}(C_{r})=\lim_{n\rightarrow\infty}\int_{{\mathbb{R}}^{2}}f_{n}(x)\,\mbox{d}\,\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\mu}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\mu}{\tmpbox}(x)=\lim_{n\rightarrow\infty}\int_{{\mathbb{R}}^{2}}\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{f_{n}}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{f_{n}}{\tmpbox}(y)\,\mbox{d}\mu(y)\,,$

for any measure $\mu\in\mathcal{S}\mathcal{M}({\mathbb{R}}^{2})$. It is easy
to see that

$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{G_{n}}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{G_{n}}{\tmpbox}(y)=\frac{\pi}{n^{2}}e^{-\pi^{2}\|y\|^{2}/n^{2}}\qquad\forall
y\in{\mathbb{R}}^{2}\,.$

Moreover, by Proposition 3.4 we have

$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\theta_{r}}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\theta_{r}}{\tmpbox}(y)=J_{0}(2\pi
r\|y\|)\qquad\forall y\in{\mathbb{R}}^{2}\,.$

Therefore, by the convolution theorem we get:

######  Lemma 5.3.

The function

$f_{n}=2nr\sqrt{\pi}G_{n}*\theta_{r}=2nr\sqrt{\pi}I_{0}(2n^{2}r\|x\|)e^{-n^{2}(\|x\|^{2}+r^{2})}\,,$

is an approximate $r$-circle with Fourier transform given by

$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{f_{n}}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{f_{n}}{\tmpbox}(y)=\frac{2r\pi\sqrt{\pi}}{n}e^{-\pi^{2}\|y\|^{2}/n^{2}}J_{0}(2\pi
r\|y\|)\qquad\forall y\in{\mathbb{R}}^{2}\,.$

∎

Combining all results from this paper we get:

######  Theorem 5.4.

Let $\mu\in\mathcal{S}\mathcal{M}({\mathbb{R}}^{2})$ and let $r>0$. Then

(5.6)
$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\mu}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\mu}{\tmpbox}(C_{r})=\lim_{n\rightarrow\infty}\frac{2r\pi\sqrt{\pi}}{n}\int_{{\mathbb{R}}^{2}}e^{-\pi^{2}\|y\|^{2}/n^{2}}J_{0}(2\pi
r\|y\|)\,\mbox{d}\mu(y)\,.$

∎

######  Remark 5.5.

When $\mu=\gamma$ is an autocorrelation measure, the above formula calculates
the diffraction measure
$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\gamma}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\gamma}{\tmpbox}(C_{r})$
for the circle $C_{r}$ in terms of the autocorrelation measure $\gamma$.

##  6\. Applications

In this section, we provide applications of Theorem 5.4 and highlight some
interesting observations that could lead to further applications. Below, we
use $\lambda$ to denote the Lebesgue measure on ${\mathbb{R}}^{2}$, and
sometimes use the notation $\mbox{d}\lambda(x)$ instead of $\mbox{d}x$ in
cases where the integral could be mistaken for a line integral.

We start with some simple examples with measures $\mu$ where
$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\mu}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\mu}{\tmpbox}(C_{r})$
is known and easily evaluated.

######  Example 6.1.

Let $\mu=\delta_{x}$ for some $x\in{\mathbb{R}}^{2}$. Then
$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\mu}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\mu}{\tmpbox}(y)=e^{-2\pi
ix\cdot y}$ and (5.6) becomes

$\int_{C_{r}}e^{-2\pi ix\cdot
y}\,\mbox{d}\lambda(y)=\lim_{n\rightarrow\infty}\frac{2r\pi\sqrt{\pi}}{n}e^{-\pi^{2}\|x\|^{2}/n^{2}}J_{0}(2\pi
r\|x\|)=0\,,$

where $\lambda$ is the Lebesgue measure in ${\mathbb{R}}^{2}$.

######  Example 6.2.

Let $\mu=\theta_{r^{\prime}}$. Recall from Proposition 3.4 that for any
$r^{\prime}>0$, we have that

$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\theta_{r^{\prime}}}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\theta_{r^{\prime}}}{\tmpbox}(y)=J_{0}(2\pi
r^{\prime}\|y\|)\qquad\forall y\in{\mathbb{R}}^{2}\,.$

Thus, in this case, (5.6) gives us:

$\int_{C_{r}}J_{0}(2\pi
r^{\prime}\|x\|)\,\mbox{d}\lambda(x)=\lim_{n\rightarrow\infty}\frac{r\pi^{\frac{1}{2}}}{n}\int_{-\pi}^{\pi}e^{-\pi^{2}(r^{\prime})^{2}/n^{2}}J_{0}(2\pi
rr^{\prime})\,\mbox{d}\theta=0\,.$

######  Remark 6.3.

More generally, if $\mu\in\mathcal{S}\mathcal{M}({\mathbb{R}}^{2})$ is such
that
$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\mu}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\mu}{\tmpbox}$
is an absolutely continuous measure, then

$\lim_{n\rightarrow\infty}\frac{2r\pi\sqrt{\pi}}{n}\int_{{\mathbb{R}}^{2}}e^{-\pi^{2}\|y\|^{2}/n^{2}}J_{0}(2\pi
r\|y\|)\,\mbox{d}\mu(y)=0\,.$

Next, let us look at the cases where $\mu$ is a character.

