# Discrete-to-Continuum Limits of Long-Range Electrical Interactions in
Nanostructures††thanks:

Prashant K. Jha Oden Institute for Computational Engineering and Sciences, The
University of Texas at Austin, Austin, USA<EMAIL_ADDRESS>Timothy
Breitzman Air Force Research Laboratory, Wright-Patterson Air Force Base, USA
<EMAIL_ADDRESS>Kaushik Dayal Department of Civil and
Environmental Engineering, Carnegie Mellon University, Pittsburgh, USA; Center
for Nonlinear Analysis, Department of Mathematical Sciences, Carnegie Mellon
University, Pittsburgh, USA; Department of Mechanical Engineering, Carnegie
Mellon University, Pittsburgh, USA<EMAIL_ADDRESS>
###### Abstract

We consider electrostatic interactions in two classes of nanostructures
embedded in a three dimensional space: (1) helical nanotubes, and (2) thin
films with uniform bending (i.e., constant mean curvature). Starting from the
atomic scale with a discrete distribution of dipoles, we obtain the continuum
limit of the electrostatic energy; the continuum energy depends on the
geometric parameters that define the nanostructure, such as the pitch and
twist of the helical nanotubes and the curvature of the thin film. We find
that the limiting energy is local in nature. This can be rationalized by
noticing that the decay of the dipole kernel is sufficiently fast when the
lattice sums run over one and two dimensions, and is also consistent with
prior work on dimension reduction of continuum micromagnetic bodies to the
thin film limit. However, an interesting contrast between the discrete-to-
continuum approach and the continuum dimension reduction approaches is that
the limit energy in the latter depends only on the normal component of the
dipole field, whereas in the discrete-to-continuum approach, both tangential
and normal components of the dipole field contribute to the limit energy.

To appear in Archive for Rational Mechanics and Analysis (DOI:
https://doi.org/10.1007/s00205-023-01869-6)

##  1 Introduction

Electrical and magnetic interactions are long-range; that is, a charge or
dipole interacts with all the other charges and dipoles in the system, and the
interactions cannot be truncated because the decay with distance is slow
[Tou56, Bro63, JM94, MD14]. We consider such electrostatic interactions in
nanostructures, specifically helical geometries and thin films with uniform
bending, in a three-dimensional ambient space. These geometries are ubiquitous
in nanotechnology; while not periodic, their structure has significant
symmetry that we exploit in this paper, using the framework of Objective
Structures [Jam06]. We exploit this symmetry to adapt periodic calculations of
the continuum energy to the setting of these nanostructures. Specifically,
starting from a discrete atomic-scale description of the electrostatic energy,
we find the limit energy when the discrete lengthscale of the nanostructures
goes to zero.

For simplicity and clarity, we assume in this paper that the charge density
can be approximated as composed of discrete dipoles. The electrostatic energy
of such a system is the sum of all pairwise dipole-dipole interactions. Unlike
short-range bonded atomic interactions that typically scale as $\displaystyle
r^{-6}$ with distance $\displaystyle r$, the dipole-dipole interactions decay
slowly with distance as $\displaystyle r^{-3}$. Consequently, we cannot simply
truncate after a few neighbors, and naive truncation can lead to qualitatively
incorrect results in numerical calculations [MD14, GD20a, GD20b]. While we use
the setting of discrete electrical dipoles, the setting of magnetic dipoles
has an identical mathematical structure and physical interpretation [Bro63,
JM94, MS02, SS09], and we borrow key ideas from that literature. A key
physical distinction between the electrical and magnetic situations is the
possibility of electrical monopoles that does not exist for magnetic case, but
we examine this elsewhere [SWB+21] and assume here that there are no free
charges. Further, we highlight that the assumption of discrete dipoles is not
very restrictive. Following the approach of [JM94], we use a background field
in our calculations, and this field enables a straightforward generalization
to the more realistic setting of a general charge density field; such an
approach was used by [Xia05] to study charge density fields in periodic
crystals.

We turn to the question of dealing with the non-periodic geometry of the
nanostructures. While neither helices nor thin films with curvature are
periodic, the framework of Objective Structures (OS) introduced in [Jam06]
provides a powerful approach to deal with such geometries. In brief, OS
provides a group-theoretic description of these nanostructures that enables a
parallel to be made with periodic lattices. This parallel to periodic lattices
has enabled the adaptation of various methods developed for lattices to the
setting of helices and thin-films, e.g. [DJ07, HTJ12, ADE13a, ADE13b]. Our
strategy in this work is to use the OS framework to adapt continuum limit
calculations from the setting of periodic lattices to the setting of
nanostructures.

Our work is focused on obtaining discrete-to-continuum limits of the energy.
This multiscale approach has proven very powerful in enabling the systematic
reduction of the very large number of degrees of freedom associated with the
discrete problem to a much more tractable continuum problem. This overall idea
has played an important role in developing models, often in conjunction with
variational tools such as $\displaystyle\Gamma$-convergence, both for bulk
crystals [BLBL02, BLBL07, Sch09, ACG08, ACG11, BBC20] as well as for thin
films and rods [Sch06, ALP18]. Further, these ideas have played a role in the
development of numerical multiscale atomistic methods such as the
quasicontinuum method [TOP96, MT02, TM11, KO01, LLO12, DLO10]. There is also a
significant literature on simultaneous dimension reduction and discrete-to-
continuum limits, e.g. [ABC08, ALP18, FJ00, LPS15, LPS17, Sch08], but these
consider interactions that decay much faster than electrostatic interactions.

All the work in the previous paragraph is restricted to the setting of short-
range bonded atomic interactions. In the context of electrical and magnetic
interactions, the calculation of continuum limit energies based on discrete-
to-continuum approaches have been examined both formally and rigorously using
discrete dipoles on a periodic 3-d lattice [Tou56, Bro63, JM94, MS02, SS09,
Sch05]. Further, this has been examined formally for periodic charge
distributions, also in 3-d, [MD14]. All of these works show that the continuum
limit energy consists of a local part and a nonlocal part. In contrast, in
this work, we consider topologically low-dimensional structures embedded in a
3-d space: a 1-d helical nanotube and a 2-d thin film with constant bending
curvature. In the limit that the discrete lengthscale characterizing the
nanotube and thin film goes to $\displaystyle 0$, we find that the limit
continuum energy is entirely local. In [CDZ09], they obtain the discrete-to-
continuum limit of an energy that includes dipole-dipole interactions in 2-d
materials using $\displaystyle\Gamma-$convergence, but they use a dipole
kernel that goes as $\displaystyle 1/r^{2}$. In contrast, in this work, while
the structures are low dimensional, they are embedded in 3-d, and, therefore,
the dipole field kernel has a $\displaystyle 1/r^{3}$ singularity.

The absence of nonlocality in the limit can be rationalized by observing that
the decay of the interactions as $\displaystyle r^{-3}$ is sufficiently fast
to enable us to obtain a local limit if summed over a (topologically) 1-d or
2-d object. We highlight a complementary body of work that applies dimension
reduction techniques to go from a 3-d continuum to a 2-d or 1-d continuum. In
the context of electrical and magnetic interactions, [GJ97] and subsequent
works [Car01, KS05, KSZ15] (for thin films) and [GH15, CH15] (for thin wires)
find, as we do, that the limit energy is not nonlocal.

The techniques employed in this work are broadly based on the rigorous results
provided in [JM94] on the continuum limit of magnetic dipole interactions on a
3-d lattice, with appropriate generalizations and modifications for our
setting. The overall strategy of [JM94] is as follows. First, the operator
that associates the discrete dipole lattice to the generated electric field is
shown to be bounded for smooth test functions; next, the pointwise limit of
the action of the operator on smooth test functions is obtained; and, finally,
using the boundedness of the operator and the density of the test functions,
the limit of the energy density is obtained. For the helical and thin film
nanostructures considered in this work, we adapt this strategy to account for
the fact that the lattice sites and dipoles are not related by a translation
transformation, but by a more general isometric transformation.

The key results of this work are as follows. First, the limit energy is
rigorously derived and found to be local. Second, the limiting energy density
depends on the macroscopic geometric parameters, such as the pitch, radius and
so on for the helical nanotube, and on the stretch and curvature for the thin
film. These parameters can be related to macroscopic measures of deformation,
and link the macroscopic deformation to the small-scale structure. Third,
while the limiting energy is local, there are energetic contributions from
both the normal and the tangential components of the dipole field. This is in
contrast to the result obtained by dimension reduction from a 3-d continuum:
in these approaches, there are no energetic contributions from the tangential
component of the dipole field. Those approaches have 3-d continuum theory as
their starting point, and are valid for situations in which the limiting thin
object has all dimensions much larger than the atomic lengthscale. In
contrast, the discrete-to-continuum approach used here is appropriate for
nanostructures in which the thin dimensions are comparable to the atomic
lengthscale.

##### Organization

In section 2, we discuss prior work, primarily on dimension reduction from a
3-d continuum to a 2-d continuum, and highlight the local nature of the
limiting energy. We then discuss heuristically the scaling of electrostatic
interactions that lead to this locality generically for topologically low-
dimensional nanostructures. In section 3, we present the main results for
helical nanotubes and thin films with constant bending curvature. We prove
various claims in section 4. In section 5, we summarize the results.

##### Notation

We denote the real line and set of integers by $\displaystyle\mathbb{R}$ and
$\displaystyle\mathbb{Z}$, respectively;
$\displaystyle\mathbb{R}^{d},\mathbb{Z}^{d}$ denote these in dimension
$\displaystyle d=1,2,3$. For any $\displaystyle c,c_{1},c_{2}\in\mathbb{R}$,
$\displaystyle c\mathbb{Z}^{d}$ denotes the set
$\displaystyle\\{cz;z\in\mathbb{Z}^{d}\\}$ and $\displaystyle
c_{1}\mathbb{Z}\times c_{2}\mathbb{Z}$ denotes the set
$\displaystyle\\{(c_{1}z_{1},c_{2}z_{2});z_{1},z_{2}\in\mathbb{Z}\\}$. The
symbols $\displaystyle{\cal L}$ and $\displaystyle U$ denote the set of
lattice sites and the lattice unit cell, respectively; $\displaystyle{\cal
L}_{\lambda}$ and $\displaystyle U_{\lambda}$ denote these in the lattice
scaled by $\displaystyle\lambda$, with $\displaystyle{\cal L}_{1},U_{1}$
denoting $\displaystyle{\cal L}_{\lambda},U_{\lambda}$ for
$\displaystyle\lambda=1$. We use
$\displaystyle\boldsymbol{x}=(x_{1},x_{2},x_{3})\in\mathbb{R}^{3}$ to denote
the point in space with components $\displaystyle x_{i}$ in the orthonormal
basis
$\displaystyle\\{\boldsymbol{e}_{1},\boldsymbol{e}_{2},\boldsymbol{e}_{3}\\}$
for $\displaystyle\mathbb{R}^{3}$. We follow the standard notation wherein
scalars are denoted by lowercase letters, vectors by bold lowercase letters,
and second order tensors by bold uppercase letters.
$\displaystyle|\boldsymbol{x}|=\sqrt{\sum_{i=1}^{n}x_{i}^{2}}$ denotes the
Euclidean norm of the vector $\displaystyle\boldsymbol{x}\in\mathbb{R}^{n}$;
$\displaystyle|\boldsymbol{A}|=\sqrt{\boldsymbol{A}\boldsymbol{:}\boldsymbol{A}}$
denotes the norm of the tensor $\displaystyle\boldsymbol{A}$; and
$\displaystyle\boldsymbol{A}\boldsymbol{:}\boldsymbol{B}=A_{ij}B_{ij}$ denotes
the inner product of the tensors $\displaystyle\boldsymbol{A}$ and
$\displaystyle\boldsymbol{B}$. For any vector
$\displaystyle\boldsymbol{a}\in\mathbb{R}^{n}$ and tensor
$\displaystyle\boldsymbol{A}$, we have
$\displaystyle|\boldsymbol{A}\boldsymbol{a}|\leq|\boldsymbol{A}||\boldsymbol{a}|$.
We use $\displaystyle|\varOmega|$ to denote the Lebesgue measure of the set
$\displaystyle\varOmega\subset\mathbb{R}^{n}$. For a set $\displaystyle
A\subset\mathbb{R}^{d}$, $\displaystyle\chi_{A}=\chi_{A}(\boldsymbol{x})$
denotes the indicator function. We use $\displaystyle L^{2}(A,B)$ to denote
the space of Lebesgue square-integrable functions $\displaystyle u\colon
A\subset\mathbb{R}^{n}\to B\subset\mathbb{R}^{m}$;
$\displaystyle(u,v)_{L^{2}(A,B)}$ for the inner product of functions
$\displaystyle u,v\in L^{2}(A,B)$; and $\displaystyle||u||_{L^{2}(A,B)}$ the
$\displaystyle L^{2}$ norm of $\displaystyle u\in L^{2}(A,B)$. When there is
no ambiguity, we will suppress $\displaystyle L^{2}(A,B)$ and write
$\displaystyle(u,v)$ and $\displaystyle||u||$. $\displaystyle
C^{\infty}_{0}(\mathbb{R}^{n},\mathbb{R}^{m})$ denotes the space of infinitely
differentiable test functions $\displaystyle
u\colon\mathbb{R}^{n}\to\mathbb{R}^{m}$ with compact support in
$\displaystyle\mathbb{R}^{n}$. For Hilbert spaces $\displaystyle V$ and
$\displaystyle W$, $\displaystyle{\cal L}(V,W)$ is the space of bounded linear
maps $\displaystyle T\colon V\to W$. The norm of the map $\displaystyle
T\in{\cal L}(V,W)$ is denoted by $\displaystyle||T||_{{\cal L}(V,W)}$, and is
given by the expression:

(1) $||T||_{{\cal L}(V,W)}=\sup_{||f||_{V}\neq
0}\frac{||Tf||_{W}}{||f||_{V}}.$

We use $\displaystyle u_{\lambda}\xrightarrow[\lambda\to 0]{}u$ to denote the
strong convergence of $\displaystyle u_{\lambda}\in V$ to $\displaystyle u\in
V$ as $\displaystyle\lambda\to 0$, i.e.,
$\displaystyle||u_{\lambda}-u||_{V}\to 0$ as $\displaystyle\lambda\to 0$.

##  2 Energy Scalings and Prior Results on Dimension Reduction

We briefly revisit the results of [GJ97] and [CH15]. Respectively, they
performed dimension reduction from the 3-d continuum to the 2-d thin film and
1-d thin wire to find the limiting magnetostatic energy.

Figure 1: The geometry of the thin film (left) and the thin wire (right).

Consider a material domain $\displaystyle\varOmega_{h}=S\times[0,h]$, where
$\displaystyle S\subset\mathbb{R}^{2}$ is a 2-d domain in the plane spanned by
$\displaystyle(\boldsymbol{e}_{1},\boldsymbol{e}_{2})$, and $\displaystyle
h>0$ is the material thickness in the normal direction
$\displaystyle\boldsymbol{e}_{3}$ (Fig. 1). Suppose
$\displaystyle\boldsymbol{d}\colon\varOmega_{h}\to\mathbb{R}^{3}$, with
$\displaystyle\boldsymbol{d}=\mathbf{0}$ on
$\displaystyle\mathbb{R}^{3}\backslash\varOmega_{h}$, is the dipole field in
the material. The electrostatic energy density is given by

$e_{h}(\boldsymbol{d})=\dfrac{1}{|\varOmega_{h}|}\int_{\varOmega_{h}}\frac{1}{2}\nabla\phi(\boldsymbol{x})\cdot\boldsymbol{d}\
\mathrm{d}\boldsymbol{x},$

where $\displaystyle|\varOmega_{h}|$ is the volume of
$\displaystyle\varOmega_{h}$, and $\displaystyle\phi$ is the electric
potential that satisfies the electrostatic equation

$\operatorname{div}(-\nabla\phi+\boldsymbol{d})=0\quad\text{on
}\mathbb{R}^{3}$

together with the constraint $\displaystyle|\boldsymbol{d}|=d$ and the decay
property

$|\nabla\phi({\mathbold x})|\to{\bf 0}\text{ as }|{\mathbold x}|\to\infty.$

Let $\displaystyle\varOmega_{1}=S\times[0,1]$, and
$\displaystyle\boldsymbol{y}(\boldsymbol{x})=(x_{1},x_{2},x_{3}/h)\in\varOmega_{1}$
for $\displaystyle\boldsymbol{x}\in\varOmega_{h}$ be the map from
$\displaystyle\varOmega_{h}$ to $\displaystyle\varOmega_{1}$. For fixed
$\displaystyle h>0$, consider the dipole field
$\displaystyle\boldsymbol{d}_{h}\colon\varOmega_{h}\to\mathbb{R}^{3}$ and
$\displaystyle\tilde{\boldsymbol{d}}_{h}\colon\varOmega_{1}\to\mathbb{R}^{3}$
such that

$\tilde{\boldsymbol{d}}_{h}(\boldsymbol{y}(\boldsymbol{x}))=\boldsymbol{d}_{h}(\boldsymbol{x}),\qquad\forall\boldsymbol{x}\in\varOmega_{h}.$

Let $\displaystyle\boldsymbol{d}_{h}$ be the sequence of dipole field for
$\displaystyle h>0$, and $\displaystyle\tilde{\boldsymbol{d}}_{h}$ is defined
as above. Assume that dipole field $\displaystyle\tilde{\boldsymbol{d}}_{h}$
is such that, first, $\displaystyle\tilde{\boldsymbol{d}}_{h}=0$ on
$\displaystyle\mathbb{R}^{3}\backslash\varOmega_{1}$, and, second, it
converges to $\displaystyle\tilde{\boldsymbol{d}}_{0}$ in $\displaystyle
L^{2}(\mathbb{R}^{3})$; then the limit of the energy density $\displaystyle
e_{h}=e_{h}(\boldsymbol{d}_{h})$ is [GJ97]:

$e_{h}(\boldsymbol{d}_{h})\to
e_{0}(\tilde{\boldsymbol{d}}_{0})=\dfrac{1}{2|\varOmega_{1}|}\int_{\varOmega_{1}}|\tilde{d}_{0_{3}}|^{2}\
\mathrm{d}\boldsymbol{x}.$

That is, the limiting energy $\displaystyle e_{0}$ is local, and only the
normal component of the dipole moment appears in the expression.

Next, consider a thin straight wire with axis along
$\displaystyle\boldsymbol{e}_{1}$, denoted by
$\displaystyle\varOmega_{h}=(-1,1)\times B_{2}(\mathbf{0},h)$, where
$\displaystyle B_{2}(\mathbf{0},h)$ is the ball of radius $\displaystyle h$
centered at $\displaystyle\mathbf{0}$ in the plane spanned by
$\displaystyle(\boldsymbol{e}_{2},\boldsymbol{e}_{3})$ (Fig. 1). Analogous to
the thin film, let
$\displaystyle\boldsymbol{d}_{h}:\varOmega_{h}\to\mathbb{R}^{3}$,
$\displaystyle\boldsymbol{d}_{h}=\mathbf{0}$ on
$\displaystyle\mathbb{R}^{3}/\varOmega_{h}$, be the dipole field in the
material; $\displaystyle\varOmega_{1}=(-1,1)\times B_{2}(\mathbf{0},1)$ be the
rescaled domain of $\displaystyle\varOmega_{h}$;
$\displaystyle\boldsymbol{y}(\boldsymbol{x})=(x_{1},x_{2}/h,x_{3}/h)\in\varOmega_{1}$
for $\displaystyle\boldsymbol{x}\in\varOmega_{h}$ be the map from
$\displaystyle\varOmega_{h}$ to $\displaystyle\varOmega_{1}$; and
$\displaystyle\tilde{\boldsymbol{d}}_{h}:\varOmega_{1}\to\mathbb{R}^{3}$ be
the rescaled dipole field defined as

(2)
$\tilde{\boldsymbol{d}}_{h}(\boldsymbol{y}(\boldsymbol{x}))=\boldsymbol{d}_{h}(\boldsymbol{x}),\qquad\forall\boldsymbol{x}\in\varOmega_{h}.$

The limiting energy density in the case of the thin wire is [CH15]:

$\frac{1}{2|\varOmega_{1}|}\int_{\varOmega_{1}}\left(|\tilde{d}_{0_{2}}|^{2}+|\tilde{d}_{0_{3}}|^{2}\right)\
\mathrm{d}\boldsymbol{x},$

where $\displaystyle\tilde{\boldsymbol{d}}_{0}:=\lim_{h\to
0}\tilde{\boldsymbol{d}}_{h}$ is the limiting dipole field. We notice that the
limiting energy is again local, and only the components of the dipole moment
perpendicular to the axis of the wire contribute.

Figure 2: A schematic of the unit cell with a dipole (a), and generic 1-d,
2-d, 3-d periodic lattices (b, c, d).

The absence of nonlocality in the limiting energy in the results above, as
well as in our results in section 3 below, can be physically understood
through the fact that these structures are 1-d or 2-d topologically. To see
this, we consider a system of discrete dipoles associated with the uniform
1-d, 2-d, and 3-d periodic lattices with the unit cell of size $\displaystyle
1$ (Figure 2). The energy of a lattice of dipoles is given by [Bro63, JM94]:

(3) $E=-\frac{1}{2}\sum_{i}\sum_{j,j\neq
i}\boldsymbol{d}(i)\cdot\boldsymbol{K}(\boldsymbol{x}_{j}-\boldsymbol{x}_{i})\boldsymbol{d}(j)=\sum_{i}|U(i)|\left[-\frac{1}{|U(i)|}\frac{1}{2}\sum_{j,j\neq
i}\boldsymbol{d}(i)\cdot\boldsymbol{K}(\boldsymbol{x}_{j}-\boldsymbol{x}_{i})\boldsymbol{d}(j)\right],$

where the sum is over the cells in the lattice, and the term inside square
bracket denotes the energy density of a cell $\displaystyle i$. Here,
$\displaystyle U(i)$ denotes the $\displaystyle i^{\text{th}}$ unit cell;
$\displaystyle|U(i)|$ the measure (volume) of the unit cell $\displaystyle
U(i)$; $\displaystyle\boldsymbol{d}(i)$ the dipole in cell $\displaystyle i$;
and, $\displaystyle\boldsymbol{x}_{i}$ the coordinate of lattice site
$\displaystyle i$. The dipole field kernel,
$\displaystyle\boldsymbol{K}=\boldsymbol{K}(\boldsymbol{x})$, is defined as:

(4)
$\boldsymbol{K}(\boldsymbol{x})=-\dfrac{1}{4\pi|\boldsymbol{x}|^{3}}\left(\boldsymbol{I}-3\dfrac{\boldsymbol{x}}{|\boldsymbol{x}|}\mathbf{\otimes}\dfrac{\boldsymbol{x}}{|\boldsymbol{x}|}\right),\quad\boldsymbol{x}\neq\mathbf{0}.$

We use these expressions to heuristically understand the scaling of the energy
for systems with different topological dimensions. For simplicity, we assume
below that the volume of the unit cell and the magnitude of the dipole are
both $\displaystyle 1$, i.e., $\displaystyle|U(i)|=1$ and
$\displaystyle|\boldsymbol{d}(i)|=1$ for each $\displaystyle i$, and some
constant factors are neglected. For the next set of bounds, assume that
$\displaystyle c_{1}$ and $\displaystyle c_{2}$ are generic positive
constants.

######  Remark 2.1 (1-d lattice).

We can estimate an upper bound on the energy density $\displaystyle e$ of a
typical unit cell as follows:

$|e|\leq
c_{1}\sum_{r=1}^{\infty}\dfrac{1}{r^{3}}\times|\boldsymbol{d}|\times(\text{number
of dipoles at }r)\leq c_{2}\sum_{r=1}^{\infty}\dfrac{1}{r^{3}}\times 1\times
1=c_{2}\sum_{r=1}^{\infty}\dfrac{1}{r^{3}}.$

We use that the total dipole moment at a distance $\displaystyle r$ from a
given unit cell is, at most, that of another dipole in the unit cell at a
distance $\displaystyle r$. This sum is well-behaved and bounded.

######  Remark 2.2 (2-d lattice).

As in the 1-d setting, we first bound the net dipole at a distance
$\displaystyle r$ from a given unit cell. Since the structure is a 2-d
lattice, the number of unit cells at a distance $\displaystyle r$ is of order
$\displaystyle 2\pi r$. Therefore, an upper bound on the energy density is:

$|e|\leq
c_{1}\sum_{r=1}^{\infty}\dfrac{1}{r^{3}}\times|\boldsymbol{d}|\times(\text{number
of dipoles at }r)\leq c_{2}\sum_{r=1}^{\infty}\dfrac{1}{r^{3}}\times 1\times
2\pi r=c_{2}2\pi\sum_{r=1}^{\infty}\dfrac{1}{r^{2}}.$

This sum is also well-behaved and bounded.

######  Remark 2.3 (3-d lattice).

Following the argument of the 2-d lattice, we now have that the net dipole at
a distance $\displaystyle r$ from a given unit cell is, at most, of the order
$\displaystyle 4\pi r^{2}$. Therefore, an upper bound on the energy density
is:

$|e|\leq
c_{1}\sum_{r=1}^{\infty}\dfrac{1}{r^{3}}\times|\boldsymbol{d}|\times(\text{number
of dipoles at }r)\leq c_{2}\sum_{r=1}^{\infty}\dfrac{1}{r^{3}}\times 1\times
4\pi r^{2}=c_{2}4\pi\sum_{r=1}^{\infty}\dfrac{1}{r}.$

This sum is divergent. However, through a more careful analysis that accounts
for the signs – not just the magnitudes – of the dipole interactions, the
energy density can be shown to be conditionally convergent111 That is, it is
convergent, while the sums of only the positive terms and only the negative
terms diverge, respectively, to $\displaystyle\pm\infty$.  [Tou56, JM94].

When the lattice sum is bounded and converges unconditionally, it is possible
to truncate after a finite distance and obtain sufficient numerical accuracy.
When the lattice sum is conditionally convergent, that can be physically
related to nonlocality; specifically, the slow convergence does not allow for
truncation, and the far-field values play an important role. [MD14] discusses
this from a physical perspective.

##  3 Results on Continuum Limits of the Electrostatic Energy

We consider two classes of nanostructures: helical nanotubes and thin films,
the latter allowing for a constant bending curvature (i.e., nonzero constant
mean curvature and zero Gauss curvature), and obtain the corresponding
continuum limit electrostatic energy. In both cases, we start with discrete
dipoles, where the discreteness is parametrized by the scale
$\displaystyle\lambda>0$, and examine the limit $\displaystyle\lambda\to 0$.
We show that the dipole-dipole interaction energy density – per unit cross-
sectional area in the case of nanotubes, and per unit thickness in the case of
films – converges to a local energy density in the limit.

###  3.1 Helical Nanotube

We consider a discrete helix with axis $\displaystyle\boldsymbol{e}_{3}$
characterized by the angle $\displaystyle\theta$ and lengthscale
$\displaystyle\delta$; the pitch of the helix is $\displaystyle
2\pi\delta/\theta$. Suppose $\displaystyle\boldsymbol{x}_{0}\in\mathbb{R}^{3}$
is a point on the helix. Then, the other points on the helix are related by an
isometric transformation of $\displaystyle\boldsymbol{x}_{0}$. Let
$\displaystyle s\in\mathbb{R}$ be the parametric coordinate of a point on the
helix. Then, the map
$\displaystyle\bar{\boldsymbol{x}}\colon\mathbb{R}\to\mathbb{R}^{3}$ that
takes a point in the parametric space to a unique point on the helix can be
expressed as

(5)
$\bar{\boldsymbol{x}}(s)=\boldsymbol{Q}(s\theta)\boldsymbol{x}_{0}+s\delta\boldsymbol{e}_{3}.$

Here $\displaystyle\boldsymbol{Q}=\boldsymbol{Q}(\alpha)$ is the rotational
tensor represented by the matrix in the orthonormal basis
$\displaystyle\\{\boldsymbol{e}_{1},\boldsymbol{e}_{2},\boldsymbol{e}_{3}\\}$
as:

(6)
$\begin{split}\boldsymbol{Q}(\alpha)&:=\left[\begin{array}[]{ccc}\cos(\alpha)&-\sin(\alpha)&0\\\
\sin(\alpha)&\cos(\alpha)&0\\\ 0&0&1\end{array}\right].\end{split}$

Note that the definition of map $\displaystyle\bar{\boldsymbol{x}}$ imply that
the helix makes a full turn in $\displaystyle s=2\pi/\theta$ from
$\displaystyle s=0$, and, therefore, the pitch of the helix is
$\displaystyle(\bar{\boldsymbol{x}}(2\pi/\theta)-\bar{\boldsymbol{x}}(0))\cdot\boldsymbol{e}_{3}=2\pi\delta/\theta$.

Without loss of generality, we assume
$\displaystyle\boldsymbol{x}_{0}=\boldsymbol{e}_{1}$. The tangent vector to
the helix at $\displaystyle s$ is given by

(7)
$\boldsymbol{t}(s)=\frac{\mathrm{d}\bar{\boldsymbol{x}}(s)}{\mathrm{d}s}=\theta\boldsymbol{Q}^{\prime}(s\theta)\boldsymbol{e}_{1}+\delta\boldsymbol{e}_{3}.$

Let
$\displaystyle\hat{\boldsymbol{t}}(s)=\boldsymbol{t}(s)/\sqrt{\theta^{2}+\delta^{2}}$
denote the unit tangent vector. We define the second order projection tensors
$\displaystyle\boldsymbol{P}_{||}=\boldsymbol{P}_{||}(s)$ and
$\displaystyle\boldsymbol{P}_{\perp}=\boldsymbol{P}_{\perp}(s)$, for
$\displaystyle s\in\mathbb{R}$, as follows

(8)
$\boldsymbol{P}_{||}(s)=\hat{\boldsymbol{t}}(s)\mathbf{\otimes}\hat{\boldsymbol{t}}(s),\qquad\boldsymbol{P}_{\perp}(s)=\boldsymbol{I}-\boldsymbol{P}_{||}(s).$

For any vector $\displaystyle\boldsymbol{a}$ and any $\displaystyle
s\in\mathbb{R}$, we have

(9)
$\boldsymbol{a}=\boldsymbol{P}_{||}(s)\boldsymbol{a}+\boldsymbol{P}_{\perp}(s)\boldsymbol{a},\qquad\text{with
}\qquad\boldsymbol{P}_{||}(s)\boldsymbol{a}\cdot\boldsymbol{P}_{\perp}(s)\boldsymbol{a}=0.$

####  3.1.1 Lattice Geometry and Dipole Moment

Let $\displaystyle{\cal L}=\mathbb{Z}$ denote the set of parametric
coordinates of the points on the helix. We consider a discrete system of
dipole moments $\displaystyle\boldsymbol{d}\colon{\cal L}\to\mathbb{R}^{3}$
associated to the points on the helix given by $\displaystyle{\cal L}$ (Figure
3). The magnitudes of the dipoles at the lattice sites are equal, but they are
oriented differently; in particular, the orientations of dipoles at lattice
sites follow the relation:

(10) ${\mathbold d}(s+1)={\mathbold Q}(\theta){\mathbold d}(s),\quad s\in{\cal
L}.$

We associate a unit cell to each lattice site. Let $\displaystyle
U(s)=[s,s+1)$ denote the unit cell in the parametric space at the site
$\displaystyle s$, for $\displaystyle s\in{\cal L}$. Let $\displaystyle S(r)$,
$\displaystyle r\in\mathbb{R}$, be given by

$\displaystyle
S(r)=\left\\{\boldsymbol{x};\,(\boldsymbol{x}-\bar{\boldsymbol{x}}(r))\cdot\boldsymbol{t}(r)=0,\,|\boldsymbol{x}-\bar{\boldsymbol{x}}(r)|^{2}<R^{2}\right\\},$

for some $\displaystyle R>0$. Note that $\displaystyle|S(r)|=|S(0)|=\pi
R^{2}$. The unit cell in real space is defined by
$\displaystyle\bar{U}(s)=\left\\{{\mathbold x}\in S(r);\,r\in U(s)\right\\}$.
We take, without loss of generality, $\displaystyle
R^{2}=1/(\pi\sqrt{\theta^{2}+\delta^{2}})$ so that
$\displaystyle|\bar{U}(s)|=\mathrm{area}(S)\times\mathrm{length}(\\{\bar{\boldsymbol{x}}(r);r\in
U(s)\\})=\pi R^{2}\sqrt{\theta^{2}+\delta^{2}}=1$.

We now consider the setting in which the cells are of size
$\displaystyle\lambda>0$, so that as $\displaystyle\lambda\to 0$ the density
of cells in the helix increases. For $\displaystyle\lambda>0$, suppose
$\displaystyle{\cal L}_{\lambda}=\lambda\mathbb{Z}$ denotes the parametric
coordinates of the sites in a scaled lattice, and
$\displaystyle\boldsymbol{d}_{\lambda}\colon{\cal
L}_{\lambda}\to\mathbb{R}^{3}$ denotes the corresponding system of dipole
moments. Associated to $\displaystyle s\in{\cal L}_{\lambda}$, let
$\displaystyle U_{\lambda}(s)=[s,s+\lambda)$ denote the cell in the parametric
space. The 3-d cell is given by
$\displaystyle\bar{U}_{\lambda}(s)=\\{\boldsymbol{x}\in S_{\lambda}(r);\,r\in
U_{\lambda}(s)\\}$, where $\displaystyle
S_{\lambda}(r)=\\{\boldsymbol{x};\,(\boldsymbol{x}-\bar{\boldsymbol{x}}(r))\cdot\boldsymbol{t}(r)=0,\,|\boldsymbol{x}-\bar{\boldsymbol{x}}(r)|^{2}<\lambda^{2}R^{2}\\}$
is the scaled cross-section. Note that
$\displaystyle\mathrm{area}(S_{\lambda}(r))=\pi\lambda^{2}R^{2}$ and
$\displaystyle|\bar{U}_{\lambda}(s)|=\pi\lambda^{2}R^{2}\times\lambda\sqrt{\theta^{2}+\delta^{2}}=\lambda^{3}$.
Let
$\displaystyle\tilde{\boldsymbol{d}}_{\lambda}\colon\mathbb{R}\to\mathbb{R}^{3}$
be the piecewise constant extension of $\displaystyle\boldsymbol{d}_{\lambda}$
given by

(11)
$\tilde{\boldsymbol{d}}_{\lambda}(s)=\frac{\boldsymbol{d}_{\lambda}(i)}{|\bar{U}_{\lambda}(s)|}=\frac{\boldsymbol{d}_{\lambda}(i)}{\lambda^{3}},\qquad\forall
s\in U_{\lambda}(i),\quad\forall i\in{\cal L}_{\lambda}.$

To compute the limit of the dipole-dipole interaction energy as
$\displaystyle\lambda\to 0$, we assume that dipole moment density field
$\displaystyle\tilde{\boldsymbol{d}}_{\lambda}$ converges to some field
$\displaystyle{\mathbold f}\in L^{2}(\mathbb{R},\mathbb{R}^{3})$ in the
$\displaystyle L^{2}$ norm. As in [JM94], instead of working with
$\displaystyle\tilde{\boldsymbol{d}}_{\lambda}$, as defined above, we could
assume that the dipole moment $\displaystyle\boldsymbol{d}_{\lambda}(i)$, for
$\displaystyle i\in{\cal L}_{\lambda}$, is due to the background dipole moment
density field $\displaystyle{\mathbold f}_{\lambda}\in
L^{2}(\mathbb{R},\mathbb{R}^{3})$ such that

(12)
$\boldsymbol{d}_{\lambda}(i)=\sqrt{\theta^{2}+\delta^{2}}\int_{U_{\lambda}(i)}\int_{S_{\lambda}(r)}{\mathbold
f}_{\lambda}(r)\ \mathrm{d}S_{\lambda}(r)\
\mathrm{d}r=\lambda^{2}\int_{U_{\lambda}(i)}{\mathbold f}_{\lambda}(r)\
\mathrm{d}r,$

where $\displaystyle\mathrm{d}S_{\lambda}(r)$ is the area measure for surface
$\displaystyle S_{\lambda}$. In the equation above, we assumed that the
background field is uniform in $\displaystyle S_{\lambda}(r)$ for all
$\displaystyle r\in\mathbb{R}$ and used $\displaystyle
R^{2}=1/(\pi\sqrt{\theta^{2}+\delta^{2}})$. The existence of one such
background field $\displaystyle{\mathbold f}_{\lambda}$ is evident: we can
define $\displaystyle{\mathbold
f}_{\lambda}=\tilde{\boldsymbol{d}}_{\lambda}$. The physical dimension of
$\displaystyle{\mathbold f}_{\lambda}$ is dipole moment per unit volume. We
have the following lemma that relates the convergence of the background dipole
moment field and the piecewise constant extension.

######  Lemma 3.1.

Let $\displaystyle{\mathbold f}_{\lambda}$, $\displaystyle\lambda>0$, be the
sequence of $\displaystyle L^{2}(\mathbb{R},\mathbb{R}^{3})$ functions, and
let $\displaystyle{\mathbold f}\in L^{2}(\mathbb{R},\mathbb{R}^{3})$ be such
that $\displaystyle{\mathbold f}_{\lambda}\to{\mathbold f}$ in $\displaystyle
L^{2}(\mathbb{R},\mathbb{R}^{3})$. Let
$\displaystyle\boldsymbol{d}_{\lambda}\colon{\cal
L}_{\lambda}\to\mathbb{R}^{3}$ be given by (12), and
$\displaystyle\tilde{\boldsymbol{d}}_{\lambda}$ be a piecewise constant
$\displaystyle L^{2}$ extension of $\displaystyle\boldsymbol{d}_{\lambda}$
given by (11). Then,
$\displaystyle\tilde{\boldsymbol{d}}_{\lambda}\to{\mathbold f}$ in
$\displaystyle L^{2}(\mathbb{R},\mathbb{R}^{3})$.

On the other hand, if $\displaystyle\boldsymbol{d}_{\lambda}\colon{\cal
L}_{\lambda}\to\mathbb{R}^{3}$ is such that
$\displaystyle\tilde{\boldsymbol{d}}_{\lambda}\to{\mathbold f}$ in
$\displaystyle L^{2}(\mathbb{R},\mathbb{R}^{3})$, then there exists a
background field $\displaystyle{\mathbold f}_{\lambda}\in
L^{2}(\mathbb{R},\mathbb{R}^{3})$ such that
$\displaystyle\boldsymbol{d}_{\lambda}$ is given by (12).

The proof is similar to the proof of Theorem 4.1 from [JM94].

######  Remark 3.2.

Since the discrete dipole field $\displaystyle\boldsymbol{d}_{\lambda}$ has
helical symmetry, from (12) we can see that $\displaystyle{\mathbold
f}_{\lambda}$ will also have helical symmetry. However, we highlight that the
background field needs to have helical symmetry only in the sense that the
effective dipole has the helical symmetry, allowing for some fluctuations from
site to site.

######  Remark 3.3.

The physical setting that we wish to examine is when dipole moments play a
significant role in the limit. Therefore, we consider an appropriate scaling
that corresponds to obtaining a finite limit for the dipole density field
$\displaystyle\tilde{\boldsymbol{d}}_{\lambda}$ (or, equivalently, the
background field, $\displaystyle{\mathbold f}_{\lambda}$). Further, we assume
convergence in $\displaystyle L^{2}$ to ensure finite energies.

(a)

(b)

Figure 3: Discrete dipole moments (red arrows) lying on the helix. (a) and (b)
show the view in the $\displaystyle(\boldsymbol{e}_{1},\boldsymbol{e}_{3})$
and $\displaystyle(\boldsymbol{e}_{1},\boldsymbol{e}_{2})$ planes
respectively. The dipole moments corresponding to different sites are related
by (10). For the parametric coordinate $\displaystyle s$,
$\displaystyle\bar{\boldsymbol{x}}(s)$ gives the coordinate of the point on
the helix.

####  3.1.2 Electrostatic Energy

For $\displaystyle\lambda>0$, the energy associated to the system of dipole
moments $\displaystyle\boldsymbol{d}_{\lambda}$ can be expressed as [Bro63,
JM94]:

$E_{\lambda}=-\dfrac{1}{2}\sum_{\begin{subarray}{c}s,s^{\prime}\in{\cal{L}}_{\lambda},\\\
s\neq
s^{\prime}\end{subarray}}\boldsymbol{d}_{\lambda}(s)\cdot\boldsymbol{K}(\bar{\boldsymbol{x}}(s^{\prime})-\bar{\boldsymbol{x}}(s))\boldsymbol{d}_{\lambda}(s^{\prime})=|S_{\lambda}|e_{\lambda},$

where $\displaystyle e_{\lambda}$ is the energy per unit area given by

(13)
$e_{\lambda}=-\dfrac{1}{2|S_{\lambda}|}\sum_{\begin{subarray}{c}s,s^{\prime}\in{\cal{L}}_{\lambda},\\\
s\neq
s^{\prime}\end{subarray}}\boldsymbol{d}_{\lambda}(s)\cdot\boldsymbol{K}(\bar{\boldsymbol{x}}(s^{\prime})-\bar{\boldsymbol{x}}(s))\boldsymbol{d}_{\lambda}(s^{\prime}).$

Substituting (12) into the expression above, and proceeding similar to Section
6 of [JM94], $\displaystyle e_{\lambda}$ can be written as

(14) $e_{\lambda}=\left({{\mathbold f}_{\lambda}},{T_{\lambda}{\mathbold
f}_{\lambda}}\right)_{L^{2}(\mathbb{R},\mathbb{R}^{3})},$

where $\displaystyle T_{\lambda}\colon L^{2}(\mathbb{R},\mathbb{R}^{3})\to
L^{2}(\mathbb{R},\mathbb{R}^{3})$ is the map given by

(15) $(T_{\lambda}{\mathbold
f})(s)=\lambda^{2}\int_{\mathbb{R}}\boldsymbol{K}_{\lambda}(s^{\prime},s){\mathbold
f}(s^{\prime})\ \mathrm{d}s^{\prime}$

and $\displaystyle\boldsymbol{K}_{\lambda}(s^{\prime},s)$, for $\displaystyle
s,s^{\prime}\in\mathbb{R}$, is the discrete dipole field kernel given by

(16)
$\boldsymbol{K}_{\lambda}(s^{\prime},s)=\sum_{\begin{subarray}{c}u,v\in{\cal{L}}_{\lambda},\\\
u\neq
v\end{subarray}}\chi_{U_{\lambda}(v)}(s^{\prime})\boldsymbol{K}(\bar{\boldsymbol{x}}(v)-\bar{\boldsymbol{x}}(u))\chi_{U_{\lambda}(u)}(s).$

##### Scaling of $\displaystyle\boldsymbol{K}_{\lambda}$

For any $\displaystyle a,b\in\mathbb{R}$, we have, using (5),

(17) $\begin{split}\bar{\boldsymbol{x}}(\lambda
a)-\bar{\boldsymbol{x}}(\lambda b)&=\boldsymbol{Q}(\lambda
a\theta)\boldsymbol{e}_{1}+\delta\lambda
a\boldsymbol{e}_{3}-\boldsymbol{Q}(\lambda
b\theta)\boldsymbol{e}_{1}-\delta\lambda b\boldsymbol{e}_{3}\\\
&=\lambda\left(\bar{\boldsymbol{x}}(a)-\bar{\boldsymbol{x}}(b)+\underbrace{\left[\dfrac{\boldsymbol{Q}(\lambda
a\theta)-\boldsymbol{Q}(\lambda
b\theta)-\left(\lambda\boldsymbol{Q}(a\theta)-\lambda\boldsymbol{Q}(b\theta)\right)}{\lambda}\right]}_{=:\boldsymbol{A}_{\lambda}(a,b)}\boldsymbol{e}_{1}\right).\end{split}$

Using the relation above, it is easy to show that

$\boldsymbol{K}_{\lambda}(s^{\prime},s)=\dfrac{1}{\lambda^{3}}\sum_{\begin{subarray}{c}u,v\in{\cal
L}_{1},\\\ u\neq
v\end{subarray}}\chi_{U_{1}(v)}(s^{\prime}/\lambda)\boldsymbol{K}(\bar{\boldsymbol{x}}(v)-\bar{\boldsymbol{x}}(u)+\boldsymbol{A}_{\lambda}(v,u)\boldsymbol{e}_{1})\chi_{U_{1}(u)}(s/\lambda),$

where we recall that $\displaystyle U_{1}(u)=[u,u+1)$, $\displaystyle
u\in{\cal L}_{1}$, is the lattice cell in the parametric space for
$\displaystyle\lambda=1$. Based on the equation above, we define a discrete
kernel $\displaystyle\boldsymbol{K}_{1,\lambda}(s,s^{\prime})$, for
$\displaystyle s,s^{\prime}\in\mathbb{R}$, as follows

(18)
$\boldsymbol{K}_{1,\lambda}(s^{\prime},s)=\sum_{\begin{subarray}{c}u,v\in{\cal
L}_{1},\\\ u\neq
v\end{subarray}}\chi_{U_{1}(v)}(s^{\prime})\boldsymbol{K}(\bar{\boldsymbol{x}}(v)-\bar{\boldsymbol{x}}(u)+\boldsymbol{A}_{\lambda}(v,u)\boldsymbol{e}_{1})\chi_{U_{1}(u)}(s).$

We then have

$\boldsymbol{K}_{\lambda}(s^{\prime},s)=\dfrac{1}{\lambda^{3}}\boldsymbol{K}_{1,\lambda}(s^{\prime}/\lambda,s/\lambda).$

####  3.1.3 Limit of the Electrostatic Energy

In this section, we obtain the limit of the energy per unit surface area
$\displaystyle e_{\lambda}$ as $\displaystyle\lambda\to 0$ assuming that the
background dipole field density $\displaystyle{\mathbold f}_{\lambda}$ (or
equivalently the dipole moment density
$\displaystyle\tilde{\boldsymbol{d}}_{\lambda}$) converges to some density
field $\displaystyle{\mathbold f}$ in $\displaystyle L^{2}$. The idea is to
first show that the map $\displaystyle T_{\lambda}$ in (15) is bounded and
obtain its limit. With that, the limit of $\displaystyle e_{\lambda}$ follows.