######  Example 6.4.

Let $\mu=\lambda$ be the Lebesgue measure on ${\mathbb{R}}^{2}$. Then
$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\mu}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\mu}{\tmpbox}=\delta_{0}$
and (5.6) tells us that for all $r>0$ we have

$\lim_{n\rightarrow\infty}\frac{2r\pi\sqrt{\pi}}{n}\int_{{\mathbb{R}}^{2}}e^{-\pi^{2}\|y\|^{2}/n^{2}}J_{0}(2\pi
r\|y\|)\,\mbox{d}y=0\,.$

######  Example 6.5.

Let $x\in{\mathbb{R}}^{2}\backslash\\{0\\}$ and let $\mu$ be the function
$y\mapsto e^{-2\pi ix\cdot y}$. Then
$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\mu}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\mu}{\tmpbox}=\delta_{x}$
and (5.6) becomes

$\lim_{n\rightarrow\infty}\frac{2r\pi\sqrt{\pi}}{n}\int_{{\mathbb{R}}^{2}}e^{-2\pi
ix\cdot y}e^{-\pi^{2}\|y\|^{2}/n^{2}}J_{0}(2\pi
r\|y\|)\,\mbox{d}y=\begin{cases}1&\|x\|=r\\\ 0&\|x\|\neq r\,.\end{cases}$

Notice that the above limit is just $\lim_{n\rightarrow\infty}f_{n}(x)$, where
$f_{n}$ is the approximate $r$-circle defined in Section 5.

Next, we consider the case where
$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\mu}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\mu}{\tmpbox}=\theta_{r^{\prime}}$
for some $r^{\prime}>0$. This example is of particular interest because it
leads to an orthogonality relation for Bessel functions of the form
$J_{0}(2\pi r\|\cdot\|)$. Applying Corollary 4.4 with $\mu=J_{0}(2\pi
r\|\cdot\|)$, we get the following:

######  Theorem 6.6.

Let $r,r^{\prime}>0$. If $\left\\{f_{n}\right\\}_{n\in\mathbb{N}}$ is any
approximate $r$-circle, then the following identity holds:

$\lim_{n\rightarrow\infty}\frac{2r\pi\sqrt{\pi}}{n}\int_{{\mathbb{R}}^{2}}\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{f_{n}}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{f_{n}}{\tmpbox}(y)J_{0}(2\pi
r^{\prime}\|y\|)\,\mbox{d}y=\begin{cases}1&r^{\prime}=r\\\ 0&r^{\prime}\neq
r\,.\end{cases}$

In particular, when $f_{n}=2nr\sqrt{\pi}G_{n}*\theta_{r}$ we have

(6.1)
$\displaystyle\lim_{n\rightarrow\infty}\frac{r\sqrt{\pi}}{n}\int_{0}^{\infty}e^{-z^{2}/4n^{2}}J_{0}(rz)J_{0}(r^{\prime}z)z\,\mbox{d}z$
$\displaystyle=\lim_{n\rightarrow\infty}\frac{4r\pi^{2}\sqrt{\pi}}{n}\int_{0}^{\infty}e^{-\pi^{2}u^{2}/n^{2}}J_{0}(2\pi
ru)J_{0}(2\pi r^{\prime}u)u\,\mbox{d}u$
$\displaystyle=\lim_{n\rightarrow\infty}\frac{2r\pi\sqrt{\pi}}{n}\int_{{\mathbb{R}}^{2}}e^{-\pi^{2}\|y\|^{2}/n^{2}}J_{0}(2\pi
r\|y\|)J_{0}(2\pi r^{\prime}\|y\|)\,\mbox{d}y=\begin{cases}1&r^{\prime}=r\\\
0&r^{\prime}\neq r\,.\end{cases}$