##### Limit of the Discrete Electric Field

Let $\displaystyle T_{1,\lambda}$ be the map with kernel
$\displaystyle\boldsymbol{K}_{1,\lambda}$. For any function
$\displaystyle{\mathbold f}\in L^{2}(\mathbb{R},\mathbb{R}^{3})$, we have

(19) $(T_{1,\lambda}{\mathbold
f})(s)=\int_{\mathbb{R}}\boldsymbol{K}_{1,\lambda}(s^{\prime},s){\mathbold
f}(s^{\prime})\ \mathrm{d}s^{\prime}.$

We have the following main result on the map $\displaystyle T_{\lambda}$.

######  Proposition 3.4.

The maps $\displaystyle T_{1,\lambda}$ and $\displaystyle T_{\lambda}$ are
bounded in $\displaystyle L^{2}(\mathbb{R},\mathbb{R}^{3})$ for all
$\displaystyle\lambda>0$ and satisfy

(20) $\left\lVert
T_{\lambda}\right\rVert_{\mathcal{L}(L^{2},L^{2})}=\left\lVert
T_{1,\lambda}\right\rVert_{\mathcal{L}(L^{2},L^{2})}.$

Further, for $\displaystyle{\mathbold f}\in
C^{\infty}_{0}(\mathbb{R},\mathbb{R}^{3})$,

$(T_{\lambda}{\mathbold f})(s)\xrightarrow[\lambda\to
0]{}-h_{0}(\boldsymbol{I}-3\boldsymbol{P}_{||}(s)){\mathbold
f}(s)=-h_{0}(\boldsymbol{P}_{\perp}(s)-2\boldsymbol{P}_{||}(s)){\mathbold
f}(s)$

pointwise, where $\displaystyle\boldsymbol{P}_{\perp}(s)$ and
$\displaystyle\boldsymbol{P}_{||}(s)$ are projection tensors that project onto
the normal plane and the tangent line to the helix respectively (see (8)).
Further, $\displaystyle h_{0}$ is a constant given by

(21) $h_{0}=\sum_{\begin{subarray}{c}v\in\mathbb{Z},\\\ v\neq
0\end{subarray}}\frac{1}{4\pi|v|^{3}(\theta^{2}+\delta^{2})^{3/2}}.$

We provide the proof of Proposition 3.4 in subsection 4.1.

##### Limit of the Energy

######  Theorem 3.5.

Let $\displaystyle{\mathbold f}_{\lambda}\in L^{2}(\mathbb{R},\mathbb{R}^{3})$
be a sequence of functions for $\displaystyle\lambda>0$ with
$\displaystyle{\mathbold f}\in L^{2}(\mathbb{R},\mathbb{R}^{3})$ such that
$\displaystyle{\mathbold f}_{\lambda}\xrightarrow[\lambda\to 0]{}{\mathbold
f}$ in $\displaystyle L^{2}$. Let the system of dipole moments
$\displaystyle\boldsymbol{d}_{\lambda}\colon{\cal
L}_{\lambda}\to\mathbb{R}^{3}$ be given by (12). Then

$e_{\lambda}\xrightarrow[\lambda\to
0]{}\dfrac{1}{2}h_{0}\left[||\boldsymbol{P}_{\perp}{\mathbold
f}||^{2}_{L^{2}(\mathbb{R},\mathbb{R}^{3})}-2||\boldsymbol{P}_{||}{\mathbold
f}||^{2}_{L^{2}(\mathbb{R},\mathbb{R}^{3})}\right],$

where $\displaystyle h_{0}$ is the constant defined in (21).

######  Proof 3.6.

Since $\displaystyle T_{\lambda}$ is bounded and $\displaystyle{\mathbold
f}_{\lambda}\to{\mathbold f}$, we have

$\lim_{\lambda\to 0}(T_{\lambda}{\mathbold f}_{\lambda})=\lim_{\lambda\to
0}(T_{\lambda}{\mathbold f})+\lim_{\lambda\to 0}(T_{\lambda}({\mathbold
f}_{\lambda}-{\mathbold f}))=\lim_{\lambda\to 0}(T_{\lambda}{\mathbold f}),$

where the decomposition of the limit into the sum of individual limits is true
if the two limits, $\displaystyle\lim_{\lambda\to 0}(T_{\lambda}{\mathbold
f})$ and $\displaystyle\lim_{\lambda\to 0}(T_{\lambda}({\mathbold
f}_{\lambda}-{\mathbold f}))$, individually exist. This is indeed true:
because $\displaystyle T_{\lambda}$ is bounded and $\displaystyle{\mathbold
f}_{\lambda}\to{\mathbold f}$ in $\displaystyle L^{2}$, we have that
$\displaystyle\lim_{\lambda\to 0}(T_{\lambda}({\mathbold
f}_{\lambda}-{\mathbold f}))=0$. Then, to show that
$\displaystyle\lim_{\lambda\to 0}(T_{\lambda}{\mathbold f})$ exist for
$\displaystyle{\mathbold f}\in L^{2}(\mathbb{R},\mathbb{R}^{3})$, we proceed
as follows.

Let $\displaystyle{\mathbold f}^{k}\in
C^{\infty}_{0}(\mathbb{R},\mathbb{R}^{3})$ be a sequence of functions such
that $\displaystyle{\mathbold f}^{k}\to{\mathbold f}$. Using Proposition 3.4,
we have

(22) $\begin{split}\lim_{\lambda\to 0}(T_{\lambda}{\mathbold
f})&=\lim_{k\to\infty}\lim_{\lambda\to 0}(T_{\lambda}{\mathbold
f}^{k})+\lim_{k\to\infty}\lim_{\lambda\to 0}(T_{\lambda}({\mathbold
f}-{\mathbold f}^{k}))=\lim_{k\to\infty}\lim_{\lambda\to
0}(T_{\lambda}{\mathbold f}^{k})\\\
&=\lim_{k\to\infty}\left(\boldsymbol{H}_{0}{\mathbold
f}^{k}\right)=\boldsymbol{H}_{0}{\mathbold f},\end{split}$

where
$\displaystyle\boldsymbol{H}_{0}=\boldsymbol{H}_{0}(s)=-h_{0}(\boldsymbol{P}_{\perp}(s)-2\boldsymbol{P}_{||}(s))$
(see Proposition 3.4).

Using the expression in (14) for $\displaystyle e_{\lambda}$, we have

(23) $\begin{split}e_{\lambda}&=-\dfrac{1}{2}\left({{\mathbold
f}_{\lambda}},{T_{\lambda}{\mathbold
f}_{\lambda}}\right)_{L^{2}(\mathbb{R},\mathbb{R}^{3})}=-\dfrac{1}{2}\left[\left({{\mathbold
f}_{\lambda}-{\mathbold f}},{T_{\lambda}{\mathbold
f}_{\lambda}}\right)_{L^{2}(\mathbb{R},\mathbb{R}^{3})}+\left({{\mathbold
f}},{T_{\lambda}{\mathbold
f}_{\lambda}}\right)_{L^{2}(\mathbb{R},\mathbb{R}^{3})}\right]\\\
&=-\dfrac{1}{2}\left[\left({{\mathbold f}_{\lambda}-{\mathbold
f}},{T_{\lambda}{\mathbold
f}_{\lambda}}\right)_{L^{2}(\mathbb{R},\mathbb{R}^{3})}+\left({{\mathbold
f}},{T_{\lambda}{\mathbold
f}}\right)_{L^{2}(\mathbb{R},\mathbb{R}^{3})}+\left({{\mathbold
f}},{T_{\lambda}({\mathbold f}_{\lambda}-{\mathbold
f})}\right)_{L^{2}(\mathbb{R},\mathbb{R}^{3})}\right].\end{split}$

The first and third terms are zero in the limit. Taking the limit of the
remaining term and using (22), we have

$\lim_{\lambda\to 0}e_{\lambda}=\lim_{\lambda\to
0}-\dfrac{1}{2}\left({{\mathbold f}_{\lambda}},{T_{\lambda}{\mathbold
f}_{\lambda}}\right)_{L^{2}(\mathbb{R},\mathbb{R}^{3})}=\dfrac{1}{2}h_{0}\left[||\boldsymbol{P}_{\perp}{\mathbold
f}||^{2}_{L^{2}(\mathbb{R},\mathbb{R}^{3})}-2||\boldsymbol{P}_{||}{\mathbold
f}||^{2}_{L^{2}(\mathbb{R},\mathbb{R}^{3})}\right].$

This completes the proof.

######  Remark 3.7.

The limiting energy only comprises of a local self-field energy. In the limit,
any point on the helix sees a uniform 1-d system of dipole moments along the
tangent line. Further, we see that both the normal components and the
tangential component of the dipole moment contribute to the energy and
electric field. This is in contrast to [CH15], where the thin wire limit of
the magnetostatic energy, obtained from dimensional reduction starting from a
3-d continuum, has contributions only from the normal component.
Heuristically, the dimension reduction starting from a 3-d continuum contains
minimal information about the detailed atomic arrangements within the
nanostructure, and hence does not capture the consequences of the helical
geometry. Because of its starting point in a 3-d continuum model, it is
appropriate for thin objects that have all dimensions being much larger than
the atomic lengthscale. On the other hand, the discrete-to-continuum approach
is appropriate for nanostructures wherein the thin dimension is comparable to
the atomic lengthscale.

###  3.2 Nanofilm with Constant Bending Curvature

Let $\displaystyle{\cal S}=(-\bar{\theta},\bar{\theta})\times\mathbb{R}$ be
the parametric space for a surface with a constant bending curvature
$\displaystyle\kappa$. The map that takes a point in the parametric space to a
unique point on the film is given by

(24) $\bar{\boldsymbol{x}}(s_{1},s_{2})={\cal
R}\boldsymbol{Q}(s_{1})\boldsymbol{e}_{1}+s_{2}\delta\boldsymbol{e}_{3},$

where $\displaystyle{\cal R}=1/\kappa$ is the inverse of curvature,
$\displaystyle\bar{\theta}>0$ is the angular size of the film, and
$\displaystyle\delta$ is the spacing in the flat direction. Here,
$\displaystyle\kappa,\delta,\bar{\theta}$ are fixed parameters for a given
film. Here, $\displaystyle\boldsymbol{Q}=\boldsymbol{Q}(\theta)$ is the
rotational tensor with the axis $\displaystyle\boldsymbol{e}_{3}$, see
definition (6). The tangent vectors at
$\displaystyle\boldsymbol{s}:=(s_{1},s_{2})\in S$ are

(25)
$\boldsymbol{t}_{1}(\boldsymbol{s})=\frac{\mathrm{d}\bar{\boldsymbol{x}}}{\mathrm{d}s_{1}}={\cal
R}\boldsymbol{Q}^{\prime}(s_{1})\boldsymbol{e}_{1},\qquad\boldsymbol{t}_{2}(\boldsymbol{s})=\frac{\mathrm{d}\bar{\boldsymbol{x}}}{\mathrm{d}s_{2}}=\delta\boldsymbol{e}_{3}$

and the normal vector is

(26) $\boldsymbol{n}(\boldsymbol{s})=\boldsymbol{Q}(s_{1})\boldsymbol{e}_{1}.$

####  3.2.1 Lattice Geometry and Dipole Moment

We consider a lattice embedded on the film
$\displaystyle\bar{\boldsymbol{x}}$. We assume that the film is one lattice
cell thick in the direction $\displaystyle\boldsymbol{n}$ normal to the film.
Suppose $\displaystyle{\cal L}\subset{\cal S}$ is the set of parametric
coordinates of the discrete lattice sites. Let $\displaystyle{\cal L}$ and the
lattice cell $\displaystyle U$ (in the parametric space $\displaystyle{\cal
S}$) be given by

(27) $\begin{split}{\cal L}&=\\{\boldsymbol{s}=(s_{1},s_{2})\in{\cal
S};\,s_{1}=i\theta_{l},s_{2}=j,\>i,j\in\mathbb{Z}\\}=\left((-\bar{\theta},\bar{\theta})\cap\theta_{l}\mathbb{Z}\right)\times\mathbb{Z}\\\
U(\boldsymbol{s})&=[s_{1},s_{1}+\theta_{l})\times[s_{2},s_{2}+1),\qquad\forall\boldsymbol{s}\in{\cal
L}.\end{split}$

Here, $\displaystyle\theta_{l}$ is the angular width of the lattice cell. We
assume that $\displaystyle\theta_{l}$ is such that the set of sites in the
angular direction,
$\displaystyle(-\bar{\theta},\bar{\theta})\cap\theta_{l}\mathbb{Z}$, is not
empty, and in fact is sufficiently large so that the continuum limit
approximation of the energy density is justified. We assume that the lattice
has unit thickness in the normal direction, and suppose that the film given by
$\displaystyle\bar{\boldsymbol{x}}$ passes through the center of the lattice
in the normal direction. Then, the unit cell for a given site
$\displaystyle\boldsymbol{s}\in{\cal L}$ is
$\displaystyle\bar{U}(\boldsymbol{s})=\\{\boldsymbol{x};\,\boldsymbol{x}=\bar{\boldsymbol{x}}(\boldsymbol{s}^{\prime})+t\boldsymbol{n}(\boldsymbol{s}^{\prime}),\,\boldsymbol{s}^{\prime}\in
U(\boldsymbol{s}),\,t\in(-1/2,1/2)\\}$ (see Fig. 4). On the lattice
$\displaystyle{\cal L}$, we define a discrete system of dipole moments
$\displaystyle\boldsymbol{d}\colon{\cal L}\to\mathbb{R}^{3}$. As in the case
of the helical nanotube, the lattice cells in real space are related by an
isometric transformation, so the magnitudes of the dipoles at the lattice
sites are equal, but they are oriented differently. In particular, we have

(28) ${\mathbold d}(\boldsymbol{s}+\boldsymbol{r})={\mathbold
Q}(r_{1}){\mathbold d}(\boldsymbol{s}),\quad\boldsymbol{s}\in{\cal L},$

where $\displaystyle\boldsymbol{r}=(r_{1},r_{2})\in{\cal L}$ such that
$\displaystyle\boldsymbol{r}+\boldsymbol{s}\in{\cal L}$ (i.e. all the
translations within $\displaystyle{\cal L}$). We see that the dipole
orientation depends only on the angular (first) parameter and is invariant
with respect to the second parameter.

(a)

(b)

Figure 4: Depiction of lattice sites in (a) parametric and (b) real space
where the orange region marks a lattice cell in the parametric and real space.

We next consider the scaling of the lattice by $\displaystyle\lambda>0$. The
scaled lattice $\displaystyle{\cal L}_{\lambda}$ and the associated lattice
cell $\displaystyle U_{\lambda}$ are defined by the natural scaling of
$\displaystyle{\cal L}$ and $\displaystyle U$ as follows

(29) $\begin{split}{\cal L}_{\lambda}&=\\{\boldsymbol{s}\in{\cal
S};\,s_{1}=i\theta_{l}\lambda,s_{2}=j\lambda,\,i,j\in\mathbb{Z}\\}=\left((-\bar{\theta},\bar{\theta})\cap\lambda\theta_{l}\mathbb{Z}\right)\times\lambda\mathbb{Z}\\\
U_{\lambda}(\boldsymbol{s})&=[s_{1},s_{1}+\lambda\theta_{l})\times[s_{2},s_{2}+\lambda),\qquad\forall\boldsymbol{s}\in{\cal
L}_{\lambda}.\end{split}$

After scaling, the thickness of the lattice cell in the normal direction is
$\displaystyle\lambda$ and the unit cell for
$\displaystyle\boldsymbol{s}\in{\cal L}_{\lambda}$ is
$\displaystyle\bar{U}_{\lambda}(\boldsymbol{s})=\\{\boldsymbol{x};\,\boldsymbol{x}=\bar{\boldsymbol{x}}(\boldsymbol{s}^{\prime})+t\boldsymbol{n}(\boldsymbol{s}^{\prime}),\,\boldsymbol{s}^{\prime}\in
U_{\lambda}(\boldsymbol{s}),\,t\in(-\lambda/2,\lambda/2)\\}$. We can show that
the unit cell in the scaled lattice has volume $\displaystyle\lambda^{3}{\cal
R}\theta_{l}$. Let $\displaystyle\boldsymbol{d}_{\lambda}\colon{\cal
L}_{\lambda}\to\mathbb{R}^{3}$ denote the discrete system of dipole moments
associated with the scaled lattice, $\displaystyle{\cal L}_{\lambda}$, and
$\displaystyle\tilde{\boldsymbol{d}}_{\lambda}\colon{\cal S}\to\mathbb{R}^{3}$
denote the piecewise constant extension of
$\displaystyle\boldsymbol{d}_{\lambda}$ given by

(30)
$\tilde{\boldsymbol{d}}_{\lambda}(\boldsymbol{s})=\frac{\boldsymbol{d}_{\lambda}(\boldsymbol{a})}{\lambda^{3}{\cal
R}\theta_{l}},\qquad\forall\boldsymbol{s}\in
U_{\lambda}(\boldsymbol{a}),\,\forall\boldsymbol{a}\in{\cal L}_{\lambda}.$

We are interested in the limit of the energy when
$\displaystyle\tilde{\boldsymbol{d}}_{\lambda}$ converges to
$\displaystyle{\mathbold f}$ in $\displaystyle L^{2}({\cal
S},\mathbb{R}^{3})$. As in the case of the helix and following [JM94], we
suppose that there exists a background dipole moment density field
$\displaystyle{\mathbold f}_{\lambda}\in L^{2}({\cal S},\mathbb{R}^{3})$ such
that the dipole moment at site $\displaystyle\boldsymbol{s}\in{\cal
L}_{\lambda}$ is given by

(31)
$\boldsymbol{d}_{\lambda}(\boldsymbol{s})=\int_{-\lambda/2}^{\lambda/2}\left[\int_{U_{\lambda}(\boldsymbol{s})}{\mathbold
f}_{\lambda}(\boldsymbol{t}){\cal R}\ \mathrm{d}t_{1}\ \mathrm{d}t_{2}\right]\
\mathrm{d}t_{3}={\cal R}\lambda\int_{U_{\lambda}(\boldsymbol{s})}{\mathbold
f}_{\lambda}(\boldsymbol{t})\ \mathrm{d}\boldsymbol{t},$

where $\displaystyle\mathrm{d}\boldsymbol{t}=\mathrm{d}t_{1}\ \mathrm{d}t_{2}$
is the area measure (note that $\displaystyle\mathrm{d}\boldsymbol{t}$ does
not include $\displaystyle{\cal R}$). The existence of one such background
field $\displaystyle{\mathbold f}_{\lambda}$ is evident: we can define
$\displaystyle{\mathbold f}_{\lambda}=\tilde{\boldsymbol{d}}_{\lambda}$.
Similar to the case of the helix, we have the following lemma that relates the
convergence of the background dipole moment field and the piecewise constant
extension.

######  Lemma 3.8.

Let $\displaystyle{\mathbold f}_{\lambda}$, $\displaystyle\lambda>0$, be a
sequence of $\displaystyle L^{2}({\cal S},\mathbb{R}^{3})$ functions and let
$\displaystyle{\mathbold f}\in L^{2}({\cal S},\mathbb{R}^{3})$ be such that
$\displaystyle{\mathbold f}_{\lambda}\to{\mathbold f}$ in $\displaystyle
L^{2}({\cal S},\mathbb{R}^{3})$. Let
$\displaystyle\boldsymbol{d}_{\lambda}\colon{\cal
L}_{\lambda}\to\mathbb{R}^{3}$ be given by (31) and let
$\displaystyle\tilde{\boldsymbol{d}}_{\lambda}$ be a piecewise constant
$\displaystyle L^{2}$ extension of $\displaystyle\boldsymbol{d}_{\lambda}$
given by (30). Then,
$\displaystyle\tilde{\boldsymbol{d}}_{\lambda}\to{\mathbold f}$ in
$\displaystyle L^{2}({\cal S},\mathbb{R}^{3})$.

On the other hand, if $\displaystyle\boldsymbol{d}_{\lambda}\colon{\cal
L}_{\lambda}\to\mathbb{R}^{3}$ is such that
$\displaystyle\tilde{\boldsymbol{d}}_{\lambda}\to{\mathbold f}$ in
$\displaystyle L^{2}({\cal S},\mathbb{R}^{3})$, then there exists a background
field $\displaystyle{\mathbold f}_{\lambda}\in L^{2}({\cal S},\mathbb{R}^{3})$
such that $\displaystyle\boldsymbol{d}_{\lambda}$ is given by (31).

The proof follows directly from the proof of Theorem 4.1 of [JM94].

######  Remark 3.9.

As different unit cells are related by isometric transformations, the dipole
moments in different unit cells are related by the rotational part of the
isometric transformation.

(a)

(b)

Figure 5: Discrete dipole moments on a nanofilm with uniform bending
curvature. (a) and (b) show the view from different perspectives.

####  3.2.2 Electrostatic Energy

As before, the energy associated to the system of dipole moments
$\displaystyle\boldsymbol{d}_{\lambda}$, for $\displaystyle\lambda>0$, is
given by

$E_{\lambda}=-\frac{1}{2}\sum_{\begin{subarray}{c}\boldsymbol{s},\boldsymbol{s}^{\prime}\in{\cal
L}_{\lambda},\\\
\boldsymbol{s}\neq\boldsymbol{s}^{\prime}\end{subarray}}\boldsymbol{d}_{\lambda}(\boldsymbol{s})\cdot\boldsymbol{K}(\bar{\boldsymbol{x}}(\boldsymbol{s}^{\prime})-\bar{\boldsymbol{x}}(\boldsymbol{s}))\boldsymbol{d}_{\lambda}(\boldsymbol{s}^{\prime})=|(-\lambda/2,\lambda/2)|\hat{e}_{\lambda},$

where $\displaystyle|(-\lambda/2,\lambda/2)|=\lambda$ is the thickness of the
lattice in normal direction, and $\displaystyle\hat{e}_{\lambda}$ is the
energy per unit length given by

(32)
$\hat{e}_{\lambda}=-\frac{1}{2\lambda}\sum_{\begin{subarray}{c}\boldsymbol{s},\boldsymbol{s}^{\prime}\in{\cal
L}_{\lambda},\\\
\boldsymbol{s}\neq\boldsymbol{s}^{\prime}\end{subarray}}\boldsymbol{d}_{\lambda}(\boldsymbol{s})\cdot\boldsymbol{K}(\bar{\boldsymbol{x}}(\boldsymbol{s}^{\prime})-\bar{\boldsymbol{x}}(\boldsymbol{s}))\boldsymbol{d}_{\lambda}(\boldsymbol{s}^{\prime}).$

For convenience, we normalize $\displaystyle\hat{e}_{\lambda}$ by
$\displaystyle{\cal R}\theta_{l}$, where $\displaystyle{\cal R}\theta_{l}$ is
independent of $\displaystyle\lambda$ and gives the size of the original
lattice in the angular direction. We let

(33) $e_{\lambda}=\frac{\hat{e}_{\lambda}}{{\cal R}\theta_{l}}\Rightarrow
E_{\lambda}=\lambda({\cal R}\theta_{l})e_{\lambda}.$

Substituting (31) and proceeding similar to the case of the helix, we can
express $\displaystyle e_{\lambda}$ as

(34) $e_{\lambda}=\left({{\mathbold f}_{\lambda}},{T_{\lambda}{\mathbold
f}_{\lambda}}\right)_{L^{2}({\cal S},\mathbb{R}^{3})},$

where $\displaystyle T_{\lambda}\colon L^{2}({\cal S},\mathbb{R}^{3})\to
L^{2}({\cal S},\mathbb{R}^{3})$ is the map defined as

(35) $(T_{\lambda}{\mathbold f})(\boldsymbol{s})=\frac{{\cal
R}}{\theta_{l}}\lambda\int_{{\cal
S}}\boldsymbol{K}_{\lambda}(\boldsymbol{s}^{\prime},\boldsymbol{s}){\mathbold
f}(\boldsymbol{s}^{\prime})\ \mathrm{d}\boldsymbol{s}^{\prime}$

and
$\displaystyle\boldsymbol{K}_{\lambda}(\boldsymbol{s}^{\prime},\boldsymbol{s})$,
for $\displaystyle\boldsymbol{s},\boldsymbol{s}^{\prime}\in{\cal S}$, is the
discrete dipole field kernel given by

(36)
$\boldsymbol{K}_{\lambda}(\boldsymbol{s}^{\prime},\boldsymbol{s})=\sum_{\begin{subarray}{c}\boldsymbol{u},\boldsymbol{v}\in{\cal{L}}_{\lambda},\\\
\boldsymbol{u}\neq\boldsymbol{v}\end{subarray}}\chi_{U_{\lambda}(\boldsymbol{v})}(\boldsymbol{s}^{\prime})\boldsymbol{K}(\bar{\boldsymbol{x}}(\boldsymbol{v})-\bar{\boldsymbol{x}}(\boldsymbol{u}))\chi_{U_{\lambda}(\boldsymbol{u})}(\boldsymbol{s}).$

##### Scaling of $\displaystyle\boldsymbol{K}_{\lambda}$

As in the case of the helix, it is convenient to first rescale the lattice
$\displaystyle{\cal L}_{\lambda}$ such that the lattice cell size is
independent of $\displaystyle\lambda$ after rescaling, and define a new map on
the rescaled lattice. This is considered next.

Let $\displaystyle{\cal
S}_{1,\lambda}=(-\bar{\theta}/\lambda,\bar{\theta}/\lambda)\times\mathbb{R}$
so that $\displaystyle\boldsymbol{s}\in S$ implies
$\displaystyle\boldsymbol{s}/\lambda\in{\cal S}_{1,\lambda}$. We define a
rescaled lattice $\displaystyle{\cal L}_{1,\lambda}$ such that
$\displaystyle\boldsymbol{s}\in{\cal L}_{\lambda}$ implies
$\displaystyle\boldsymbol{s}/\lambda\in{\cal L}_{1,\lambda}$. It is given by

(37) ${\cal L}_{1,\lambda}=\\{\boldsymbol{s}\in{\cal
S}_{1,\lambda};\,s_{1}=i\theta_{l},\,s_{2}=j,\,i,j\in\mathbb{Z}\\}=\left((-\bar{\theta}/\lambda,\bar{\theta}/\lambda)\cap\theta_{l}\mathbb{Z}\right)\times\mathbb{Z}.$

The lattice cell for $\displaystyle\boldsymbol{s}\in{\cal L}_{1,\lambda}$ is
given by $\displaystyle U_{1}(\boldsymbol{s})$, where $\displaystyle
U_{1}(\boldsymbol{s})$ is defined in (27) (using $\displaystyle\lambda=1$ in
$\displaystyle U_{\lambda}$). For
$\displaystyle\boldsymbol{a},\boldsymbol{b}\in{\cal S}_{1,\lambda}$, we have

(38)
$\bar{\boldsymbol{x}}(\lambda\boldsymbol{a})-\bar{\boldsymbol{x}}(\lambda\boldsymbol{b})=\lambda\left(\bar{\boldsymbol{x}}(\boldsymbol{a})-\bar{\boldsymbol{x}}(\boldsymbol{b})+\boldsymbol{A}_{\lambda}(\boldsymbol{a},\boldsymbol{b})\boldsymbol{e}_{1}\right),$

where

(39) $\boldsymbol{A}_{\lambda}(\boldsymbol{a},\boldsymbol{b})=\frac{{\cal
R}}{\lambda}\left[\boldsymbol{Q}(\lambda a_{1})-\boldsymbol{Q}(\lambda
b_{1})-\lambda\boldsymbol{Q}(a_{1})+\lambda\boldsymbol{Q}(b_{1})\right].$

Keeping in mind these definitions, for $\displaystyle\boldsymbol{u}\in{\cal
L}_{1,\lambda}$, we also note

(40)
$\begin{split}\chi_{U_{\lambda}(\lambda\boldsymbol{u})}(\boldsymbol{s})&=\begin{cases}1\qquad\text{if
}\boldsymbol{s}\in U_{\lambda}(\lambda\boldsymbol{u}),\\\
0\qquad\text{otherwise}\end{cases}=\quad\begin{cases}1\qquad\text{if
}\boldsymbol{s}/\lambda\in U_{1}(\boldsymbol{u}),\\\
0\qquad\text{otherwise}\end{cases}=\chi_{U_{1}(\boldsymbol{u})}(\boldsymbol{s}/\lambda).\end{split}$

Using the above relation and (38), we can show, for any
$\displaystyle\boldsymbol{s},\boldsymbol{s}^{\prime}\in{\cal S}$,

$\boldsymbol{K}_{\lambda}(\boldsymbol{s}^{\prime},\boldsymbol{s})=\frac{1}{\lambda^{3}}\sum_{\begin{subarray}{c}\boldsymbol{u},\boldsymbol{v}\in{\cal
L}_{1,\lambda},\\\
\boldsymbol{u}\neq\boldsymbol{v}\end{subarray}}\chi_{U_{1}(\boldsymbol{v})}(\boldsymbol{s}^{\prime}/\lambda)\boldsymbol{K}(\bar{\boldsymbol{x}}(\boldsymbol{v})-\bar{\boldsymbol{x}}(\boldsymbol{u})+\boldsymbol{A}_{\lambda}(\boldsymbol{v},\boldsymbol{u})\boldsymbol{e}_{1})\chi_{U_{1}(\boldsymbol{u})}(\boldsymbol{s}/\lambda).$

If we introduce the discrete dipole field kernel
$\displaystyle\boldsymbol{K}_{1,\lambda}(\boldsymbol{s}^{\prime},\boldsymbol{s})$,
for $\displaystyle\boldsymbol{s},\boldsymbol{s}^{\prime}\in{\cal
S}_{1,\lambda}$, defined on $\displaystyle{\cal L}_{1,\lambda}$ as:

(41)
$\boldsymbol{K}_{1,\lambda}(\boldsymbol{s}^{\prime},\boldsymbol{s})=\sum_{\begin{subarray}{c}\boldsymbol{u},\boldsymbol{v}\in{\cal
L}_{1,\lambda},\\\
\boldsymbol{u}\neq\boldsymbol{v}\end{subarray}}\chi_{U_{1}(\boldsymbol{v})}(\boldsymbol{s}^{\prime})\boldsymbol{K}(\bar{\boldsymbol{x}}(\boldsymbol{v})-\bar{\boldsymbol{x}}(\boldsymbol{u})+\boldsymbol{A}_{\lambda}(\boldsymbol{v},\boldsymbol{u})\boldsymbol{e}_{1})\chi_{U_{1}(\boldsymbol{u})}(\boldsymbol{s}),$

we have shown that:

$\boldsymbol{K}_{\lambda}(\boldsymbol{s}^{\prime},\boldsymbol{s})=\frac{1}{\lambda^{3}}\boldsymbol{K}_{1,\lambda}(\boldsymbol{s}^{\prime}/\lambda,\boldsymbol{s}/\lambda),\qquad\forall\boldsymbol{s},\boldsymbol{s}^{\prime}\in{\cal
S}.$

####  3.2.3 Limit of the Electrostatic Energy

In this section, we obtain the limit of the energy per unit length
$\displaystyle e_{\lambda}$. The broad strategy is similar to the helical
nanotube. We first show that the map $\displaystyle T_{\lambda}$ is bounded
and obtain its limit. The continuum limit of the energy density $\displaystyle
e_{\lambda}$ then follows easily.

##### Limit of the Discrete Electric Field

Let $\displaystyle T_{1,\lambda}\colon L^{2}({\cal
S}_{1,\lambda},\mathbb{R}^{3})\to L^{2}({\cal S}_{1,\lambda},\mathbb{R}^{3})$
be the map with kernel $\displaystyle\boldsymbol{K}_{1,\lambda}$. For any
function $\displaystyle{\mathbold f}\in L^{2}({\cal
S}_{1,\lambda},\mathbb{R}^{3})$, we have

(42) $(T_{1,\lambda}{\mathbold f})(\boldsymbol{s})=\frac{{\cal
R}}{\theta_{l}}\int_{{\cal
S}_{1,\lambda}}\boldsymbol{K}_{1,\lambda}(\boldsymbol{s}^{\prime},\boldsymbol{s}){\mathbold
f}(\boldsymbol{s}^{\prime})\
\mathrm{d}\boldsymbol{s}^{\prime},\qquad\forall\boldsymbol{s}\in{\cal
S}_{1,\lambda}.$

Let
$\displaystyle\boldsymbol{H}_{\lambda}=\boldsymbol{H}_{\lambda}(\boldsymbol{s})$
be the zeroth order moment (with respect to the first argument) of kernel
$\displaystyle\boldsymbol{K}_{\lambda}$ given by

(43) $\boldsymbol{H}_{\lambda}(\boldsymbol{s})=\frac{{\cal
R}\lambda}{\theta_{l}}\int_{\boldsymbol{s}^{\prime}\in{\cal
S}}\boldsymbol{K}_{\lambda}(\boldsymbol{s}^{\prime},\boldsymbol{s})\
\mathrm{d}\boldsymbol{s}^{\prime},\quad\forall\boldsymbol{s}\in{\cal S}.$

We now state the limit result of $\displaystyle T_{\lambda}$.

######  Proposition 3.10.

Suppose $\displaystyle 0<\theta<\pi/4$. The maps $\displaystyle T_{1,\lambda}$
and $\displaystyle T_{\lambda}$ are bounded in $\displaystyle L^{2}$ for all
$\displaystyle\lambda>0$ and satisfy

$\left\lVert T_{\lambda}\right\rVert_{\mathcal{L}(L^{2}({\cal
S},\mathbb{R}^{3}),L^{2}({\cal S},\mathbb{R}^{3}))}=\left\lVert
T_{1,\lambda}\right\rVert_{\mathcal{L}(L^{2}({\cal
S}_{1,\lambda},\mathbb{R}^{3}),L^{2}({\cal S}_{1,\lambda},\mathbb{R}^{3}))}.$

Further, for $\displaystyle{\mathbold f}\in
C^{\infty}_{0}(\mathbb{R},\mathbb{R}^{3})$,

$(T_{\lambda}{\mathbold f})(\boldsymbol{s})\xrightarrow[\lambda\to
0]{}\boldsymbol{H}_{0}(\boldsymbol{s}){\mathbold f}(\boldsymbol{s}),$

pointwise, where $\displaystyle\boldsymbol{H}_{0}(\boldsymbol{s})$, for
$\displaystyle\boldsymbol{s}\in{\cal S}$, is given by

(44) $\boldsymbol{H}_{0}(\boldsymbol{s})=\lim_{\lambda\to
0}\boldsymbol{H}_{\lambda}(\boldsymbol{s})={\cal
R}\sum_{\begin{subarray}{c}\boldsymbol{u}=(u_{1},u_{2})\in\theta_{l}\mathbb{Z}\times\mathbb{Z},\\\
\boldsymbol{u}\neq\mathbf{0}\end{subarray}}\boldsymbol{K}\left(u_{1}\boldsymbol{t}_{1}(\boldsymbol{s})+u_{2}\boldsymbol{t}_{2}(\boldsymbol{s})\right)$

and
$\displaystyle\boldsymbol{t}_{i}(\boldsymbol{s})=\frac{\mathrm{d}\bar{\boldsymbol{x}}(\boldsymbol{s})}{\mathrm{d}s_{i}}$,
$\displaystyle i=1,2$, are the tangent vectors to the film.

We provide the proof of Proposition 3.10 in subsection 4.2. Based on the
proposition above, we state the main result for the thin film.

##### Limit of the Energy

######  Theorem 3.11.

Let $\displaystyle{\mathbold f}_{\lambda}\in L^{2}({\cal S},\mathbb{R}^{3})$
be a sequence of functions for $\displaystyle\lambda>0$ with
$\displaystyle{\mathbold f}\in L^{2}({\cal S},\mathbb{R}^{3})$ such that
$\displaystyle{\mathbold f}_{\lambda}\to{\mathbold f}$ in $\displaystyle
L^{2}({\cal S},\mathbb{R}^{3})$. Let the system of dipole moments
$\displaystyle\boldsymbol{d}_{\lambda}\colon{\cal
L}_{\lambda}\to\mathbb{R}^{3}$ be given by (31). Let $\displaystyle
e_{\lambda}$, given by (34), be the energy per unit length normalized by
$\displaystyle{\cal R}\theta_{l}$. Then

$e_{\lambda}\xrightarrow[\lambda\to 0]{}-\dfrac{1}{2}\left({{\mathbold
f}},{\boldsymbol{H}_{0}{\mathbold f}}\right)_{L^{2}({\cal
S},\mathbb{R}^{3})},$

where $\displaystyle\boldsymbol{H}_{0}=\boldsymbol{H}_{0}(\boldsymbol{s})$ is
defined in Proposition 3.10, see (44).

The proof of Theorem 3.11 follows from the proof of Theorem 3.5 and using
Proposition 3.10.

######  Remark 3.12.

Note that, for $\displaystyle\boldsymbol{s}\in{\cal S}$,

(45)
$\boldsymbol{H}_{0}(\boldsymbol{s})=\boldsymbol{Q}(s_{1})\boldsymbol{H}_{0}(\mathbf{0})\boldsymbol{Q}(-s_{1}).$

Thus, if the limiting dipole moment field $\displaystyle{\mathbold f}$ is
uniform in the $\displaystyle\boldsymbol{e}_{3}$ direction, the electric field
$\displaystyle\boldsymbol{H}_{0}(\boldsymbol{s}){\mathbold f}(\boldsymbol{s})$
will be independent of the $\displaystyle\boldsymbol{e}_{3}$-coordinate. It is
easy to see from the expression of $\displaystyle\boldsymbol{H}_{0}$ that both
the normal component and the tangential components of the dipole field
contribute to the electric field and energy. This is in contrast to [GJ97],
where the thin film limit of the magnetostatic energy, obtained from
dimensional reduction starting from a 3-d continuum, has contributions only
from the normal component. However, it is consistent with the result for the
helical nanostructure studied in this paper, in that the components of the
dipole field aligned with the thin direction contribute to the limit energy.
As we argued there, the dimensional reduction from a 3-d continuum model is
appropriate for thin films that have thickness that is much larger than the
atomic lengthscale, whereas the discrete-to-continuum approach is appropriate
for nanostructures that have thickness that is comparable to the atomic
lengthscale.

##  4 Proof of Assertions

###  4.1 Helical Nanotube

In this section, we prove Proposition 3.4. First, we collect some important
results, and then show that $\displaystyle T_{\lambda}$ is bounded and extends
from $\displaystyle{\mathbold f}\in C^{\infty}_{0}(\mathbb{R},\mathbb{R}^{3})$
to $\displaystyle L^{2}(\mathbb{R},\mathbb{R}^{3})$. We then obtain the limit
of the map $\displaystyle T_{\lambda}$.

######  Lemma 4.1.

  * 1.

For any $\displaystyle a,b\in\mathbb{R}$,

(46)
$\bar{\boldsymbol{x}}(b)-\bar{\boldsymbol{x}}(a)=\boldsymbol{Q}(a\theta)[(\boldsymbol{Q}((b-a)\theta)-\boldsymbol{I})\boldsymbol{e}_{1}+\delta(b-a)\boldsymbol{e}_{3}],$

where $\displaystyle\bar{\boldsymbol{x}}$ is the map (5),
$\displaystyle\boldsymbol{Q}$ is the rotational tensor (6),
$\displaystyle\theta$ and $\displaystyle\delta$ define the helix.

  * 2.

For any $\displaystyle\theta\in(0,\pi)$,

(47) $\delta\leq\min_{a,b\in{\cal L}_{1},a\neq
b}|\bar{\boldsymbol{x}}(b)-\bar{\boldsymbol{x}}(a)|,$

where $\displaystyle{\cal L}_{1}$ is $\displaystyle{\cal
L}_{\lambda}=\lambda\mathbb{Z}$ for $\displaystyle\lambda=1$.

  * 3.

For any $\displaystyle a,b\in{\cal L}_{1}$ and $\displaystyle\lambda>0$,

(48)
$\delta|a-b|\leq|\bar{\boldsymbol{x}}(a)-\bar{\boldsymbol{x}}(b)+\boldsymbol{A}_{\lambda}(a,b)\boldsymbol{e}_{1}|,$

where $\displaystyle\boldsymbol{A}_{\lambda}(a,b)$ is given by

$\boldsymbol{A}_{\lambda}(a,b)=\frac{\boldsymbol{Q}(\lambda
a\theta)-\boldsymbol{Q}(\lambda
b\theta)-(\lambda\boldsymbol{Q}(a\theta)-\lambda\boldsymbol{Q}(b\theta))}{\lambda}.$

  * 4.

For any $\displaystyle s,s^{\prime}\in\mathbb{R}$ such that
$\displaystyle|s-s^{\prime}|\geq 1$, suppose $\displaystyle a,b\in{\cal
L}_{1}$ are such that $\displaystyle s\in[a,a+1),s^{\prime}\in[b,b+1)$, then

(49) $\frac{|s-s^{\prime}|}{|a-b|}<3.$

######  Proof 4.2.

  1. 1.

For any $\displaystyle\alpha,\beta\in\mathbb{R}$, we have the identities

(50)
$\boldsymbol{Q}^{T}(\alpha)=\boldsymbol{Q}(-\alpha),\qquad\boldsymbol{Q}(\alpha)\boldsymbol{Q}(\beta)=\boldsymbol{Q}(\alpha+\beta),\qquad\boldsymbol{Q}(\alpha)\boldsymbol{e}_{3}=\boldsymbol{e}_{3},$

where the last relation shows that $\displaystyle\boldsymbol{e}_{3}$ is the
axis of $\displaystyle\boldsymbol{Q}$. By noting the definition of
$\displaystyle\bar{\boldsymbol{x}}$ in (5) and using the identities above,
(46) follows.

  2. 2.

To show (47), we use (46) to get

$|\bar{\boldsymbol{x}}(b)-\bar{\boldsymbol{x}}(a)|^{2}=|(\boldsymbol{Q}((b-a)\theta)-\boldsymbol{I})\boldsymbol{e}_{1}|^{2}+\delta^{2}|b-a|^{2}\geq\delta^{2}|b-a|^{2}\geq\delta^{2},$

where we used the fact that $\displaystyle|b-a|\geq 1$ for $\displaystyle
a,b\in{\cal L}_{1},a\neq b$.

  3. 3.

To show (48), we substitute the definition of
$\displaystyle\boldsymbol{A}_{\lambda}$ to get

(51)
$\bar{\boldsymbol{x}}(a)-\bar{\boldsymbol{x}}(b)+\boldsymbol{A}_{\lambda}(a,b)\boldsymbol{e}_{1}=\frac{\boldsymbol{Q}(\lambda
a\theta)-\boldsymbol{Q}(\lambda
b\theta)}{\lambda}\boldsymbol{e}_{1}+(a-b)\delta\boldsymbol{e}_{3}.$

Since $\displaystyle\boldsymbol{Q}(\alpha)\boldsymbol{e}_{1}$ is orthogonal to
$\displaystyle\boldsymbol{e}_{3}$ for any $\displaystyle\alpha$, we have

$|\bar{\boldsymbol{x}}(a)-\bar{\boldsymbol{x}}(b)+\boldsymbol{A}_{\lambda}(a,b)\boldsymbol{e}_{1}|\geq\delta|a-b|.$

  4. 4.

To show (49), we note that for $\displaystyle s,s^{\prime}\in\mathbb{R}$ such
that $\displaystyle|s-s^{\prime}|\geq 1$ with $\displaystyle a,b\in{\cal
L}_{1}$ and $\displaystyle s\in[a,a+1),s^{\prime}\in[b,b+1)$, we can write
$\displaystyle s=a+\Delta s$ and $\displaystyle s^{\prime}=b+\Delta
s^{\prime}$ with $\displaystyle 0\leq\Delta s,\Delta s^{\prime}<1$. Thus

$\frac{|s-s^{\prime}|}{|a-b|}=\frac{|a-b+(\Delta s-\Delta
s^{\prime})|}{|a-b|}\leq\frac{|a-b|+|\Delta s-\Delta
s^{\prime}|}{|a-b|}<1+\frac{2}{|a-b|}\leq 3,$

where in the last step we used the fact that $\displaystyle|a-b|\geq 1$ for
$\displaystyle a,b\in{\cal L}_{1},a\neq b$ (which is ensured when
$\displaystyle|s-s^{\prime}|\geq 1$).

This completes the proof.