###### Proof.

Corollary 4.4 applied to $\mu=J_{0}(2\pi r\|\cdot\|)$ gives

$\lim_{n\rightarrow\infty}\frac{2r\pi\sqrt{\pi}}{n}\int_{{\mathbb{R}}^{2}}e^{-\pi^{2}\|y\|^{2}/n^{2}}J_{0}(2\pi
r\|y\|)J_{0}(2\pi r^{\prime}\|y\|)\,\mbox{d}y=\begin{cases}1&r^{\prime}=r\\\
0&r^{\prime}\neq r\,.\end{cases}\,.$

Next, switching to polar coordinates gives

$\int_{{\mathbb{R}}^{2}}e^{-\pi^{2}\|y\|^{2}/n^{2}}J_{0}(2\pi
r\|y\|)J_{0}(2\pi
r^{\prime}\|y\|)\,\mbox{d}y=2\pi\int_{0}^{\infty}e^{-\pi^{2}u^{2}/n^{2}}J_{0}(2\pi
ru)J_{0}(2\pi r^{\prime}u)u\,\mbox{d}u\,.$

The substitution $z=2\pi u$ yields the remaining formula. ∎

######  Remark 6.7.

The integrals that appear in (6.1) can be evaluated using standard tables of
integrals (e.g. [GR00, 6.633 (2)]).

The above examples show that Corollary 4.4 and Theorem 5.4 can lead to
interesting integral identities. In the next section, we will focus on pure
point measures; however, before doing so, let us state here the following
result, which follows immediately from the rotational invariance of the
pinwheel diffraction [MPS06].

######  Proposition 6.8.

Let $\gamma$ be the autocorrelation of the pinwheel tiling and let $r>0$.
Then, there exists a circle of radius $r$ in the pinwheel diffraction if and
only if

$\lim_{n\rightarrow\infty}\frac{2r\pi\sqrt{\pi}}{n}\int_{{\mathbb{R}}^{2}}e^{-\pi^{2}\|y\|^{2}/n^{2}}J_{0}(2\pi
r\|y\|)\,\mbox{d}\gamma(y)\neq 0\,.$

∎

######  Remark 6.9.

The same also holds for the autocorrelation and diffraction of any tiling with
statistical circular symmetry [Fre08], as any such tiling has rotationally
invariant diffraction.

###  6.1. Measures with pure point Fourier transform.

Next we consider some examples with pure point diffraction. In this case, the
left side of (5.6) is the sum of the intensities of the Bragg peaks that lie
on the circle $C_{r}$. More precisely, we get the following:

######  Proposition 6.10.

Let $r>0$ and let $\mu\in\mathcal{S}\mathcal{M}({\mathbb{R}}^{2})$ be such
that
$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\mu}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\mu}{\tmpbox}$
is pure point, i.e.
$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\mu}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\mu}{\tmpbox}$
is concentrated on a countable subset of ${\mathbb{R}}^{2}$. Let $f_{n}$ be
any approximate $r$-circle. Then

$\lim_{n\rightarrow\infty}\int_{{\mathbb{R}}^{2}}\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{f_{n}}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{f_{n}}{\tmpbox}(y)\,\mbox{d}\mu(y)=\sum_{\begin{subarray}{c}x\in{\mathbb{R}}^{2}\\\
\|x\|=r\end{subarray}}\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\mu}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\mu}{\tmpbox}(\\{x\\})\,.$

In particular, when $f_{n}=2nr\sqrt{\pi}G_{n}*\theta_{r}$ we have

$\lim_{n\rightarrow\infty}\frac{2r\pi\sqrt{\pi}}{n}\int_{{\mathbb{R}}^{2}}e^{-\pi^{2}\|y\|^{2}/n^{2}}J_{0}(2\pi
r\|y\|)\,\mbox{d}\mu(y)=\sum_{\begin{subarray}{c}x\in{\mathbb{R}}^{2}\\\
\|x\|=r\end{subarray}}\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\mu}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\mu}{\tmpbox}(\\{x\\})\,.$

∎

This result leads to an interesting formula for the sum of two squares
function from number theory. Let us first recall its definition.

######  Definition 6.11.

The sum of two squares function $r_{2}$ is the function
$r_{2}:\mathbb{N}\to\mathbb{N}$ defined as

$r_{2}(n):={\rm card}\\{(k,l)\in{\mathbb{Z}}^{2}:k^{2}+l^{2}=n\\}\,.$

######  Remark 6.12.

The sum of two squares function has the following properties:

  * (a)

$r_{2}$ is an arithmetic function, meaning that $r_{2}(mn)=r_{2}(m)r_{2}(n)$
for all relatively prime $m,n$.

  * (b)

$r_{2}$ has the alternate formula

$r_{2}(n)=4(d_{1}(n)-d_{3}(n))\,,$

where $d_{j}(n)$ is the number of divisors of $n$ which equal $j\pmod{4}$.