####  4.1.1 Boundedness

We next show that $\displaystyle T_{\lambda}$ is a bounded map. Let
$\displaystyle S_{\lambda}\colon L^{2}(\mathbb{R},\mathbb{R}^{3})\to
L^{2}(\mathbb{R},\mathbb{R}^{3})$ be an isometry defined as

$(S_{\lambda}{\mathbold f})(s):=\lambda^{1/2}{\mathbold f}(\lambda s).$

It is easy to see that $\displaystyle||S_{\lambda}{\mathbold
f}||_{L^{2}(\mathbb{R},\mathbb{R}^{3})}=||{\mathbold
f}||_{L^{2}(\mathbb{R},\mathbb{R}^{3})}$. The inverse of $\displaystyle
S_{\lambda}$ is given by

(52) $(S^{-1}_{\lambda}{\mathbold f})(s)=\lambda^{-1/2}{\mathbold
f}(s/\lambda).$

Using $\displaystyle S_{\lambda}$, we can show – noting the definition of
$\displaystyle T_{\lambda}$ in (15) – for $\displaystyle{\mathbold f}\in
L^{2}(\mathbb{R},\mathbb{R}^{3})$,

(53) $\begin{split}(T_{\lambda}{\mathbold
f})(s)&=\lambda^{2}\int_{\mathbb{R}}\boldsymbol{K}_{\lambda}(s^{\prime},s){\mathbold
f}(s^{\prime})\
\mathrm{d}s^{\prime}=\lambda^{2}\int_{\mathbb{R}}\dfrac{1}{\lambda^{3}}\boldsymbol{K}_{1,\lambda}(s^{\prime}/\lambda,s/\lambda)\left(\lambda^{-1/2}(S_{\lambda}{\mathbold
f})(s^{\prime}/\lambda)\right)\ \mathrm{d}s^{\prime}\\\
&=\lambda^{-3/2}\int_{\mathbb{R}}\boldsymbol{K}_{1,\lambda}(s^{\prime},s/\lambda)(S_{\lambda}{\mathbold
f})(s^{\prime})\lambda\
\mathrm{d}s^{\prime}=\lambda^{-1/2}(T_{1,\lambda}(S_{\lambda}{\mathbold
f}))(s/\lambda)\\\ &=(S_{\lambda}^{-1}T_{1,\lambda}S_{\lambda}{\mathbold
f})(s),\end{split}$

where we used a change of variables and (52). It follows from the above
equation that

(54) $\begin{split}\left\lVert
T_{\lambda}\right\rVert_{\mathcal{L}(L^{2},L^{2})}&=\sup_{\left\lVert{\mathbold
f}\right\rVert\neq 0}\dfrac{\left\lVert T_{\lambda}{\mathbold
f}\right\rVert_{L^{2}(\mathbb{R},\mathbb{R}^{3})}}{\left\lVert{\mathbold
f}\right\rVert_{L^{2}(\mathbb{R},\mathbb{R}^{3})}}=\sup_{\left\lVert{\mathbold
f}\right\rVert\neq 0}\dfrac{\left\lVert
S_{\lambda}^{-1}(T_{1,\lambda}S_{\lambda}{\mathbold
f})\right\rVert_{L^{2}(\mathbb{R},\mathbb{R}^{3})}}{\left\lVert{\mathbold
f}\right\rVert_{L^{2}(\mathbb{R},\mathbb{R}^{3})}}\\\
&=\sup_{\left\lVert{\mathbold f}\right\rVert\neq 0}\dfrac{\left\lVert
T_{1,\lambda}(S_{\lambda}{\mathbold
f})\right\rVert_{L^{2}(\mathbb{R},\mathbb{R}^{3})}}{\left\lVert{\mathbold
f}\right\rVert_{L^{2}(\mathbb{R},\mathbb{R}^{3})}}=\sup_{\left\lVert
S_{\lambda}{\mathbold f}\right\rVert\neq 0}\dfrac{\left\lVert
T_{1,\lambda}(S_{\lambda}{\mathbold
f})\right\rVert_{L^{2}(\mathbb{R},\mathbb{R}^{3})}}{\left\lVert
S_{\lambda}{\mathbold f}\right\rVert_{L^{2}(\mathbb{R},\mathbb{R}^{3})}}\\\
&=\left\lVert
T_{1,\lambda}\right\rVert_{\mathcal{L}(L^{2},L^{2})},\end{split}$

where we have used that $\displaystyle||{\mathbold
f}||_{L^{2}(\mathbb{R},\mathbb{R}^{3})}=||S_{\lambda}{\mathbold
f}||_{L^{2}(\mathbb{R},\mathbb{R}^{3})}$. This completes the proof of (20) in
Proposition 3.4. Next, we show that $\displaystyle T_{1,\lambda}$ is a bounded
map to prove the boundedness of $\displaystyle T_{\lambda}$. We first analyze
the discrete dipole field kernel $\displaystyle\boldsymbol{K}_{1,\lambda}$,
which is defined as

(55)
$\boldsymbol{K}_{1,\lambda}(s^{\prime},s)=\sum_{\begin{subarray}{c}u,v\in{\cal
L}_{1},\\\ u\neq
v\end{subarray}}\chi_{U_{1}(v)}(s^{\prime})\boldsymbol{K}(\bar{\boldsymbol{x}}(v)-\bar{\boldsymbol{x}}(u)+\boldsymbol{A}_{\lambda}(v,u)\boldsymbol{e}_{1})\chi_{U_{1}(u)}(s),$

where $\displaystyle U_{\lambda}(s)=[s,s+\lambda)$ for $\displaystyle
s\in{\cal L}_{\lambda}$, and $\displaystyle\boldsymbol{A}_{\lambda}(a,b)$ is
given by (17).

Consider some typical $\displaystyle s,s^{\prime}\in\mathbb{R}$ and the
corresponding $\displaystyle a,b\in{\cal L}_{1}$ such that $\displaystyle
s\in[a,a+1),s^{\prime}\in[b,b+1)$. From (55), we have, for all $\displaystyle
s,s^{\prime}\in\mathbb{R}$ such that $\displaystyle|s-s^{\prime}|<1$,

  * •

If $\displaystyle a=b$, then
$\displaystyle\boldsymbol{K}_{1,\lambda}(s,s^{\prime})=\mathbf{0}$.

  * •

If $\displaystyle a\neq b$, then from (47), we have

$\displaystyle|\boldsymbol{K}_{1,\lambda}(s,s^{\prime})|\leq\sqrt{6}/(4\pi\delta^{3})$

using
$\displaystyle|\boldsymbol{A}\boldsymbol{a}|\leq|\boldsymbol{A}|\,|\boldsymbol{a}|$
and
$\displaystyle|\boldsymbol{I}-3(\boldsymbol{x}/|\boldsymbol{x}|)\mathbf{\otimes}(\boldsymbol{x}/|\boldsymbol{x}|)|\leq\sqrt{6}$,
$\displaystyle\forall\boldsymbol{x}\neq\mathbf{0}$ .

Combining the two cases above,
$\displaystyle|\boldsymbol{K}_{1,\lambda}(s,s^{\prime})|\leq\sqrt{6}/(4\pi\delta^{3})$.

We now consider the case when $\displaystyle|s-s^{\prime}|\geq 1$. Noting that
for this case, $\displaystyle a\neq b$. We proceed as follows

(56)
$\begin{split}|\boldsymbol{K}_{1,\lambda}(s,s^{\prime})|&\leq\dfrac{\sqrt{6}}{4\pi|s-s^{\prime}|^{3}}\dfrac{|s-s^{\prime}|^{3}}{|a-b|^{3}}\dfrac{|a-b|^{3}}{|\bar{\boldsymbol{x}}(a)-\bar{\boldsymbol{x}}(b)+\boldsymbol{A}_{\lambda}(a,b)\boldsymbol{e}_{1}|^{3}}\\\
&\leq\dfrac{\sqrt{6}}{4\pi|s-s^{\prime}|^{3}}3^{3}\dfrac{|a-b|^{3}}{\delta^{3}|a-b|^{3}}=\frac{3^{3}\sqrt{6}}{4\pi\delta^{3}}\frac{1}{|s-s^{\prime}|^{3}},\end{split}$

where we used the bounds (48) and (49). Combining the above bound for
$\displaystyle|s-s^{\prime}|\geq 1$ with the bound for
$\displaystyle|s-s^{\prime}|<1$, and renaming the constants, we can write

(57)
$|\boldsymbol{K}_{1,\lambda}(s,s^{\prime})|\leq\dfrac{C_{1}}{C_{2}+|s-s^{\prime}|^{3}}.$

Next, note that, since the kernel $\displaystyle\boldsymbol{K}_{1,\lambda}$
satisfies (57), we have

(58) $\int_{\mathbb{R}}|\boldsymbol{K}_{1,\lambda}(s^{\prime},s)|\
\mathrm{d}s^{\prime}\leq
C_{3},\quad\int_{\mathbb{R}}|\boldsymbol{K}_{1,\lambda}(s^{\prime},s)|\
\mathrm{d}s\leq C_{3},$

for some fixed $\displaystyle C_{3}<\infty$ independent of
$\displaystyle\lambda$. Using the above bound, we can show that, for all
$\displaystyle{\mathbold f}\in C^{\infty}_{0}(\mathbb{R},\mathbb{R}^{3})$,

(59) $||T_{1,\lambda}{\mathbold f}||_{L^{2}(\mathbb{R},\mathbb{R}^{3})}\leq
C_{3}||{\mathbold f}||_{L^{2}(\mathbb{R},\mathbb{R}^{3})},$

which establishes that $\displaystyle T_{1,\lambda}$ is a bounded linear map
on $\displaystyle C^{\infty}_{0}(\mathbb{R},\mathbb{R}^{3})$. Since
$\displaystyle C^{\infty}_{0}(\mathbb{R},\mathbb{R}^{3})$ is dense in
$\displaystyle L^{2}(\mathbb{R},\mathbb{R}^{3})$, it follows that
$\displaystyle T_{1,\lambda}$ is also bounded in $\displaystyle
L^{2}(\mathbb{R},\mathbb{R}^{3})$, and extends as a bounded linear map from
$\displaystyle C^{\infty}_{0}(\mathbb{R},\mathbb{R}^{3})$ to $\displaystyle
L^{2}(\mathbb{R},\mathbb{R}^{3})$. This argument together with (54) completes
the proof of boundedness of maps $\displaystyle T_{\lambda}$ and
$\displaystyle T_{1,\lambda}$. Now, it remains to show (59) for
$\displaystyle{\mathbold f}\in C^{\infty}_{0}(\mathbb{R},\mathbb{R}^{3})$. Let
$\displaystyle{\mathbold f}\in C^{\infty}_{0}(\mathbb{R},\mathbb{R}^{3})$ and
proceed as follows:

(60) $\begin{split}||T_{1,\lambda}{\mathbold
f}||_{L^{2}(\mathbb{R},\mathbb{R}^{3})}^{2}&=\int_{\mathbb{R}}\left|(T_{1,\lambda}{\mathbold
f})(s)\right|^{2}\
\mathrm{d}s=\int_{\mathbb{R}}\left(\int_{\mathbb{R}}\boldsymbol{K}_{1,\lambda}(s^{\prime},s){\mathbold
f}(s^{\prime})\ \mathrm{d}s^{\prime}\right)^{2}\ \mathrm{d}s\\\
&=\int_{\mathbb{R}}\left[\int_{\mathbb{R}}\int_{\mathbb{R}}|\boldsymbol{K}_{1,\lambda}(s^{\prime},s)|\,|\boldsymbol{K}_{1,\lambda}(t^{\prime},s)|\,|{\mathbold
f}(s^{\prime})|\,|{\mathbold f}(t^{\prime})|\,\ \mathrm{d}s^{\prime}\
\mathrm{d}t^{\prime}\right]\ \mathrm{d}s\\\
&\leq\int_{\mathbb{R}}\left[\int_{\mathbb{R}}\int_{\mathbb{R}}|\boldsymbol{K}_{1,\lambda}(s^{\prime},s)|\,|\boldsymbol{K}_{1,\lambda}(t^{\prime},s)|\,\left(\frac{|{\mathbold
f}(s^{\prime})|^{2}}{2}+\frac{|{\mathbold f}(t^{\prime})|^{2}}{2}\right)\
\mathrm{d}s^{\prime}\ \mathrm{d}t^{\prime}\right]\ \mathrm{d}s\\\
&=\int_{\mathbb{R}}\left[\frac{1}{2}2\left(\int_{\mathbb{R}}|\boldsymbol{K}_{1,\lambda}(s^{\prime},s)|\
\mathrm{d}s^{\prime}\right)\left(\int_{\mathbb{R}}|\boldsymbol{K}_{1,\lambda}(t^{\prime},s)|\,|{\mathbold
f}(t^{\prime})|^{2}\ \mathrm{d}t^{\prime}\right)\right]\ \mathrm{d}s\\\ &\leq
C_{3}\underbrace{\int_{\mathbb{R}}\left[\int_{\mathbb{R}}|\boldsymbol{K}_{1,\lambda}(t^{\prime},s)|\,|{\mathbold
f}(t^{\prime})|^{2}\ \mathrm{d}t^{\prime}\right]\
\mathrm{d}s}_{=:I},\end{split}$

where, in the third line we have used that $\displaystyle|{\mathbold
f}(s^{\prime})|\,|{\mathbold f}(t^{\prime})|\leq|{\mathbold
f}(s^{\prime})|^{2}/2+|{\mathbold f}(t^{\prime})|^{2}/2$; in the fourth line,
we have used symmetry to extract a factor of $\displaystyle 2$; in the last
line, we have used the bound (58). Since $\displaystyle{\mathbold f}\in
C^{\infty}_{0}(\mathbb{R},\mathbb{R}^{3})$, there exist $\displaystyle R>0$
such that the support of $\displaystyle{\mathbold f}$ is a subset of
$\displaystyle(-R,R)$. So, for any $\displaystyle\rho>0$, we have

(61)
$\begin{split}I_{\rho}&:=\int_{-\rho}^{\rho}\left[\int_{\mathbb{R}}|\boldsymbol{K}_{1,\lambda}(t^{\prime},s)|\,|{\mathbold
f}(t^{\prime})|^{2}\ \mathrm{d}t^{\prime}\right]\
\mathrm{d}s=\int_{-\rho}^{\rho}\left[\int_{-R}^{R}|\boldsymbol{K}_{1,\lambda}(t^{\prime},s)|\,|{\mathbold
f}(t^{\prime})|^{2}\ \mathrm{d}t^{\prime}\right]\ \mathrm{d}s\\\
&=\int_{-R}^{R}|{\mathbold
f}(t^{\prime})|^{2}\left(\int_{-\rho}^{\rho}|\boldsymbol{K}_{1,\lambda}(t^{\prime},s)|\
\mathrm{d}s\right)\ \mathrm{d}t^{\prime}\leq C_{3}||{\mathbold
f}||_{L^{2}(\mathbb{R},\mathbb{R}^{3})}^{2},\end{split}$

where we have applied Fubini’s theorem to switch the order of integration –
this is allowed because
$\displaystyle|\boldsymbol{K}_{1,\lambda}(t^{\prime},s)|\,|{\mathbold
f}(t^{\prime})|^{2}$ is integrable in
$\displaystyle(s,t^{\prime})\in(-\rho,\rho)\times(-R,R)$ – and then used (58).
Thus, we have shown that $\displaystyle I_{\rho}$ is bounded, with the bound
independent of $\displaystyle\rho$. This, together with the fact that
$\displaystyle I_{\rho}$ is monotonically increasing with $\displaystyle\rho$,
shows that $\displaystyle\lim_{\rho\to\infty}I_{\rho}$ exists and is equal to
$\displaystyle I$, where $\displaystyle I$ is defined in (60). The limit
$\displaystyle I$ is also bounded, i.e., $\displaystyle I\leq
C_{3}||{\mathbold f}||_{L^{2}(\mathbb{R},\mathbb{R}^{3})}^{2}$. Combining this
observation with (60) completes the proof of (59).

####  4.1.2 Limit of the Map $\displaystyle T_{\lambda}$

Let $\displaystyle{\mathbold f}\in C^{\infty}_{0}(\mathbb{R},\mathbb{R}^{3})$.
Write $\displaystyle T_{\lambda}{\mathbold f}$ as:

(62) $T_{\lambda}({\mathbold
f})(s)=\lambda^{2}\int_{\mathbb{R}}\boldsymbol{K}_{\lambda}(s^{\prime},s){\mathbold
f}(s^{\prime})\
\mathrm{d}s^{\prime}=\underbrace{\left[\lambda^{2}\int_{\mathbb{R}}\boldsymbol{K}_{\lambda}(s^{\prime},s)\
\mathrm{d}s^{\prime}\right]}_{=:\boldsymbol{H}_{\lambda}(s)}{\mathbold
f}(s)+\lambda^{2}\int_{\mathbb{R}}\boldsymbol{K}_{\lambda}(s^{\prime},s)({\mathbold
f}(s^{\prime})-{\mathbold f}(s))\ \mathrm{d}s^{\prime}.$

The second term above is zero in the limit $\displaystyle\lambda\to 0$. To see
this, we first obtain two useful inequalities. For any $\displaystyle R>0$, we
have, using the bound on $\displaystyle|\boldsymbol{K}_{1,\lambda}|$ from
(57),

(63) $\begin{split}\int_{\left\lvert s-s^{\prime}\right\rvert\leq
R\lambda}|\boldsymbol{K}_{1,\lambda}(s^{\prime}/\lambda,s/\lambda)|\
\mathrm{d}s^{\prime}\leq\int_{\left\lvert s-s^{\prime}\right\rvert\leq
R\lambda}\dfrac{C_{1}}{C_{2}+|s-s^{\prime}|^{3}/\lambda^{3}}\
\mathrm{d}s^{\prime}\leq\frac{C_{1}}{C_{2}}\int_{\left\lvert
s-s^{\prime}\right\rvert\leq R\lambda}\
\mathrm{d}s^{\prime}=\frac{C_{1}}{C_{2}}R\lambda\end{split}$

and

(64) $\begin{split}&\dfrac{1}{\lambda}\int_{\left\lvert
s-s^{\prime}\right\rvert\geq
R\lambda}|\boldsymbol{K}_{1,\lambda}(s^{\prime}/\lambda,s/\lambda)|\
\mathrm{d}s^{\prime}\leq\dfrac{1}{\lambda}\int_{\left\lvert
s-s^{\prime}\right\rvert\geq R\lambda}\dfrac{C_{1}}{C_{2}+\left\lvert
s/\lambda-s^{\prime}/\lambda\right\rvert^{3}}\
\mathrm{d}s^{\prime}=\int_{\left\lvert t\right\rvert\geq
R}\dfrac{C_{1}}{C_{2}+\left\lvert t\right\rvert^{3}}\ \mathrm{d}t,\end{split}$

where the last equality in the equation above follows from the change of
variables $\displaystyle t=(s^{\prime}-s)/\lambda$. Next, noting that
$\displaystyle\lambda^{2}\boldsymbol{K}_{\lambda}(s^{\prime},s)=\frac{1}{\lambda}\boldsymbol{K}_{1,\lambda}(s^{\prime}/\lambda,s/\lambda)$,
we obtain the following bound on the second term of (62):

(65)
$\begin{split}&\left|\lambda^{2}\int_{\mathbb{R}}\boldsymbol{K}_{\lambda}(s^{\prime},s)({\mathbold
f}(s^{\prime})-{\mathbold f}(s))\ \mathrm{d}s^{\prime}\right|\\\
&=\left|\dfrac{1}{\lambda}\int_{\mathbb{R}}\boldsymbol{K}_{1,\lambda}(s^{\prime}/\lambda,s/\lambda)({\mathbold
f}(s^{\prime})-{\mathbold f}(s))\ \mathrm{d}s^{\prime}\right|\\\
&=\left|\dfrac{1}{\lambda}\int_{\left\lvert s-s^{\prime}\right\rvert\geq
R\lambda}\boldsymbol{K}_{1,\lambda}(s^{\prime}/\lambda,s/\lambda){\mathbold
f}(s^{\prime})\ \mathrm{d}s^{\prime}-\dfrac{1}{\lambda}\int_{\left\lvert
s-s^{\prime}\right\rvert\geq
R\lambda}\boldsymbol{K}_{1,\lambda}(s^{\prime}/\lambda,s/\lambda)\
\mathrm{d}s^{\prime}{\mathbold f}(s)\right.\\\
&\quad\left.+\dfrac{1}{\lambda}\int_{\left\lvert s-s^{\prime}\right\rvert\leq
R\lambda}\boldsymbol{K}_{1,\lambda}(s^{\prime}/\lambda,s/\lambda)({\mathbold
f}(s^{\prime})-{\mathbold f}(s))\ \mathrm{d}s^{\prime}\right|\\\
&\leq\left|\dfrac{1}{\lambda}\int_{\left\lvert s-s^{\prime}\right\rvert\geq
R\lambda}\boldsymbol{K}_{1,\lambda}(s^{\prime}/\lambda,s/\lambda){\mathbold
f}(s^{\prime})\
\mathrm{d}s^{\prime}\right|+\left|\dfrac{1}{\lambda}\int_{\left\lvert
s-s^{\prime}\right\rvert\geq
R\lambda}\boldsymbol{K}_{1,\lambda}(s^{\prime}/\lambda,s/\lambda)\
\mathrm{d}s^{\prime}{\mathbold f}(s)\right|\\\
&\quad+\left|\dfrac{1}{\lambda}\int_{\left\lvert s-s^{\prime}\right\rvert\leq
R\lambda}\boldsymbol{K}_{1,\lambda}(s^{\prime}/\lambda,s/\lambda)({\mathbold
f}(s^{\prime})-{\mathbold f}(s))\ \mathrm{d}s^{\prime}\right|\\\
&\leq\sup_{t}|{\mathbold
f}^{\prime}(t)|\frac{R\lambda}{\lambda}\int_{\left\lvert
s-s^{\prime}\right\rvert\leq
R\lambda}|\boldsymbol{K}_{1,\lambda}(s^{\prime}/\lambda,s/\lambda)|\
\mathrm{d}s^{\prime}+2\sup_{t}|{\mathbold
f}(t)|\frac{1}{\lambda}\int_{\left\lvert s-s^{\prime}\right\rvert\geq
R\lambda}|\boldsymbol{K}_{1,\lambda}(s^{\prime}/\lambda,s/\lambda)|\
\mathrm{d}s^{\prime}\\\ &\leq\sup_{t}|{\mathbold
f}^{\prime}(t)|\frac{C_{1}R^{2}}{C_{2}}\lambda+2\sup_{t}|{\mathbold
f}(t)|\int_{\left\lvert t\right\rvert\geq R}\dfrac{C_{1}}{C_{2}+\left\lvert
t\right\rvert^{3}}\ \mathrm{d}t,\end{split}$

where we have used the fact that $\displaystyle{\mathbold f}\in
C^{\infty}_{0}(\mathbb{R},\mathbb{R}^{3})$, and therefore,
$\displaystyle|{\mathbold f}(s^{\prime})|\leq\sup_{t}|{\mathbold f}(t)|$, and
$\displaystyle|{\mathbold f}(s^{\prime})-{\mathbold f}(s)|\leq
R\lambda\sup_{t}|{\mathbold f}^{\prime}(t)|$ for $\displaystyle
s^{\prime}\in\\{t:|s-t|\leq R\lambda\\}$. We have also used the inequalities
(63) and (64) in the last step.