  * (c)

In some places $r_{2}(n)$ is called the shelling function, see for example
[BGJR00], [BG03].

By the Poisson summation formula, we have that
$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\delta_{{\mathbb{Z}}^{2}}}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\delta_{{\mathbb{Z}}^{2}}}{\tmpbox}=\delta_{{\mathbb{Z}}^{2}}$.
Therefore, we get the following formula for $r_{2}(n)$ in terms of Bessel
functions. We only list the result using the approximate $r$-circle from
Section 5, but similar formulas can be deduced for any approximate $r$-circle.

######  Proposition 6.13.

The sum of two squares function satisfies the following:

  * (a)

For all $k\in\mathbb{N}$ we have

$r_{2}(k)=\lim_{n\rightarrow\infty}\frac{2\pi\sqrt{\pi
k}}{n}\sum_{m=0}^{\infty}r_{2}(m)e^{-\frac{\pi^{2}m}{n^{2}}}J_{0}(2\pi\sqrt{km})\,.$

  * (b)

For all $r>0$ such that $r^{2}\notin\mathbb{N}$ we have

$\lim_{n\rightarrow\infty}\frac{2r\pi\sqrt{\pi}}{n}\sum_{m=0}^{\infty}r_{2}(m)e^{-\frac{\pi^{2}m\textbf{}}{n^{2}}}J_{0}(2\pi
r\sqrt{m})=0\,.$

###### Proof.

By Proposition 6.10, for each $r>0$ we have

$\delta_{{\mathbb{Z}}^{2}}(C_{r})=\lim_{n\rightarrow\infty}\frac{2r\pi\sqrt{\pi}}{n}\sum_{y\in{\mathbb{Z}}^{2}}e^{-\pi^{2}\|y\|^{2}/n^{2}}J_{0}(2\pi
r\|y\|)\,\,.$

Next, grouping the elements in ${\mathbb{Z}}^{2}$ by their norm, we get

$\displaystyle\sum_{y\in{\mathbb{Z}}^{2}}e^{-\pi^{2}\|y\|^{2}/n^{2}}J_{0}(2\pi
r\|y\|)=\sum_{m=0}^{\infty}\sum_{\begin{subarray}{c}y\in{\mathbb{Z}}^{2}\\\
\|y\|^{2}=m\end{subarray}}e^{-\pi^{2}\|y\|^{2}/n^{2}}J_{0}(2\pi r\|y\|)$
$\displaystyle=\sum_{m=0}^{\infty}\sum_{\begin{subarray}{c}y\in{\mathbb{Z}}^{2}\\\
\|y\|^{2}=m\end{subarray}}e^{-\pi^{2}m/n^{2}}J_{0}(2\pi
r\sqrt{m})=\sum_{m=0}^{\infty}r_{2}(m)e^{-\pi^{2}m/n^{2}}J_{0}(2\pi
r\sqrt{m})\,.$

Now, if $r^{2}=k\in\mathbb{N}$ we have

$\delta_{{\mathbb{Z}}^{2}}(C_{r})=r_{2}(k)\,,$

while when $r^{2}\notin\mathbb{N}$ we have

$\delta_{{\mathbb{Z}}^{2}}(C_{r})=0\,.$

∎

This can easily be extended to arbitrary lattices in ${\mathbb{R}}^{2}$.
Indeed, for a locally finite point set $\Lambda\subseteq{\mathbb{R}}^{2}$
denote by $r_{\Lambda}:[0,\infty)\to{\mathbb{R}}$ the function

$r_{\Lambda}(k):={\rm
card}\\{(s,t)\in\Lambda:s^{2}+t^{2}=k\\}=\delta_{\Lambda}(C_{\sqrt{k}})\,.$

Then, exactly as in Proposition 6.13 we have:

######  Proposition 6.14.

Let $L\subseteq{\mathbb{R}}^{2}$ be a lattice with dual lattice $L^{0}$. Then,
for all $k>0$ we have

$r_{L}(k)=\frac{1}{\det(L)}\lim_{n\rightarrow\infty}\frac{2\pi\sqrt{\pi
k}}{n}\sum_{m\in[0,\infty)}r_{L^{0}}(m)e^{-\frac{\pi^{2}m}{n^{2}}}J_{0}(2\pi\sqrt{mk})\,.$

∎

##  Appendix A A family of approximate $r$-circles

In this appendix we offer an alternate approach to constructing approximate
$r$-circles via convolution with the annulus

$C_{r,\frac{1}{n}}:=\left\\{x\in{\mathbb{R}}^{2}\,:\,r-\frac{1}{n}<\|x\|<r+\frac{1}{n}\right\\}\qquad\forall
n\in\mathbb{N}\,,$

instead of with $\theta_{r}$.

We provide general conditions on a sequence of functions $g_{n}$ that suffice
for the convolution $g_{n}*1_{C_{r,\frac{1}{n}}}$ to be an approximate
$r$-circle. The resulting formula for
$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\mu}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\mu}{\tmpbox}(C_{r})$
is more complicated than that given in Theorem 5.4; however, the approach
yields a large family of approximate $r$-circles, whereas the approach taken
in Section 5 is specific to the convolution of $\theta_{r}$ with the Gaussian
functions $G_{n}$.

######  Lemma A.1.

Let $r>0$, and let $\left\\{g_{n}\right\\}_{n\in\mathbb{N}}$ be any sequence
in ${\mathcal{S}}({\mathbb{R}}^{2})$ with the following properties:

  1. (i)

$g_{n}\geq 0$ for all $n\in\mathbb{N}$;

  2. (ii)

$\lim\limits_{n\rightarrow\infty}\int_{B_{\frac{1}{n}}}g_{n}(x)\,\mbox{d}x=1$;

  3. (iii)

there exists some $\varphi\in{\mathcal{S}}({\mathbb{R}}^{2})$ such that

$g_{n}(x)\leq\varphi(x)\qquad\forall\|x\|\geq\frac{1}{n}\,.$

Then the sequence

$f_{n}:=g_{n}*1_{C_{r,\frac{1}{n}}}\qquad\forall n\in\mathbb{N}\,,$

is an approximate $r$-circle.

###### Proof.

By Theorem 2.8, since $g_{n}$ is a Schwartz function and
$1_{C_{r,\frac{1}{n}}}$ is compactly supported, we have that
$f_{n}=g_{n}*1_{C_{r,\frac{1}{n}}}\in{\mathcal{S}}({\mathbb{R}}^{2})$ for each
$n\in\mathbb{N}$. Next, we need to show that the sequence
$\left\\{f_{n}\right\\}_{n\in\mathbb{N}}$ satisfies conditions (C1) and (C2)
of Definition 4.1. We do will this in three steps. First, let us define:

$u_{n}:=(g_{n}\cdot 1_{B_{\frac{1}{n}}})*1_{C_{r,\frac{1}{n}}}\,,\qquad
v_{n}:=f_{n}-u_{n}=(g_{n}\cdot 1_{{\mathbb{R}}^{2}\backslash
B_{\frac{1}{n}}})*1_{C_{r,\frac{1}{n}}}\,.$

Then $f_{n}=u_{n}+v_{n}$.

Step 1. We show that there exists a constant $C>0$ and some sequence $a_{n}\to
0$ such that for all $n\in\mathbb{N}$ and $x\in{\mathbb{R}}^{2}$ we have

$\displaystyle 0\leq u_{n}(x)\leq C\,,\qquad 0\leq v_{n}(x)\leq a_{n}\,.$

Observe that:

$u_{n}(x)=\int_{{\mathbb{R}}^{2}}g_{n}(x-y)1_{B_{\frac{1}{n}}}(x-y)1_{C_{r,\frac{1}{n}}}(y)\,\mbox{d}y=\int_{x-C_{r,\frac{1}{n}}}g_{n}(y)1_{B_{\frac{1}{n}}}(y)\,\mbox{d}y\,,$

where

$x-C_{r,\frac{1}{n}}=\left\\{y\in{\mathbb{R}}^{2}\,:\,r-\frac{1}{n}<\|x-y\|<r+\frac{1}{n}\right\\}\,.$

From this and our assumptions we get

$0\stackrel{{\scriptstyle(i)}}{{\leq}}u_{n}(x)\leq\int_{B_{\frac{1}{n}}}g_{n}(y)\,\mbox{d}y\stackrel{{\scriptstyle(ii)}}{{\leq}}C<\infty\qquad\forall
x\in{\mathbb{R}}^{2}\,,$

where the last inequality uses that
$\int_{B_{\frac{1}{n}}}g_{n}(x)\,\mbox{d}x$ is a convergent sequence, and
hence is bounded by some $C>0$.

Next, by assumptions (i) and (iii) we have:

$0\stackrel{{\scriptstyle(i)}}{{\leq}}v_{n}(x)\stackrel{{\scriptstyle(iii)}}{{\leq}}\|\varphi\|_{\infty}\cdot\mbox{\text{Area}}(C_{r,\frac{1}{n}})=:a_{n}\rightarrow
0\,.$

Step 2. We show that $u_{n}(x)\rightarrow 1_{C_{r}}(x)$ for all
$x\in{\mathbb{R}}^{2}$. To prove this, we split the problem into two cases.

  * •

_Case 1:_ $\|x\|\neq r$.