We note that the final inequality in (65) is true for any $\displaystyle R>0$.
Further, the two terms on the right side have a limit as
$\displaystyle\lambda\to 0$ – the second term clearly is independent of
$\displaystyle\lambda$ – for any $\displaystyle R>0$. Therefore, the limit
$\displaystyle\lambda\to 0$ of the two terms individually is equal to the
limit of the sum, keeping $\displaystyle R$ fixed. Thus, taking the limit
$\displaystyle\lambda\to 0$ in (65), we have

(66) $\begin{split}&\lim_{\lambda\to
0}\,\left|\lambda^{2}\int_{\mathbb{R}}\boldsymbol{K}_{\lambda}(s^{\prime},s)({\mathbold
f}(s^{\prime})-{\mathbold f}(s))\ \mathrm{d}s^{\prime}\right|\\\
&\leq\lim_{\lambda\to 0}\left[\sup_{t}|{\mathbold
f}^{\prime}(t)|\frac{C_{1}}{C_{2}}R\lambda\right]+\lim_{\lambda\to
0}\left[2\sup_{t}|{\mathbold f}(t)|\int_{\left\lvert t\right\rvert\geq
R}\dfrac{C_{1}}{C_{2}+\left\lvert t\right\rvert^{3}}\ \mathrm{d}t\right]\\\
&=2\sup_{t}|{\mathbold f}(t)|\int_{\left\lvert t\right\rvert\geq
R}\dfrac{C_{1}}{C_{2}+\left\lvert t\right\rvert^{3}}\ \mathrm{d}t,\end{split}$

for any $\displaystyle R>0$. Since the bound above is true for any
$\displaystyle R>0$, and the left side is independent of $\displaystyle R$, we
can take the limit $\displaystyle R\to\infty$, where the limit exists and is
$\displaystyle 0$ for the right side, to get

(67) $\begin{split}\lim_{\lambda\to
0}\,\left|\lambda^{2}\int_{\mathbb{R}}\boldsymbol{K}_{\lambda}(s^{\prime},s)({\mathbold
f}(s^{\prime})-{\mathbold f}(s))\
\mathrm{d}s^{\prime}\right|\leq\lim_{R\to\infty}\,2\sup_{t}|{\mathbold
f}(t)|\int_{\left\lvert t\right\rvert\geq R}\dfrac{C_{1}}{C_{2}+\left\lvert
t\right\rvert^{3}}\ \mathrm{d}t=0.\end{split}$

Thus, we have from (62) that

(68) $\lim_{\lambda\to 0}T_{\lambda}({\mathbold f})(s)=\left[\lim_{\lambda\to
0}\boldsymbol{H}_{\lambda}(s)\right]{\mathbold f}(s).$

We next compute the limit of $\displaystyle\boldsymbol{H}_{\lambda}(s)$. Fix
$\displaystyle s\in\mathbb{R}$ and suppose $\displaystyle a\in{\cal
L}_{\lambda}$ such that $\displaystyle s\in U_{\lambda}(a)$. Using the
definition of $\displaystyle\boldsymbol{K}_{\lambda}(s^{\prime},s)$, we have

(69)
$\boldsymbol{H}_{\lambda}(s)=\lambda^{2}\int_{\mathbb{R}}\boldsymbol{K}(s^{\prime},s)\
\mathrm{d}s^{\prime}=\lambda^{2}\sum_{\begin{subarray}{c}u\in{\cal L}_{1},\\\
u\neq
a\end{subarray}}\boldsymbol{K}(\bar{\boldsymbol{x}}(u)-\bar{\boldsymbol{x}}(a))\int_{U_{\lambda}(u)}\
\mathrm{d}t=\lambda^{3}\sum_{\begin{subarray}{c}u\in\lambda\mathbb{Z},\\\
u\neq
a\end{subarray}}\boldsymbol{K}(\bar{\boldsymbol{x}}(u)-\bar{\boldsymbol{x}}(a)).$

From (50), we have
$\displaystyle\bar{\boldsymbol{x}}(u)-\bar{\boldsymbol{x}}(a)=\boldsymbol{Q}(a\theta)((\boldsymbol{Q}((u-a)\theta)-\boldsymbol{I})\boldsymbol{e}_{1}+(u-a)\delta\boldsymbol{e}_{3})$.
Using the identity
$\displaystyle\boldsymbol{K}(\boldsymbol{Q}\boldsymbol{x})=\boldsymbol{Q}\boldsymbol{K}(\boldsymbol{x})\boldsymbol{Q}^{T}$
and
$\displaystyle\boldsymbol{K}(\lambda\boldsymbol{x})=\boldsymbol{K}(\boldsymbol{x})/\lambda^{3}$,
we get

(70)
$\begin{split}\boldsymbol{H}_{\lambda}(s)&=\boldsymbol{Q}(a\theta)\left[\sum_{u\in\lambda\mathbb{Z},u\neq
a}\boldsymbol{K}((\boldsymbol{Q}((u-a)\theta)-\boldsymbol{I})/\lambda\boldsymbol{e}_{1}+(u-a)\delta/\lambda\boldsymbol{e}_{3})\right]\boldsymbol{Q}(-a\theta)\\\
&=\boldsymbol{Q}(a\theta)\left[\sum_{i\in\mathbb{Z},i\neq
0}\boldsymbol{K}((\boldsymbol{Q}(i\lambda\theta)-\boldsymbol{I})/\lambda\boldsymbol{e}_{1}+i\delta\boldsymbol{e}_{3})\right]\boldsymbol{Q}(-a\theta),\end{split}$

where we have changed variables $\displaystyle i=(u-a)/\lambda$. Note that
$\displaystyle a\in\lambda\mathbb{Z}$, and, therefore,
$\displaystyle(u-a)\in\lambda\mathbb{Z}$ for $\displaystyle
u\in\lambda\mathbb{Z}$, which implies $\displaystyle i\in\mathbb{Z}$. Since
$\displaystyle s$ is related to $\displaystyle a$ by $\displaystyle s\in
U_{\lambda}(a)$, we have $\displaystyle a\to s$ in the limit
$\displaystyle\lambda\to 0$. Therefore, we get

$\boldsymbol{H}_{0}(s):=\lim_{\lambda\to
0}\boldsymbol{H}_{\lambda}(s)=\boldsymbol{Q}(s\theta)\left[\lim_{\lambda\to
0}\sum_{i\in\mathbb{Z}-\\{0\\}}\boldsymbol{K}((\boldsymbol{Q}(i\lambda\theta)-\boldsymbol{I})/\lambda\boldsymbol{e}_{1}+i\delta\boldsymbol{e}_{3})\right]\boldsymbol{Q}(-s\theta).$

To take the limit inside the summation, we show that the sum is absolutely
convergent for all $\displaystyle\lambda>0$ as follows:

(71)
$\begin{split}a_{\lambda}&:=\sum_{i\in\mathbb{Z}-\\{0\\}}\left|\boldsymbol{K}\left((\boldsymbol{Q}(i\lambda\theta)-\boldsymbol{I})/\lambda\boldsymbol{e}_{1}+i\delta\boldsymbol{e}_{3}\right)\right|\leq\sum_{i\in\mathbb{Z}-\\{0\\}}\dfrac{c}{\left|(\boldsymbol{Q}(i\lambda\theta)-\boldsymbol{I})/\lambda\boldsymbol{e}_{1}+i\delta\boldsymbol{e}_{3})\right|^{3}}\\\
&=\sum_{i\in\mathbb{Z}-\\{0\\}}\dfrac{c}{(4\sin^{2}(i\lambda\theta/2)/\lambda^{2}+i^{2}\delta^{2})^{3/2}}\leq\sum_{i\in\mathbb{Z}-\\{0\\}}\dfrac{c}{|i|^{3}}<\infty,\qquad\forall\lambda>0.\end{split}$

Now, we can write

$\boldsymbol{H}_{0}(s)=\boldsymbol{Q}(s\theta)\left[\sum_{i\in\mathbb{Z}-\\{0\\}}\lim_{\lambda\to
0}\boldsymbol{K}\left(\dfrac{\boldsymbol{Q}(i\lambda\theta)-\boldsymbol{I}}{i\lambda\theta}(i\theta\boldsymbol{e}_{1})+i\delta\boldsymbol{e}_{3}\right)\right]\boldsymbol{Q}(-s\theta).$

Note that for a fixed $\displaystyle i\in\mathbb{Z}$

$\lim_{\lambda\to
0}\dfrac{(\boldsymbol{Q}(i\lambda\theta)-\boldsymbol{I})}{i\lambda\theta}i\theta\boldsymbol{e}_{1}+i\delta\boldsymbol{e}_{3}=\lim_{h=i\lambda\theta\to
0}\dfrac{(\boldsymbol{Q}(h)-\boldsymbol{I})}{h}i\theta\boldsymbol{e}_{1}+i\delta\boldsymbol{e}_{3}=i\theta\boldsymbol{Q}^{\prime}(0)\boldsymbol{e}_{1}+i\delta\boldsymbol{e}_{3},$

where
$\displaystyle\boldsymbol{Q}^{\prime}(0)=\mathrm{d}/\mathrm{d}x\boldsymbol{Q}(x)|_{x=0}$.
Now, using the equation above, and the fact that
$\displaystyle\boldsymbol{K}(\boldsymbol{x})$ is smooth away from
$\displaystyle\boldsymbol{x}=\mathbf{0}$ (which is ensured in the summation),
we get

$\boldsymbol{H}_{0}(s)=\boldsymbol{Q}(s\theta)\left[\sum_{i\in\mathbb{Z}-\\{0\\}}\boldsymbol{K}\left(i\theta\boldsymbol{Q}^{\prime}(0)\boldsymbol{e}_{1}+i\delta\boldsymbol{e}_{3}\right)\right]\boldsymbol{Q}(-s\theta)=\boldsymbol{Q}(s\theta)\boldsymbol{H}_{0}(0)\boldsymbol{Q}(-s\theta).$

Combining this with (68), we get

$\lim_{\lambda\to 0}T_{\lambda}({\mathbold
f})(s)=\boldsymbol{H}_{0}(s){\mathbold
f}(s)=\boldsymbol{Q}(s\theta)\boldsymbol{H}_{0}(0)\boldsymbol{Q}(-s\theta){\mathbold
f}(s).$

Next, we simplify $\displaystyle\boldsymbol{H}_{0}(s)$. Using
$\displaystyle\boldsymbol{Q}\boldsymbol{K}(\boldsymbol{x})\boldsymbol{Q}^{T}=\boldsymbol{K}(\boldsymbol{Q}\boldsymbol{x})$
and
$\displaystyle\boldsymbol{Q}(s\theta)\boldsymbol{Q}^{\prime}(0)=\boldsymbol{Q}^{\prime}(s\theta)$,
we can show

(72)
$\boldsymbol{H}_{0}(s)=\sum_{i\in\mathbb{Z}-\\{0\\}}\boldsymbol{K}(i\theta\boldsymbol{Q}^{\prime}(s\theta)\boldsymbol{e}_{1}+i\delta\boldsymbol{e}_{3})=\sum_{i\in\mathbb{Z}-\\{0\\}}\boldsymbol{K}\left(i\;|\boldsymbol{t}(s)|\;\hat{\boldsymbol{t}}(s)\right),$

where
$\displaystyle\boldsymbol{t}(s)=\theta\boldsymbol{Q}^{\prime}(s\theta)\boldsymbol{e}_{1}+\delta\boldsymbol{e}_{3}$
is the tangent vector, and
$\displaystyle\hat{\boldsymbol{t}}(s)=\boldsymbol{t}(s)/|\boldsymbol{t}(s)|$
with $\displaystyle|\boldsymbol{t}(s)|=\sqrt{\theta^{2}+\delta^{2}}$. In (72),
by noting the definition of the dipole field kernel
$\displaystyle\boldsymbol{K}$, it is easy to show that

$\boldsymbol{H}_{0}(s)=-h_{0}\left[\boldsymbol{I}-3\hat{\boldsymbol{t}}(s)\mathbf{\otimes}\hat{\boldsymbol{t}}(s)\right]=-h_{0}\left[\boldsymbol{P}_{\perp}{\mathbold
f}(s)-2\boldsymbol{P}_{||}(s)\right],$

with $\displaystyle h_{0}$ defined as

(73)
$h_{0}=\sum_{i\in\mathbb{Z}-\\{0\\}}\frac{1}{4\pi|i|^{3}(\theta^{2}+\delta^{2})^{3/2}}$

and projection tensors
$\displaystyle\boldsymbol{P}_{||}(s)=\hat{\boldsymbol{t}}(s)\mathbf{\otimes}\hat{\boldsymbol{t}}(s)$
and
$\displaystyle\boldsymbol{P}_{\perp}(s)=\boldsymbol{I}-\boldsymbol{P}_{||}(s)$.
This finishes the proof of Proposition 3.4.

###  4.2 Nanofilm with Uniform Bending

In this section, we prove Proposition 3.10. The outline of the proof is
similar to the case of the helix in subsection 4.1.

######  Lemma 4.3.

  * 1.

Suppose $\displaystyle\boldsymbol{s},\boldsymbol{s}^{\prime}\in{\cal
S}_{1,\lambda}=(-\bar{\theta}/\lambda,\bar{\theta}/\lambda)\times\mathbb{R}$
such that $\displaystyle\boldsymbol{a},\boldsymbol{b}\in{\cal
L}_{1,\lambda}=(-\bar{\theta}/\lambda,\bar{\theta}/\lambda)\cap\theta_{l}\mathbb{Z}\times\mathbb{Z}$
with $\displaystyle\boldsymbol{s}\in
U_{1}(\boldsymbol{a})=[a_{1},a_{1}+\theta_{l})\times[a_{2},a_{2}+1),\boldsymbol{s}^{\prime}\in
U_{1}(\boldsymbol{b})$. When
$\displaystyle|\boldsymbol{s}-\boldsymbol{s}^{\prime}|\geq\min\\{\theta_{l},1\\}$,
we have $\displaystyle\boldsymbol{a}\neq\boldsymbol{b}$ and

(74)
$\frac{|\boldsymbol{s}-\boldsymbol{s}^{\prime}|}{|\boldsymbol{a}-\boldsymbol{b}|}<1+\frac{\theta_{l}+1}{\min\\{\theta_{l},1\\}}=:c_{L}.$

  * 2.

For any $\displaystyle\boldsymbol{a},\boldsymbol{b}\in{\cal L}_{1,\lambda}$,
we have

(75)
$c_{A}|\boldsymbol{a}-\boldsymbol{b}|\leq|\bar{\boldsymbol{x}}(\boldsymbol{a})-\bar{\boldsymbol{x}}(\boldsymbol{b})+\boldsymbol{A}_{\lambda}(\boldsymbol{a},\boldsymbol{b})\boldsymbol{e}_{1}|,$

where $\displaystyle\bar{\boldsymbol{x}}$ is given by (24) and
$\displaystyle\boldsymbol{A}_{\lambda}(\boldsymbol{a},\boldsymbol{b})$ is
defined as

$\boldsymbol{A}_{\lambda}(\boldsymbol{a},\boldsymbol{b})=\frac{{\cal
R}}{\lambda}\left[\boldsymbol{Q}(\lambda a_{1})-\boldsymbol{Q}(\lambda
b_{1})-\lambda\boldsymbol{Q}(a_{1})+\lambda\boldsymbol{Q}(b_{1})\right].$

Here $\displaystyle c_{A}=\min\\{\delta,{\cal
R}\sqrt{1-\bar{\theta}^{2}/3}\\}$ is the constant independent of
$\displaystyle\lambda$; recall that $\displaystyle\delta$ is the parameter in
the map $\displaystyle\bar{\boldsymbol{x}}$, see (24). Note that
$\displaystyle c_{A}>0$ for $\displaystyle 0<\bar{\theta}<\pi/2$.

######  Proof 4.4.

To show (74), we proceed as follows. For
$\displaystyle\boldsymbol{s},\boldsymbol{s}^{\prime}\in{\cal S}_{1,\lambda}$
and corresponding $\displaystyle\boldsymbol{a},\boldsymbol{b}\in{\cal
L}_{1,\lambda}$, there exists
$\displaystyle\Delta\boldsymbol{s},\Delta\boldsymbol{s}^{\prime}$ such that
$\displaystyle\boldsymbol{s}=\boldsymbol{a}+\Delta\boldsymbol{s},\boldsymbol{s}^{\prime}=\boldsymbol{b}+\Delta\boldsymbol{b}$
with $\displaystyle 0\leq\Delta s_{1},\Delta s^{\prime}_{1}<\theta_{l}$,
$\displaystyle 0\leq\Delta s_{2},\Delta s^{\prime}_{2}<1$. We have the bound

(76)
$\frac{|\boldsymbol{s}-\boldsymbol{s}^{\prime}|}{|\boldsymbol{a}-\boldsymbol{b}|}\leq
1+\frac{|\Delta s_{1}-\Delta s^{\prime}_{1}|+|\Delta s_{2}-\Delta
s^{\prime}_{2}|}{|\boldsymbol{a}-\boldsymbol{b}|}<1+\frac{\theta_{l}+1}{|\boldsymbol{a}-\boldsymbol{b}|}\leq
1+\frac{\theta_{l}+1}{\min\\{\theta_{l},1\\}},$

where in the last step we used the fact that any
$\displaystyle\boldsymbol{a},\boldsymbol{b}\in{\cal L}_{1,\lambda}$,
satisfying $\displaystyle\boldsymbol{a}\neq\boldsymbol{b}$, are at least
$\displaystyle\min\\{\theta_{l},1\\}$ distance apart.

We next show (75). Using

$\bar{\boldsymbol{x}}(\boldsymbol{a})-\bar{\boldsymbol{x}}(\boldsymbol{b})+\boldsymbol{A}_{\lambda}(\boldsymbol{a},\boldsymbol{b})\boldsymbol{e}_{1}=\frac{{\cal
R}}{\lambda}(\boldsymbol{Q}(\lambda a_{1})-\boldsymbol{Q}(\lambda
b_{1}))\boldsymbol{e}_{1}+\delta(a_{2}-b_{2})\boldsymbol{e}_{3}$

and

$|(\boldsymbol{Q}(\theta_{1})-\boldsymbol{Q}(\theta_{2}))\boldsymbol{e}_{1}|^{2}=(\cos\theta_{1}-\cos\theta_{2})^{2}+(\sin\theta_{1}-\sin\theta_{2})^{2}=2(1-\cos(\theta_{1}-\theta_{2})),$

we have that

(77)
$|\bar{\boldsymbol{x}}(\boldsymbol{a})-\bar{\boldsymbol{x}}(\boldsymbol{b})+\boldsymbol{A}_{\lambda}(\boldsymbol{a},\boldsymbol{b})\boldsymbol{e}_{1}|^{2}=\delta^{2}|a_{2}-b_{2}|^{2}+\frac{2{\cal
R}^{2}}{\lambda^{2}}(1-\cos(\lambda a_{1}-\lambda b_{1})).$

Let $\displaystyle r=a_{1}-b_{1}$. Then, using a Taylor expansion and the mean
value theorem, there exists $\displaystyle\xi$ such that

$1-\cos(\lambda
r)=\frac{1}{2}\lambda^{2}r^{2}-\frac{1}{24}\lambda^{4}r^{4}\cos(\xi).$

Since $\displaystyle-1\leq\cos(\xi)\leq 1$, it follows

$1-\cos(\lambda
r)\geq\frac{1}{2}\lambda^{2}r^{2}-\frac{1}{24}\lambda^{4}r^{4}.$

Substituting the relation above in (77), we get

$|\bar{\boldsymbol{x}}(\boldsymbol{a})-\bar{\boldsymbol{x}}(\boldsymbol{b})+\boldsymbol{A}_{\lambda}(\boldsymbol{a},\boldsymbol{b})\boldsymbol{e}_{1}|^{2}\geq\delta^{2}|a_{2}-b_{2}|^{2}+{\cal
R}^{2}r^{2}\left(1-\frac{1}{12}\lambda^{2}r^{2}\right).$

Since $\displaystyle\boldsymbol{a},\boldsymbol{b}\in{\cal L}_{1,\lambda}$, we
have $\displaystyle-2\bar{\theta}<\lambda r<2\bar{\theta}$, and

$1-\frac{1}{12}\lambda^{2}r^{2}\geq
1-\frac{1}{12}\bar{\theta}^{2}4=1-\frac{\bar{\theta}^{2}}{3}.$

Using the two equations above and defining the constant $\displaystyle c_{A}$
as in Lemma 4.3(2), (75) can be easily shown.

####  4.2.1 Boundedness

Let $\displaystyle S_{\lambda}\colon L^{2}({\cal S},\mathbb{R}^{3})\to
L^{2}({\cal S}_{1,\lambda},\mathbb{R}^{3})$ be a map such that, for any
$\displaystyle{\mathbold f}\in L^{2}({\cal S},\mathbb{R}^{3})$,

$(S_{\lambda}{\mathbold f})(\boldsymbol{s})=\lambda{\mathbold
f}(\lambda\boldsymbol{s}),\qquad\forall\boldsymbol{s}\in{\cal S}_{1,\lambda}.$

It is easy to see that $\displaystyle S_{\lambda}$ is an isometry. The inverse
of $\displaystyle S_{\lambda}$ is given by

$(S^{-1}_{\lambda}{\mathbold f})(\boldsymbol{s})=\lambda^{-1}{\mathbold
f}(\boldsymbol{s}/\lambda),\qquad\forall\boldsymbol{s}\in{\cal S}.$

Following the similar steps in subsubsection 4.1.1, we can show that

$||T_{\lambda}||_{{\cal L}(L^{2},L^{2})}=||T_{1,\lambda}||_{{\cal
L}(L^{2},L^{2})}.$

Thus, to show that $\displaystyle T_{\lambda}$ is a bounded map, it is
sufficient to show that $\displaystyle T_{1,\lambda}$ is bounded. Towards that
goal, we first establish that

(78)
$|\boldsymbol{K}_{1,\lambda}(\boldsymbol{s},\boldsymbol{s}^{\prime})|\leq\frac{C_{1}}{C_{2}+|\boldsymbol{s}-\boldsymbol{s}^{\prime}|^{3}},\qquad\forall\boldsymbol{s}^{\prime},\boldsymbol{s}\in{\cal
S}_{1,\lambda},$

where $\displaystyle C_{1},C_{2}$ are constants that may depend on the
parameters $\displaystyle R,\theta,\delta$ defining the surface
$\displaystyle{\cal S}$, but are independent of $\displaystyle\lambda$.