If $\|x\|\neq r$, then there exists an $N\in\mathbb{N}$ such that
$\overline{B_{\frac{1}{n}}}\cap(x-C_{r,\frac{1}{n}})=\emptyset$ for all $n\geq
N$. From this and the observation that $g_{n}1_{B_{\frac{1}{n}}}$ is supported
in $\overline{B_{\frac{1}{n}}}$, we get

$u_{n}(x)=\int_{x-C_{r,\frac{1}{n}}}g_{n}(y)1_{B_{\frac{1}{n}}}(y)\,\mbox{d}y=\int_{x-C_{r,\frac{1}{n}}}0\,\mbox{d}y=0\,,$

for all $n\geq N$, so $u_{n}(x)\rightarrow 0$.

  * •

_Case 2:_ $\|x\|=r$.

If $\|x\|=r$, then we have $B_{\frac{1}{n}}\subseteq x-C_{r,\frac{1}{n}}$ for
every $n\in\mathbb{N}$. From this, we see that

$u_{n}(x)=\int_{x-C_{r,\frac{1}{n}}}g_{n}(y)1_{B_{\frac{1}{n}}}(y)\,\mbox{d}y=\int_{B_{\frac{1}{n}}}g_{n}(y)\,\mbox{d}y\,,$

so by (ii), we get $u_{n}(x)\rightarrow 1$.

Moreover, by Step 1, we have that $\lim_{n\rightarrow\infty}v_{n}(x)=0$ for
all $x$, so it follows that condition (C1) is satisfied.

Step 3. We prove (C2). Observe that $u_{n}$ is compactly supported with

$\mbox{\rm supp}(u_{n})\subseteq\mbox{\rm supp}(g_{n}\cdot
1_{B_{\frac{1}{n}}})+\mbox{\rm
supp}(1_{C_{r,\frac{1}{n}}})\subseteq\overline{B_{\frac{1}{n}}}+\overline{B_{r+\frac{1}{n}}}\subseteq\overline{B_{r+2}}\qquad\forall
n\in\mathbb{N}\,.$

Fix some $\psi\in C_{\mathsf{c}}^{\infty}({\mathbb{R}}^{2})$ such that
$\psi\geq 1_{\overline{B_{r+2}}}$. Then, using positivity, we have

$0\stackrel{{\scriptstyle(i)}}{{\leq}}u_{n}\leq C1_{\overline{B_{r+2}}}\leq
C\psi\in{\mathcal{S}}({\mathbb{R}}^{2})\qquad\forall n\in\mathbb{N}\,,$

and

$\displaystyle
0\stackrel{{\scriptstyle(i)}}{{\leq}}v_{n}\leq\varphi*\psi\in{\mathcal{S}}({\mathbb{R}}^{2})\,.$

It follows that

$0\leq f_{n}=u_{n}+v_{n}\leq
C\psi+\varphi*\psi\in{\mathcal{S}}({\mathbb{R}}^{2})\qquad\forall
n\in\mathbb{N}\,,$

Therefore, (C2) is satisfied. ∎

######  Remark A.2.

The same proof shows that Lemma A.1 also holds in ${\mathbb{R}}^{d}$.

Next, we would like to use this to derive a result similar to Theorem 5.4 that
gives a formula for
$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\mu}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\mu}{\tmpbox}(C_{r})$
for all $\mu\in\mathcal{S}\mathcal{M}({\mathbb{R}}^{2})$. We will need the
Fourier transform of the characteristic function $1_{B_{R}}$. It turns out
that this can be expressed in terms of the following Bessel function $J_{1}$:

######  Definition A.3.

For each $t\geq 0$, define

$J_{1}(t):=\begin{cases}\frac{1}{t}\int_{0}^{t}sJ_{0}(s)\,\mbox{d}s&t>0\\\
0&t=0\,.\end{cases}$

This function is the _Bessel function of the first kind and order 1_.

The above definition of $J_{1}$ is equivalent to the integral formula

$\int_{0}^{x}uJ_{0}(u)\mbox{d}u=xJ_{1}(x)\qquad\forall x\in{\mathbb{R}}\,,$

which we will use below.

######  Remark A.4.

$J_{1}$ is usually defined as

$J_{1}(t)=\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{i(\theta-t\sin\theta)}\,\mbox{d}\theta\,.$

The equivalence of these definitions is most easily seen from the well-known
series representations for $J_{0}$ and $J_{1}$ [Wat44, p. 45].