To show (78), we recall that $\displaystyle\bar{\theta}$ is the fixed angular
extent of the film and satisfies the bound $\displaystyle
0<\bar{\theta}<\pi/2$ (in fact we restrict it such that $\displaystyle
0<\bar{\theta}<\pi/4$). Let
$\displaystyle\boldsymbol{s},\boldsymbol{s}^{\prime}\in{\cal S}_{1,\lambda}$
be any two generic points, and let
$\displaystyle\boldsymbol{a},\boldsymbol{b}\in{\cal L}_{1,\lambda}$ be such
that $\displaystyle\boldsymbol{s}\in U_{1}(\boldsymbol{a})$ and
$\displaystyle\boldsymbol{s}^{\prime}\in U_{1}(\boldsymbol{b})$. We refer to
subsubsection 3.2.1 and subsubsection 3.2.2 for the notation appearing in this
section.

First, consider $\displaystyle\boldsymbol{s},\boldsymbol{s}^{\prime}$ such
that
$\displaystyle|\boldsymbol{s}-\boldsymbol{s}^{\prime}|\geq\min\\{\theta_{l},1\\}$.
For this case, we have $\displaystyle\boldsymbol{a}\neq\boldsymbol{b}$. Noting
that
$\displaystyle|\boldsymbol{I}-3(\boldsymbol{x}/|\boldsymbol{x}|)\mathbf{\otimes}(\boldsymbol{x}/|\boldsymbol{x}|)|=\sqrt{6}$,
$\displaystyle\forall\boldsymbol{x}\neq\mathbf{0}$, we have

(79)
$\begin{split}|\boldsymbol{K}_{1,\lambda}(\boldsymbol{s},\boldsymbol{s}^{\prime})|&\leq\frac{\sqrt{6}}{4\pi|\bar{\boldsymbol{x}}(\boldsymbol{a})-\bar{\boldsymbol{x}}(\boldsymbol{b})+\boldsymbol{A}_{\lambda}(\boldsymbol{a},\boldsymbol{b})\boldsymbol{e}_{1}|^{3}}\\\
&=\frac{\sqrt{6}}{4\pi|\boldsymbol{s}-\boldsymbol{s}^{\prime}|^{3}}\frac{|\boldsymbol{s}-\boldsymbol{s}^{\prime}|^{3}}{|\boldsymbol{a}-\boldsymbol{b}|^{3}}\frac{|\boldsymbol{a}-\boldsymbol{b}|^{3}}{|\bar{\boldsymbol{x}}(\boldsymbol{a})-\bar{\boldsymbol{x}}(\boldsymbol{b})+\boldsymbol{A}_{\lambda}(\boldsymbol{a},\boldsymbol{b})\boldsymbol{e}_{1}|^{3}}\\\
&\leq\frac{\sqrt{6}}{4\pi|\boldsymbol{s}-\boldsymbol{s}^{\prime}|^{3}}c_{L}^{3}\frac{1}{c_{A}^{3}},\end{split}$

where we have used the bounds (74) and (75).

Next, we consider the case when
$\displaystyle|\boldsymbol{s}-\boldsymbol{s}^{\prime}|<\min\\{\theta_{l},1\\}$.
This can be further divided in two cases:

  * •

Case 1: $\displaystyle\boldsymbol{a}=\boldsymbol{b}$ which implies
$\displaystyle|\boldsymbol{K}_{1,\lambda}(\boldsymbol{s}^{\prime},\boldsymbol{s})|=0$.

  * •

Case 2: $\displaystyle\boldsymbol{a}\neq\boldsymbol{b}$. For this case, we
have

(80)
$|\boldsymbol{K}_{1,\lambda}(\boldsymbol{s},\boldsymbol{s}^{\prime})|\leq\frac{\sqrt{6}}{4\pi|\bar{\boldsymbol{x}}(\boldsymbol{a})-\bar{\boldsymbol{x}}(\boldsymbol{b})+\boldsymbol{A}_{\lambda}(\boldsymbol{a},\boldsymbol{b})\boldsymbol{e}_{1}|^{3}}\leq\frac{\sqrt{6}}{4\pi
c_{A}^{3}|\boldsymbol{a}-\boldsymbol{b}|^{3}}.$

Note that when $\displaystyle\boldsymbol{a}\neq\boldsymbol{b}$, we can have
either $\displaystyle a_{1}=b_{1},a_{2}=b_{2}\pm 1$; or $\displaystyle
a_{1}=b_{1}\pm\theta_{l},a_{2}=b_{2}$; or $\displaystyle
a_{1}=b_{1}\pm\theta_{l},a_{2}=b_{2}\pm 1$. For all of these cases, the
denominator in (80) is bounded from below because
$\displaystyle|\boldsymbol{a}-\boldsymbol{b}|\geq\min\\{\theta_{l},1\\}$.
Thus, we have

(81)
$|\boldsymbol{K}_{1,\lambda}(\boldsymbol{s},\boldsymbol{s}^{\prime})|\leq\frac{\sqrt{6}}{4\pi
c_{A}^{3}(\min\\{\theta_{l},1\\})^{3}}.$

In summary (81) holds for any
$\displaystyle\boldsymbol{s},\boldsymbol{s}^{\prime}$ such that
$\displaystyle|\boldsymbol{s}-\boldsymbol{s}^{\prime}|<\min\\{\theta_{l},1\\}$.

Combining the bound for the case
$\displaystyle|\boldsymbol{s}-\boldsymbol{s}^{\prime}|<\min\\{\theta_{l},1\\}$
with the bound for the case
$\displaystyle|\boldsymbol{s}-\boldsymbol{s}^{\prime}|\geq\min\\{\theta_{l},1\\}$,
we can write

(82)
$|\boldsymbol{K}_{1,\lambda}(\boldsymbol{s},\boldsymbol{s}^{\prime})|\leq\frac{C_{1}}{C_{2}+|\boldsymbol{s}-\boldsymbol{s}^{\prime}|^{3}},$

where we have renamed the constants for convenience. This completes the proof
of (78).

Next, we show $\displaystyle T_{1,\lambda}$ is a bounded map on $\displaystyle
L^{2}({\cal S}_{1,\lambda},\mathbb{R}^{3})$. Since
$\displaystyle\boldsymbol{K}_{1,\lambda}$ satisfies (78), it can be shown that

(83) $\frac{{\cal R}}{\theta_{l}}\int_{{\cal
S}_{1,\lambda}}|\boldsymbol{K}_{1,\lambda}(\boldsymbol{s}^{\prime},\boldsymbol{s})|\
\mathrm{d}\boldsymbol{s}^{\prime}\leq C_{3},\quad\frac{{\cal
R}}{\theta_{l}}\int_{{\cal
S}_{1,\lambda}}|\boldsymbol{K}_{1,\lambda}(\boldsymbol{s}^{\prime},\boldsymbol{s})|\
\mathrm{d}\boldsymbol{s}\leq C_{3},$

for some fixed $\displaystyle C_{3}<\infty$ independent of
$\displaystyle\lambda$. Following the steps in obtaining inequality (60), it
is easy to obtain, for $\displaystyle{\mathbold f}\in C^{\infty}_{0}({\cal
S}_{1,\lambda},\mathbb{R}^{3})$,

(84) $||T_{1,\lambda}{\mathbold f}||_{L^{2}({\cal
S}_{1,\lambda},\mathbb{R}^{3})}^{2}\leq C_{3}\underbrace{\int_{{\cal
S}_{1,\lambda}}\left[\frac{{\cal R}}{\theta_{l}}\int_{{\cal
S}_{1,\lambda}}|\boldsymbol{K}_{1,\lambda}(\boldsymbol{t}^{\prime},\boldsymbol{s})|\,|{\mathbold
f}(\boldsymbol{t}^{\prime})|^{2}\ \mathrm{d}\boldsymbol{t}^{\prime}\right]\
\mathrm{d}\boldsymbol{s}}_{=:I}.$

Let $\displaystyle\rho>0$, and let $\displaystyle I_{\rho}$ is defined as

(85) $I_{\rho}:=\int_{B_{2}(\mathbf{0},\rho)\cap{\cal
S}_{1,\lambda}}\left[\frac{{\cal R}}{\theta_{l}}\int_{{\cal
S}_{1,\lambda}}|\boldsymbol{K}_{1,\lambda}(\boldsymbol{t}^{\prime},\boldsymbol{s})|\,|{\mathbold
f}(\boldsymbol{t}^{\prime})|^{2}\ \mathrm{d}\boldsymbol{t}^{\prime}\right]\
\mathrm{d}\boldsymbol{s},$

where $\displaystyle
B_{2}(\mathbf{0},\rho)=\\{(s_{1},s_{2})\in\mathbb{R}^{2}:\sqrt{s_{1}^{2}+s_{2}^{2}}\leq\rho\\}$
is the two-dimensional ball of radius $\displaystyle\rho$ centered at
$\displaystyle\mathbf{0}$. Based on the arguments in the last paragraph of
subsubsection 4.1.1, we find that $\displaystyle\lim_{\rho\to\infty}I_{\rho}$
exist and it is given by $\displaystyle\lim_{\rho\to\infty}I_{\rho}=I$, where
$\displaystyle I$ is defined in (84), and that the limit $\displaystyle I$
satisfies the bound $\displaystyle I\leq C_{3}||{\mathbold f}||_{L^{2}({\cal
S}_{1,\lambda},\mathbb{R}^{3})}^{2}$. Combining this with (84), we have shown
that, for $\displaystyle{\mathbold f}\in C^{\infty}_{0}({\cal
S}_{1,\lambda},\mathbb{R}^{3})$,

$||T_{1,\lambda}{\mathbold f}||_{L^{2}({\cal
S}_{1,\lambda},\mathbb{R}^{3})}\leq C_{3}||{\mathbold f}||_{L^{2}({\cal
S}_{1,\lambda},\mathbb{R}^{3})}.$

Arguing as in the case of the helix, the map $\displaystyle T_{1,\lambda}$ is
a bounded linear map on $\displaystyle L^{2}({\cal
S}_{1,\lambda},\mathbb{R}^{3})$.

####  4.2.2 Limit of the Map $\displaystyle T_{\lambda}$

Let $\displaystyle{\mathbold f}\in C^{\infty}_{0}({\cal S},\mathbb{R}^{3})$.
We write $\displaystyle T_{\lambda}{\mathbold f}$ as follows

(86) $\begin{split}(T_{\lambda}{\mathbold f})(\boldsymbol{s})&=\frac{{\cal
R}\lambda}{\theta_{l}}\int_{{\cal
S}}\boldsymbol{K}_{\lambda}(\boldsymbol{s}^{\prime},\boldsymbol{s}){\mathbold
f}(\boldsymbol{s}^{\prime})\ \mathrm{d}\boldsymbol{s}^{\prime}\\\
&=\underbrace{\left[\frac{{\cal R}\lambda}{\theta_{l}}\int_{{\cal
S}}\boldsymbol{K}_{\lambda}(\boldsymbol{s}^{\prime},\boldsymbol{s})\
\mathrm{d}\boldsymbol{s}^{\prime}\right]}_{=:\boldsymbol{H}_{\lambda}(\boldsymbol{s})}{\mathbold
f}(\boldsymbol{s})+\frac{{\cal R}\lambda}{\theta_{l}}\int_{{\cal
S}}\boldsymbol{K}_{\lambda}(\boldsymbol{s}^{\prime},\boldsymbol{s})({\mathbold
f}(\boldsymbol{s}^{\prime})-{\mathbold f}(\boldsymbol{s}))\
\mathrm{d}\boldsymbol{s}^{\prime}.\end{split}$

We next show that the second term in (86) is zero in the limit
$\displaystyle\lambda\to 0$. Fix $\displaystyle\bar{R}>0$, then we have

(87) $\begin{split}I&:=\left|\lambda\int_{{\cal
S}}\boldsymbol{K}_{\lambda}(\boldsymbol{s}^{\prime},\boldsymbol{s})({\mathbold
f}(\boldsymbol{s}^{\prime})-{\mathbold f}(\boldsymbol{s}))\
\mathrm{d}\boldsymbol{s}^{\prime}\right|\\\ &\leq\left|\lambda\int_{{\cal
S}}\chi_{|\boldsymbol{s}-\boldsymbol{s}^{\prime}|\geq\bar{R}\lambda}(\boldsymbol{s}^{\prime})\boldsymbol{K}_{\lambda}(\boldsymbol{s}^{\prime},\boldsymbol{s})({\mathbold
f}(\boldsymbol{s}^{\prime})-{\mathbold f}(\boldsymbol{s}))\
\mathrm{d}\boldsymbol{s}^{\prime}\right|\\\ &\quad+\left|\lambda\int_{{\cal
S}}\chi_{|\boldsymbol{s}-\boldsymbol{s}^{\prime}|\leq\bar{R}\lambda}(\boldsymbol{s}^{\prime})\boldsymbol{K}_{\lambda}(\boldsymbol{s}^{\prime},\boldsymbol{s})({\mathbold
f}(\boldsymbol{s}^{\prime})-{\mathbold f}(\boldsymbol{s}))\
\mathrm{d}\boldsymbol{s}^{\prime}\right|\\\ &\leq\lambda\int_{{\cal
S}}\chi_{|\boldsymbol{s}-\boldsymbol{s}^{\prime}|\geq\bar{R}\lambda}(\boldsymbol{s}^{\prime})|\boldsymbol{K}_{\lambda}(\boldsymbol{s}^{\prime},\boldsymbol{s})|\,|{\mathbold
f}(\boldsymbol{s}^{\prime})-{\mathbold f}(\boldsymbol{s})|\
\mathrm{d}\boldsymbol{s}^{\prime}\\\ &\quad+\lambda\int_{{\cal
S}}\chi_{|\boldsymbol{s}-\boldsymbol{s}^{\prime}|\leq\bar{R}\lambda}(\boldsymbol{s}^{\prime})|\boldsymbol{K}_{\lambda}(\boldsymbol{s}^{\prime},\boldsymbol{s})|\,|{\mathbold
f}(\boldsymbol{s}^{\prime})-{\mathbold f}(\boldsymbol{s})|\
\mathrm{d}\boldsymbol{s}^{\prime}\\\
&\leq\left(2\sup_{\boldsymbol{s}^{\prime}}|{\mathbold
f}(\boldsymbol{s}^{\prime})|\right)\underbrace{\lambda\int_{{\cal
S}}\chi_{|\boldsymbol{s}-\boldsymbol{s}^{\prime}|\geq\bar{R}\lambda}(\boldsymbol{s}^{\prime})|\boldsymbol{K}_{\lambda}(\boldsymbol{s}^{\prime},\boldsymbol{s})|\
\mathrm{d}\boldsymbol{s}^{\prime}}_{=:I_{1}}\\\
&\quad+\left(\bar{R}\sup_{\boldsymbol{s}^{\prime}}|\nabla{\mathbold
f}(\boldsymbol{s}^{\prime})|\right)\underbrace{\lambda^{2}\int_{{\cal
S}}\chi_{|\boldsymbol{s}-\boldsymbol{s}^{\prime}|\leq\bar{R}\lambda}(\boldsymbol{s}^{\prime})|\boldsymbol{K}_{\lambda}(\boldsymbol{s}^{\prime},\boldsymbol{s})|\
\mathrm{d}\boldsymbol{s}^{\prime}}_{=:I_{2}},\end{split}$

where, in the last step, we have used that $\displaystyle|{\mathbold
f}(\boldsymbol{s}^{\prime})-{\mathbold f}(\boldsymbol{s})|\leq
2\sup_{\boldsymbol{t}}|{\mathbold f}(\boldsymbol{t})|$, and, for
$\displaystyle\boldsymbol{s}^{\prime}\in\\{\boldsymbol{t}:|\boldsymbol{t}-\boldsymbol{s}|\leq\bar{R}\lambda\\}$,
we have $\displaystyle|{\mathbold f}(\boldsymbol{s}^{\prime})-{\mathbold
f}(\boldsymbol{s})|\leq\bar{R}\lambda\sup_{\boldsymbol{t}}|\nabla{\mathbold
f}(\boldsymbol{t})|$. To bound $\displaystyle I_{1}$ and $\displaystyle
I_{2}$, first observe that, due to
$\displaystyle\boldsymbol{K}_{\lambda}(\boldsymbol{s}^{\prime},\boldsymbol{s})=\boldsymbol{K}_{1,\lambda}(\boldsymbol{s}^{\prime}/\lambda,\boldsymbol{s}/\lambda)/\lambda^{3}$
and the bound on $\displaystyle\boldsymbol{K}_{1,\lambda}$ in (82), there
holds

(88)
$|\boldsymbol{K}_{\lambda}(\boldsymbol{s}^{\prime},\boldsymbol{s})|=\frac{1}{\lambda^{3}}|\boldsymbol{K}_{1,\lambda}(\boldsymbol{s}^{\prime}/\lambda,\boldsymbol{s}/\lambda)|\leq\frac{1}{\lambda^{3}}\frac{C_{1}}{C_{2}+|\boldsymbol{s}^{\prime}-\boldsymbol{s}|^{3}/\lambda^{3}},$

and as a direct consequence, we have

(89) $\begin{split}I_{1}&\leq\lambda\frac{1}{\lambda^{3}}\int_{{\cal
S}}\chi_{|\boldsymbol{s}-\boldsymbol{s}^{\prime}|\geq\bar{R}\lambda}(\boldsymbol{s}^{\prime})\frac{C_{1}}{C_{2}+|\boldsymbol{s}^{\prime}-\boldsymbol{s}|^{3}/\lambda^{3}}\
\mathrm{d}\boldsymbol{s}^{\prime}\\\
&\leq\frac{1}{\lambda^{2}}\int_{|\boldsymbol{s}-\boldsymbol{s}^{\prime}|\geq\bar{R}\lambda}\frac{C_{1}}{C_{2}+|\boldsymbol{s}^{\prime}-\boldsymbol{s}|^{3}/\lambda^{3}}\
\mathrm{d}\boldsymbol{s}^{\prime}\\\
&=\int_{|\boldsymbol{t}|\geq\bar{R}}\frac{C_{1}}{C_{2}+|\boldsymbol{t}|^{3}}\
\mathrm{d}\boldsymbol{t},\end{split}$

where, in the second step, the domain of integration was enlarged; and, in the
final step, the change of variables
$\displaystyle\boldsymbol{t}=(\boldsymbol{s}^{\prime}-\boldsymbol{s})/\lambda$
(so that $\displaystyle\lambda^{2}\ \mathrm{d}\boldsymbol{t}=\
\mathrm{d}\boldsymbol{s}^{\prime}$) was introduced. Next, using the definition
of $\displaystyle I_{2}$ in (87) and (88), we get

(90) $\begin{split}I_{2}&\leq\lambda^{2}\frac{1}{\lambda^{3}}\int_{{\cal
S}}\chi_{|\boldsymbol{s}-\boldsymbol{s}^{\prime}|\leq\bar{R}\lambda}(\boldsymbol{s}^{\prime})\frac{C_{1}}{C_{2}+|\boldsymbol{s}^{\prime}-\boldsymbol{s}|^{3}/\lambda^{3}}\
\mathrm{d}\boldsymbol{s}^{\prime}\\\
&\leq\frac{1}{\lambda}\int_{|\boldsymbol{s}-\boldsymbol{s}^{\prime}|\leq\bar{R}\lambda}\frac{C_{1}}{C_{2}+|\boldsymbol{s}^{\prime}-\boldsymbol{s}|^{3}/\lambda^{3}}\
\mathrm{d}\boldsymbol{s}^{\prime}\\\
&\leq\lambda\int_{|\boldsymbol{t}|\leq\bar{R}}\frac{C_{1}}{C_{2}+|\boldsymbol{t}|^{3}}\
\mathrm{d}\boldsymbol{t}\\\
&\leq\left[\frac{C_{1}}{C_{2}}\pi\bar{R}^{2}\right]\lambda,\end{split}$

where, in the second step, the domain of integration was enlarged; in the
third step, the change of variables
$\displaystyle\boldsymbol{t}=(\boldsymbol{s}^{\prime}-\boldsymbol{s})/\lambda$
was introduced; and, finally, in the last step, $\displaystyle
C_{1}/(C_{2}+|\boldsymbol{t}|^{3})\leq C_{1}/C_{2}$ was used. Combining the
results of (89) and (90) with (87), we have shown that

(91) $\begin{split}I&\leq\left(2\sup_{\boldsymbol{s}^{\prime}}|{\mathbold
f}(\boldsymbol{s}^{\prime})|\right)\,I_{1}+\left(\bar{R}\sup_{\boldsymbol{s}^{\prime}}|\nabla{\mathbold
f}(\boldsymbol{s}^{\prime})|\right)\,I_{2}\\\
&\leq\left(2\sup_{\boldsymbol{s}^{\prime}}|{\mathbold
f}(\boldsymbol{s}^{\prime})|\right)\,\int_{|\boldsymbol{t}|\geq\bar{R}}\frac{C_{1}}{C_{2}+|\boldsymbol{t}|^{3}}\
\mathrm{d}\boldsymbol{t}+\left(\bar{R}\sup_{\boldsymbol{s}^{\prime}}|\nabla{\mathbold
f}(\boldsymbol{s}^{\prime})|\right)\,\left[\frac{C_{1}}{C_{2}}\pi\bar{R}^{2}\right]\lambda,\end{split}$

for any $\displaystyle\bar{R}>0$. Arguing as in the case of helix – see the
discussion associated with (67) – we have

$\lim_{\lambda\to 0}I\leq\left(2\sup_{\boldsymbol{s}^{\prime}}|{\mathbold
f}(\boldsymbol{s}^{\prime})|\right)\,\int_{|\boldsymbol{t}|\geq\bar{R}}\frac{C_{1}}{C_{2}+|\boldsymbol{t}|^{3}}\
\mathrm{d}\boldsymbol{t},$

for any $\displaystyle\bar{R}>0$. We can take the limit
$\displaystyle\bar{R}\to\infty$ of both sides above since the inequality is
valid for any $\displaystyle\bar{R}>0$, and the left side is independent of
$\displaystyle\bar{R}$ and the limit of $\displaystyle\bar{R}\to\infty$ of the
right side is well-defined and equal to $\displaystyle 0$. We have therefore
shown

$\lim_{\lambda\to 0}I=\lim_{\lambda\to 0}\left|\lambda\int_{{\cal
S}}\boldsymbol{K}_{\lambda}(\boldsymbol{s}^{\prime},\boldsymbol{s})({\mathbold
f}(\boldsymbol{s}^{\prime})-{\mathbold f}(\boldsymbol{s}))\
\mathrm{d}\boldsymbol{s}^{\prime}\right|\leq\lim_{\bar{R}\to\infty}\left[\left(2\sup_{\boldsymbol{s}^{\prime}}|{\mathbold
f}(\boldsymbol{s}^{\prime})|\right)\,\int_{|\boldsymbol{t}|\geq\bar{R}}\frac{C_{1}}{C_{2}+|\boldsymbol{t}|^{3}}\
\mathrm{d}\boldsymbol{t}\right]=0.$

Thus, from (86), we have

(92) $\lim_{\lambda\to 0}(T_{\lambda}{\mathbold
f})(\boldsymbol{s})=\left[\lim_{\lambda\to
0}\boldsymbol{H}_{\lambda}(\boldsymbol{s})\right]{\mathbold
f}(\boldsymbol{s}).$