Note that, since $1_{B_{R}}$ has compact support, its Fourier transform
$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{1_{B_{R}}}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{1_{B_{R}}}{\tmpbox}$
must be a continuous function. We get the following formula:

######  Lemma A.5.

For any $R>0$,
$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{1_{B_{R}}}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{1_{B_{R}}}{\tmpbox}$
is the continuous function given by

(A.1)
$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{1_{B_{R}}}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{1_{B_{R}}}{\tmpbox}(y)=\begin{cases}\frac{R}{\|y\|}J_{1}(2\pi
R\|y\|)&y\neq 0\\\ \pi R^{2}&y=0\end{cases}\qquad\forall
y\in{\mathbb{R}}^{2}\,.$

###### Proof.

Let $R>0$ and $y\in{\mathbb{R}}^{2}$. When $y=0$, we have

$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{1_{B_{R}}}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{1_{B_{R}}}{\tmpbox}(0)=\int_{{\mathbb{R}}^{2}}1_{B_{R}}(x)\,\mbox{d}x=\pi
R^{2}\,.$

Next consider the case when $y\neq 0$. Since $1_{B_{R}}$ is rotational
invariant we get

as claimed.

∎

Next, observe that whenever $n>\frac{1}{r}$, we have

$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{1_{C_{r,\frac{1}{n}}}}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{1_{C_{r,\frac{1}{n}}}}{\tmpbox}=\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{1_{B_{r+\frac{1}{n}}}}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{1_{B_{r+\frac{1}{n}}}}{\tmpbox}-\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{1_{B_{r-\frac{1}{n}}}}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{1_{B_{r-\frac{1}{n}}}}{\tmpbox}\,.$

This lets us evaluate
$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{1_{C_{r,\frac{1}{n}}}}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{1_{C_{r,\frac{1}{n}}}}{\tmpbox}$
using Lemma A.5:

######  Corollary A.6.

For any $r>0$ and $n>\frac{1}{r}$, we have

(A.2)
$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{1_{C_{r,\frac{1}{n}}}}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{1_{C_{r,\frac{1}{n}}}}{\tmpbox}(y)=\begin{cases}\frac{nr+1}{n\|y\|}J_{1}\left(\frac{2\pi\|y\|(nr+1)}{n}\right)-\frac{nr-1}{n\|y\|}J_{1}\left(\frac{2\pi\|y\|(nr-1)}{n}\right)&y\neq
0\\\
\pi\left(r+\frac{1}{n}\right)^{2}-\pi\left(r-\frac{1}{n}\right)^{2}&y=0\end{cases}\qquad\forall
y\in{\mathbb{R}}^{2}\,.$

∎

As shown above, we have an explicit formula for the Fourier transform
$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{1_{C_{r,\frac{1}{n}}}}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{1_{C_{r,\frac{1}{n}}}}{\tmpbox}$.
Thus, given a Schwartz function $g_{n}$ as in Lemma A.1, we can use the
convolution theorem to get

$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{g_{n}*1_{C_{r,\frac{1}{n}}}}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{g_{n}*1_{C_{r,\frac{1}{n}}}}{\tmpbox}=\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{g_{n}}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{g_{n}}{\tmpbox}\cdot\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{1_{C_{r,\frac{1}{n}}}}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{1_{C_{r,\frac{1}{n}}}}{\tmpbox}\,,$

which would allow us to compute explicitly the Fourier transform
$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{g_{n}*1_{C_{r,\frac{1}{n}}}}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{g_{n}*1_{C_{r,\frac{1}{n}}}}{\tmpbox}$
provided that
$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{g_{n}}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{g_{n}}{\tmpbox}$
can also be computed explicitly. In particular, combining Lemma 4.2, Lemma
A.1, and Corollary A.6 gives us the following general theorem:

######  Theorem A.7.

Let $r>0$. If $\left\\{g_{n}\right\\}_{n\in\mathbb{N}}$ is a sequence in
${\mathcal{S}}({\mathbb{R}}^{2})$ satisfying the conditions of Lemma A.1 and
$\mu\in\mathcal{S}\mathcal{M}({\mathbb{R}}^{2})$, then we have

(A.3)
$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\mu}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\mu}{\tmpbox}(C_{r})=\lim_{n\rightarrow\infty}\int_{{\mathbb{R}}^{2}}\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{g_{n}}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{g_{n}}{\tmpbox}(y)\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{1_{C_{r,\frac{1}{n}}}}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{1_{C_{r,\frac{1}{n}}}}{\tmpbox}(y)\,\mbox{d}\mu(y)\,,$

where
$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{1_{C_{r,\frac{1}{n}}}}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{1_{C_{r,\frac{1}{n}}}}{\tmpbox}(y)$
is given by (A.2). ∎

######  Remark A.8.

The above formula (A.3) also holds when
$\left\\{g_{n}\right\\}_{n\in\mathbb{N}}$ is a sequence of compactly supported
functions in $L^{1}({\mathbb{R}}^{2})\cap L^{2}({\mathbb{R}}^{2})$ satisfying

$\mbox{\rm supp}(g_{n})\subseteq\overline{B_{\frac{1}{n}}}\qquad\forall
n\in\mathbb{N}\,,$

and conditions (i) and (ii) of Lemma A.1, provided that
$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\mu}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\mu}{\tmpbox}$
is translation bounded. This can be seen from [RS17a, Theorem 3.12] and the
Fourier transformability of $\mu$ as a measure (see Remark 2.6).