##### Limit of $\displaystyle\boldsymbol{H}_{\lambda}$

Consider a typical $\displaystyle\boldsymbol{s}\in{\cal S}$ such that
$\displaystyle\boldsymbol{s}\in U_{\lambda}(\boldsymbol{a})$ where
$\displaystyle\boldsymbol{a}\in{\cal L}_{\lambda}$. Recall that
$\displaystyle{\cal L}_{\lambda}$ is the lattice for $\displaystyle\lambda>0$
and $\displaystyle
U_{\lambda}(\boldsymbol{a})=[a_{1},a_{1}+\theta_{l}\lambda)\times[a_{2},a_{2}+\lambda)$
is the lattice cell. In the definition of
$\displaystyle\boldsymbol{H}_{\lambda}$, we substitute
$\displaystyle\boldsymbol{K}_{\lambda}$, to get

(93) $\boldsymbol{H}_{\lambda}(\boldsymbol{s})=\frac{{\cal
R}\lambda}{\theta_{l}}\sum_{\boldsymbol{u}\in{\cal
L}_{\lambda},\boldsymbol{u}\neq\boldsymbol{a}}\boldsymbol{K}(\bar{\boldsymbol{x}}(\boldsymbol{u})-\bar{\boldsymbol{x}}(\boldsymbol{a}))\int_{U_{\lambda}(\boldsymbol{u})}\
\mathrm{d}\boldsymbol{s}^{\prime}={\cal
R}\lambda^{3}\sum_{\boldsymbol{u}\in{\cal
L}_{\lambda},\boldsymbol{u}\neq\boldsymbol{a}}\boldsymbol{K}(\bar{\boldsymbol{x}}(\boldsymbol{u})-\bar{\boldsymbol{x}}(\boldsymbol{a})).$

Substituting the definition of transformation
$\displaystyle\bar{\boldsymbol{x}}$ in (24), we can show for
$\displaystyle\boldsymbol{a},\boldsymbol{u}\in{\cal L}_{\lambda}$ that

$\bar{\boldsymbol{x}}(\boldsymbol{u})-\bar{\boldsymbol{x}}(\boldsymbol{a})=\boldsymbol{Q}(a_{1}\theta)\left[{\cal
R}(\boldsymbol{Q}(u_{1}-a_{1})-\boldsymbol{I})\boldsymbol{e}_{1}+(u_{2}-a_{2})\delta\boldsymbol{e}_{3}\right].$

Using the identities
$\displaystyle\boldsymbol{K}(\boldsymbol{Q}(t)\boldsymbol{x})=\boldsymbol{Q}(t)\boldsymbol{K}(\boldsymbol{x})\boldsymbol{Q}^{T}(t)$
and
$\displaystyle\boldsymbol{K}(\lambda\boldsymbol{x})=\boldsymbol{K}(\boldsymbol{x})/\lambda^{3}$,
from (93), we have

(94) $\boldsymbol{H}_{\lambda}(\boldsymbol{s})={\cal
R}\boldsymbol{Q}(a_{1})\underbrace{\left[\sum_{\boldsymbol{u}\in{\cal
L}_{\lambda},\boldsymbol{u}\neq\boldsymbol{a}}\boldsymbol{K}\left({\cal
R}(\boldsymbol{Q}(u_{1}-a_{1})-\boldsymbol{I})/\lambda\boldsymbol{e}_{1}+(u_{2}-a_{2})\delta/\lambda\boldsymbol{e}_{3}\right)\right]}_{=:\bar{\boldsymbol{H}}_{\lambda}(\boldsymbol{s})}\boldsymbol{Q}(-a_{1}).$

We analyze $\displaystyle\bar{\boldsymbol{H}}_{\lambda}$ as follows. First, we
expand the sum $\displaystyle\boldsymbol{u}\in{\cal L}_{\lambda}$

(95)
$\begin{split}\bar{\boldsymbol{H}}_{\lambda}(\boldsymbol{s})&=\sum_{u_{2}\in\lambda\mathbb{Z}}\left[\sum_{\begin{subarray}{c}u_{1}\in\lambda\theta_{l}\mathbb{Z}\cap(-\bar{\theta},\bar{\theta}),\\\
(u_{1},u_{2})\neq\boldsymbol{a}\end{subarray}}\boldsymbol{K}\left({\cal
R}\frac{\boldsymbol{Q}(u_{1}-a_{1})-\boldsymbol{I}}{\lambda}\boldsymbol{e}_{1}+\delta\frac{u_{2}-a_{2}}{\lambda}\boldsymbol{e}_{3}\right)\right]\\\
&=\sum_{t_{2}^{\prime}\in\mathbb{Z}}\left[\sum_{\begin{subarray}{c}u_{1}\in\lambda\theta_{l}\mathbb{Z}\cap(-\bar{\theta},\bar{\theta}),\\\
(u_{1},t_{2}^{\prime})\neq(a_{1},0)\end{subarray}}\boldsymbol{K}\left({\cal
R}\frac{\boldsymbol{Q}(u_{1}-a_{1})-\boldsymbol{I}}{\lambda}\boldsymbol{e}_{1}+\delta
t_{2}^{\prime}\boldsymbol{e}_{3}\right)\right],\end{split}$

where we introduced the new variable $\displaystyle
t_{2}^{\prime}=(u_{2}-a_{2})/\lambda$. Since $\displaystyle
u_{2},a_{2}\in\lambda\mathbb{Z}$, we have $\displaystyle
t_{2}^{\prime}\in\mathbb{Z}$. Using a Taylor expansion and the mean value
theorem, we have the identity

(96)
$\boldsymbol{Q}(u_{1}-a_{1})-\boldsymbol{I}=\boldsymbol{Q}^{\prime}(\xi)(u_{1}-a_{1}),$

where $\displaystyle\xi=\xi(u_{1}-a_{1})\in(-\bar{\theta},\bar{\theta})$
depends on $\displaystyle u_{1}-a_{1}$. Formula (96) suggests that we
decompose (95) as:

(97)
$\begin{split}\bar{\boldsymbol{H}}_{\lambda}(\boldsymbol{s})=&\underbrace{\sum_{t_{2}^{\prime}\in\mathbb{Z}}\left[\sum_{\begin{subarray}{c}u_{1}\in\lambda\theta_{l}\mathbb{Z}\cap(-\bar{\theta},\bar{\theta}),\\\
(u_{1},t_{2}^{\prime})\neq(a_{1},0)\end{subarray}}\boldsymbol{K}\left({\cal
R}\boldsymbol{Q}^{\prime}(0)\frac{u_{1}-a_{1}}{\lambda}\boldsymbol{e}_{1}+\delta
t_{2}^{\prime}\boldsymbol{e}_{3}\right)\right]}_{=:\bar{\boldsymbol{H}}_{\lambda}^{(1)}(\boldsymbol{s})}\\\
&+\underbrace{\sum_{t_{2}^{\prime}\in\mathbb{Z}}\left[\sum_{\begin{subarray}{c}u_{1}\in\lambda\theta_{l}\mathbb{Z}\cap(-\bar{\theta},\bar{\theta}),\\\
(u_{1},t_{2}^{\prime})\neq(a_{1},0)\end{subarray}}\left\\{\boldsymbol{K}\left({\cal
R}\frac{\boldsymbol{Q}(u_{1}-a_{1})-\boldsymbol{I}}{\lambda}\boldsymbol{e}_{1}+\delta
t_{2}^{\prime}\boldsymbol{e}_{3}\right)-\boldsymbol{K}\left({\cal
R}\boldsymbol{Q}^{\prime}(0)\frac{u_{1}-a_{1}}{\lambda}\boldsymbol{e}_{1}+\delta
t_{2}^{\prime}\boldsymbol{e}_{3}\right)\right\\}\right]}_{=:\bar{\boldsymbol{H}}_{\lambda}^{(2)}(\boldsymbol{s})}.\end{split}$

Step 1: We show $\displaystyle\bar{\boldsymbol{H}}_{\lambda}^{(2)}$ goes to
zero in the limit $\displaystyle\lambda\to 0$. Let

(98) $\boldsymbol{x}_{1}={\cal
R}\frac{\boldsymbol{Q}(u_{1}-a_{1})-\boldsymbol{I}}{\lambda}\boldsymbol{e}_{1},\quad\boldsymbol{x}_{2}={\cal
R}\boldsymbol{Q}^{\prime}(0)\frac{u_{1}-a_{1}}{\lambda}\boldsymbol{e}_{1},\quad\boldsymbol{z}=\delta
t_{2}^{\prime}\boldsymbol{e}_{3}.$

Consider a function $\displaystyle\boldsymbol{y}\colon[0,1]\to\mathbb{R}^{3}$
defined as

(99)
$\boldsymbol{y}(r)=\boldsymbol{x}_{1}+r(\boldsymbol{x}_{2}-\boldsymbol{x}_{1}).$

Note that, since $\displaystyle(u_{1},t_{2}^{\prime})\neq(a_{1},0)$ and
$\displaystyle t_{2}^{\prime}\in\mathbb{Z}$, we have

(100)
$|\boldsymbol{y}(r)+\boldsymbol{z}|\geq\min\\{\delta,\min_{r\in[0,1],u_{1}\in(\lambda\theta_{l}\mathbb{Z}-\\{a_{1}\\})\cap(-\bar{\theta},\bar{\theta})}|\boldsymbol{y}(r)|\\}.$

We show that
$\displaystyle\min_{r\in[0,1],u_{1}\in(\lambda\theta_{l}\mathbb{Z}-\\{a_{1}\\})\cap(-\bar{\theta},\bar{\theta})}|\boldsymbol{y}(r)|>0$
and the lower bound is independent of $\displaystyle\lambda$. For convenience,
let $\displaystyle t=u_{1}-a_{1}$. Since $\displaystyle
u_{1},a_{1}\in\lambda\theta_{l}\mathbb{Z}\cap(-\bar{\theta},\bar{\theta})$,
and $\displaystyle u_{1}\neq a_{1}$, we have $\displaystyle
t\in(\lambda\theta_{l}\mathbb{Z}-\\{0\\})\cap(-2\bar{\theta},2\bar{\theta})$.
The hypothesis of Proposition 3.10 restricts $\displaystyle\bar{\theta}$ such
that

(101) $0<\bar{\theta}<\pi/4\quad\Rightarrow 0<\cos(2\bar{\theta})<1.$

With $\displaystyle t=u_{1}-a_{1}$, writing out the action of
$\displaystyle\boldsymbol{Q}(t)$ and $\displaystyle\boldsymbol{Q}^{\prime}(0)$
on $\displaystyle\boldsymbol{e}_{1}$, we get

(102) $\begin{split}\boldsymbol{x}_{1}&={\cal
R}\frac{\boldsymbol{Q}(t)-\boldsymbol{I}}{\lambda}\boldsymbol{e}_{1}=\frac{{\cal
R}}{\lambda}[(\cos(t)-1)\boldsymbol{e}_{1}+\sin(t)\boldsymbol{e}_{2}],\\\
\boldsymbol{x}_{2}&={\cal
R}\boldsymbol{Q}^{\prime}(0)\frac{t}{\lambda}\boldsymbol{e}_{1}=\frac{{\cal
R}t}{\lambda}[-\sin(t)\boldsymbol{e}_{1}+\cos(t)\boldsymbol{e}_{2}].\end{split}$

Through elementary calculations, we can show

(103)
$|\boldsymbol{y}(r)|^{2}=|\boldsymbol{x}_{1}+r(\boldsymbol{x}_{2}-\boldsymbol{x}_{1})|^{2}=\frac{{\cal
R}^{2}}{\lambda^{2}}\left[2(1-r)^{2}(1-\cos(t))+r^{2}t^{2}+2r(1-r)t\sin(t)\right].$

Using a Taylor expansion and noting that $\displaystyle
t\in(\lambda\theta_{l}\mathbb{Z}-\\{0\\})\cap(-2\bar{\theta},2\bar{\theta})$,
there exists $\displaystyle\xi_{1},\xi_{2}\in(-2\bar{\theta},2\bar{\theta})$
with $\displaystyle\xi_{1}=\xi_{1}(t),\xi_{2}=\xi_{2}(t)$ such that

(104) $1-\cos(t)=\cos(\xi_{1})t^{2}/2,\quad\sin(t)=t\cos(\xi_{2}).$

Thus

(105) $\begin{split}|\boldsymbol{y}(r)|^{2}&=\frac{{\cal
R}^{2}}{\lambda^{2}}\left[2(1-r)^{2}\cos(\xi_{1})t^{2}/2+r^{2}t^{2}+2r(1-r)t\cos(\xi_{2})t\right]\\\
&=\frac{{\cal
R}^{2}t^{2}}{\lambda^{2}}\left[(1-r)^{2}\cos(\xi_{1})+r^{2}+2r(1-r)\cos(\xi_{2})\right]\\\
&\geq\frac{{\cal
R}^{2}t^{2}}{\lambda^{2}}\left[(1-r)^{2}\min_{\xi\in(-2\bar{\theta},2\bar{\theta})}\cos(\xi)+r^{2}+2r(1-r)\min_{\xi\in(-2\bar{\theta},2\bar{\theta})}\cos(\xi)\right]\\\
&=\frac{{\cal
R}^{2}t^{2}}{\lambda^{2}}\left[(1-r)^{2}\cos(2\bar{\theta})+r^{2}+2r(1-r)\cos(2\bar{\theta})\right]\\\
&\geq\frac{{\cal
R}^{2}t^{2}}{\lambda^{2}}\min_{r\in[0,1]}\left[(1-r)^{2}\cos(2\bar{\theta})+r^{2}+2r(1-r)\cos(2\bar{\theta})\right]\\\
&=\frac{{\cal R}^{2}t^{2}}{\lambda^{2}}\cos(2\bar{\theta}),\end{split}$

where we used the fact that
$\displaystyle\min_{\xi\in(-2\bar{\theta},2\bar{\theta})}\cos(\xi)=\cos(2\bar{\theta})$
in the fourth line, and $\displaystyle\cos(2\bar{\theta})$ is the minimum with
respect to $\displaystyle r\in[0,1]$ of the function in the square bracket in
the fifth line. Further, since $\displaystyle
t\in(\lambda\theta_{l}\mathbb{Z}-\\{0\\})\cap(-2\bar{\theta},2\bar{\theta})$,
we have

(106) $0<C_{y}:=\frac{{\cal
R}^{2}\lambda^{2}\theta_{l}^{2}}{\lambda^{2}}\cos(2\bar{\theta})=({\cal
R}\theta_{l})^{2}\cos(2\bar{\theta})\leq|\boldsymbol{y}(r)|^{2},$

for any $\displaystyle
t\in(\lambda\theta_{l}\mathbb{Z}-\\{0\\})\cap(-2\bar{\theta},2\bar{\theta})$
and $\displaystyle r\in[0,1]$. The lower bound on
$\displaystyle|\boldsymbol{y}(r)|$ is independent of $\displaystyle\lambda$
and $\displaystyle r$. Finally, combining (106) with (100), we get

(107) $0<C_{yz}:=\min\\{\delta,{\cal
R}\theta_{l}\sqrt{\cos(2\bar{\theta})}\\}\leq|\boldsymbol{y}(r)+\boldsymbol{z}|.$

Proceeding further, we have, from the fundamental theorem of calculus,

(108)
$\begin{split}\boldsymbol{K}(\boldsymbol{x}_{1}+\boldsymbol{z})-\boldsymbol{K}(\boldsymbol{x}_{2}+\boldsymbol{z})&=\int_{0}^{1}\frac{\
\mathrm{d}}{\ \mathrm{d}r}\boldsymbol{K}(\boldsymbol{y}(r)+\boldsymbol{z})\
\mathrm{d}r=\int_{0}^{1}\nabla\boldsymbol{K}(\boldsymbol{y}(r)+\boldsymbol{z})\frac{\
\mathrm{d}}{\ \mathrm{d}r}\boldsymbol{y}(r)\ \mathrm{d}r\\\
&=\int_{0}^{1}\nabla\boldsymbol{K}(\boldsymbol{y}(r)+\boldsymbol{z})(\boldsymbol{x}_{2}-\boldsymbol{x}_{1})\
\mathrm{d}r.\end{split}$

Note that because of (107),
$\displaystyle\nabla\boldsymbol{K}(\boldsymbol{y}(r)+\boldsymbol{z})$ exists
and is bounded. From the definition of
$\displaystyle\bar{\boldsymbol{H}}_{\lambda}^{(2)}$ in (97), a change of
variable $\displaystyle t=u_{1}-a_{1}$, the definition of
$\displaystyle\boldsymbol{x}_{1},\boldsymbol{x}_{2},\boldsymbol{z}$ in (98)
and (102), and noting the identity (108), we have

(109)
$\begin{split}|\bar{\boldsymbol{H}}_{\lambda}^{(2)}(\boldsymbol{s})|&\leq\sum_{t_{2}^{\prime}\in\mathbb{Z}}\left[\sum_{\begin{subarray}{c}u_{1}\in\lambda\theta_{l}\mathbb{Z}\cap(-\bar{\theta},\bar{\theta}),\\\
(u_{1},t_{2}^{\prime})\neq(a_{1},0)\end{subarray}}\left|\boldsymbol{K}\left({\cal
R}\frac{\boldsymbol{Q}(u_{1}-a_{1})-\boldsymbol{I}}{\lambda}\boldsymbol{e}_{1}+\delta
t_{2}^{\prime}\boldsymbol{e}_{3}\right)-\boldsymbol{K}\left({\cal
R}\boldsymbol{Q}^{\prime}(0)\frac{u_{1}-a_{1}}{\lambda}\boldsymbol{e}_{1}+\delta
t_{2}^{\prime}\boldsymbol{e}_{3}\right)\right|\right]\\\
&\leq\sum_{t_{2}^{\prime}\in\mathbb{Z}}\left[\sum_{t\in\lambda\theta_{l}\mathbb{Z}-\\{0\\}\cap(-2\bar{\theta},2\bar{\theta})}\left|\boldsymbol{K}\left(\boldsymbol{x}_{1}+\boldsymbol{z}\right)-\boldsymbol{K}\left(\boldsymbol{x}_{2}+\boldsymbol{z}\right)\right|\right]\\\
&\leq\sum_{t_{2}^{\prime}\in\mathbb{Z}}\left[\sum_{t\in\lambda\theta_{l}\mathbb{Z}-\\{0\\}\cap(-2\bar{\theta},2\bar{\theta})}\int_{0}^{1}\left|\nabla\boldsymbol{K}(\boldsymbol{y}(r)+\boldsymbol{z})\right|\;|\boldsymbol{x}_{2}-\boldsymbol{x}_{1}|\
\mathrm{d}r\right]\\\
&\leq\sum_{t_{2}^{\prime}\in\mathbb{Z}}\left[\sum_{t\in\lambda\theta_{l}\mathbb{Z}-\\{0\\}\cap(-2\bar{\theta},2\bar{\theta})}\int_{0}^{1}\frac{C}{|\boldsymbol{y}(r)+\boldsymbol{z}|^{4}}|\boldsymbol{x}_{2}-\boldsymbol{x}_{1}|\
\mathrm{d}r\right]\\\
&=\sum_{t_{2}^{\prime}\in\mathbb{Z}}\left[\sum_{t\in\lambda\theta_{l}\mathbb{Z}-\\{0\\}\cap(-2\bar{\theta},2\bar{\theta})}\int_{0}^{1}\frac{C}{(|\boldsymbol{y}(r)|^{2}+|\boldsymbol{z}|^{2})^{2}}|\boldsymbol{x}_{2}-\boldsymbol{x}_{1}|\
\mathrm{d}r\right],\end{split}$

where we used the bound on the gradient of $\displaystyle\boldsymbol{K}$ with
constant $\displaystyle C>0$ fixed.

Next, we get an upper bound on
$\displaystyle|\boldsymbol{x}_{1}-\boldsymbol{x}_{2}|$ in terms of
$\displaystyle t$. From (102), we have

$\displaystyle\boldsymbol{x}_{2}-\boldsymbol{x}_{1}=\frac{{\cal
R}}{\lambda}\left[t\boldsymbol{Q}^{\prime}(0)-\boldsymbol{Q}(t)+\boldsymbol{I}\right]\boldsymbol{e}_{1}.$

By a Taylor expansion and the mean value theorem, we have
$\displaystyle\boldsymbol{Q}(t)=\boldsymbol{I}+t\boldsymbol{Q}^{\prime}(0)+(t^{2}/2)\boldsymbol{Q}^{\prime\prime}(\xi)$
where $\displaystyle\xi=\xi(t)\in(-2\bar{\theta},2\bar{\theta})$ depends on
$\displaystyle t$. Substituting this and using the bound
$\displaystyle|\boldsymbol{Q}^{\prime\prime}_{ij}(\xi)|\leq 1$, we obtain

(110) $|\boldsymbol{x}_{2}-\boldsymbol{x}_{1}|=\frac{{\cal
R}}{\lambda}\frac{|t|^{2}}{2}|\boldsymbol{Q}^{\prime\prime}(\xi)|\leq\frac{{\cal
R}}{\lambda}\frac{|t|^{2}}{2}.$

Combining the equation above with (109), we get

(111)
$\begin{split}|\bar{\boldsymbol{H}}_{\lambda}^{(2)}(\boldsymbol{s})|&\leq\sum_{t_{2}^{\prime}\in\mathbb{Z}}\left[\sum_{t\in\lambda\theta_{l}\mathbb{Z}-\\{0\\}\cap(-2\bar{\theta},2\bar{\theta})}\int_{0}^{1}\frac{C}{(|\boldsymbol{y}(r)|^{2}+|\boldsymbol{z}|^{2})^{2}}\frac{{\cal
R}}{\lambda}\frac{|t|^{2}}{2}\ \mathrm{d}r\right]\\\
&=\sum_{t_{2}^{\prime}\in\mathbb{Z}}\left[\sum_{t^{\prime}\in\theta_{l}\mathbb{Z}-\\{0\\}\cap(-2\bar{\theta}/\lambda,2\bar{\theta}/\lambda)}\int_{0}^{1}\frac{C}{(|\boldsymbol{y}(r)|^{2}+|\boldsymbol{z}|^{2})^{2}}\frac{{\cal
R}}{\lambda}\frac{\lambda^{2}|t^{\prime}|^{2}}{2}\ \mathrm{d}r\right]\\\
&\leq\lambda\left\\{\sum_{t_{2}^{\prime}\in\mathbb{Z}}\left[\sum_{t^{\prime}\in\theta_{l}\mathbb{Z}-\\{0\\}\cap(-2\bar{\theta}/\lambda,2\bar{\theta}/\lambda)}\int_{0}^{1}\frac{C}{(|\boldsymbol{y}(r)|^{2}+|\boldsymbol{z}|^{2})^{2}}\frac{{\cal
R}|t^{\prime}|^{2}}{2}\ \mathrm{d}r\right]\right\\},\end{split}$