We conclude with an example using a sequence of Gaussian functions and Theorem
A.7 to produce an alternate formula (A.7) for the intensity of a circle.

######  Example A.9.

Consider the sequence of Gaussian functions

$g_{n}(x):=\frac{n^{4}}{\pi}e^{-n^{4}\|x\|^{2}}\qquad\forall
x\in{\mathbb{R}}^{2},\ \forall n\in\mathbb{N}\,.$

Then $g_{n}\geq 0$ and

$\displaystyle\int_{B_{\frac{1}{n}}}g_{n}(x)\,\mbox{d}x$
$\displaystyle=\frac{n^{4}}{\pi}\int_{\|x\|\leq\frac{1}{n}}e^{-n^{4}\|x\|^{2}}\,\mbox{d}x$
$\displaystyle\stackrel{{\scriptstyle
y=n^{2}x}}{{=\joinrel=\joinrel=\joinrel=\joinrel=}}\frac{1}{\pi}\int_{\|y\|\leq
n}e^{-\|y\|^{2}}\,\mbox{d}y\rightarrow\frac{1}{\pi}\int_{{\mathbb{R}}^{2}}e^{-\|y\|^{2}}\,\mbox{d}y=1\,,$

so $\\{g_{n}\\}_{n\in\mathbb{N}}$ satisfies conditions (i) and (ii) of Lemma
A.1.

Next, to verify condition (iii), we demonstrate that, for the Schwartz
function

$\varphi(x):=\frac{4}{\pi}e^{1-\|x\|^{2}}\qquad\forall
x\in{\mathbb{R}}^{2}\,,$

we have $g_{n}(x)\leq\varphi(x)$ whenever $x\in{\mathbb{R}}^{2}\backslash
B_{\frac{1}{n}}$. Equivalently, we want to verify that

(A.4) $\frac{n^{4}}{4}\leq e^{1+(n^{4}-1)t^{2}}\qquad\forall
t\geq\frac{1}{n}\,.$

Indeed, for $t\geq\frac{1}{n}$, we have

$1+(n^{4}-1)t^{2}\geq 1+n^{2}-\frac{1}{n^{2}}\geq n^{2}\,,$

so

$\ln\left(\frac{n^{4}}{4}\right)=2\ln\left(\frac{n^{2}}{2}\right)\leq
2\cdot\frac{n^{2}}{2}=n^{2}\leq
1+(n^{4}-1)t^{2}=\ln\left(e^{1+(n^{4}-1)t^{2}}\right)\,.$

This verifies (A.4).

Thus, we have shown that $\left\\{g_{n}\right\\}_{n\in\mathbb{N}}$ satisfies
the conditions of Lemma A.1, so given a measure
$\mu\in\mathcal{S}\mathcal{M}({\mathbb{R}}^{2})$ and some $r>0$, we can apply
Theorem A.7 to obtain the following formula for
$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\mu}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\mu}{\tmpbox}(C_{r})$:

(A.7)
$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{\gamma}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-505.89pt]{4.30554pt}{505.89pt}}}{}}{0.5ex}}\stackon[1pt]{\gamma}{\tmpbox}(C_{r})=\lim_{n\rightarrow\infty}\int_{{\mathbb{R}}^{2}}e^{-\pi^{2}\|y\|^{2}/n^{4}}\\\
\times\left[\frac{nr+1}{n\|y\|}J_{1}\left(\frac{2\pi\|y\|(nr+1)}{n}\right)-\frac{nr-1}{n\|y\|}J_{1}\left(\frac{2\pi\|y\|(nr-1)}{n}\right)\right]\,\mbox{d}\gamma(y)\,.$

for any $\mu\in\mathcal{S}\mathcal{M}({\mathbb{R}}^{2})$ and any $r>0$.

## Acknowledgements

E.R.K. was supported by NSERC via the NSERC CGS D scholarship. N.S. was
supported by the NSERC Discovery grants 2020-00038 and 2024-04853. The authors
are grateful to Michael Baake and Noel Murasko for some discussions and
suggestions which improved the quality of the manuscript.

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