# Nonlocal Mean Curvature with Integrable Kernel

Animesh Biswas  Department of Mathematics
University of Nebraska-Lincoln
210 Avery Hall, Lincoln
NE 68588, United States of America<EMAIL_ADDRESS>,  Mikil Foss
<EMAIL_ADDRESS>and  Petronela Radu<EMAIL_ADDRESS>
###### Abstract.

We study the prescribed constant mean curvature problem in the nonlocal
setting where the nonlocal curvature has been defined as

$H^{J}_{\Omega}(x):=\int_{\mathbb{R}^{n}}J(x-y)(\chi_{\Omega^{c}}(y)-\chi_{\Omega}(y))dy,$

where $x\in\mathbb{R}^{n}$, $\Omega\subset\mathbb{R}^{n}$, $\chi$ is the
characteristic function for a set, $J$ is a radially symmetric, nonegative,
nonincreasing convolution kernel. Several papers have studied the case of
nonlocal curvature with nonintegrable singularity, a generalization of the
classical curvature concept, which requires the regularity of the boundary to
be above $C^{2}$. Nonlocal curvature of this form appears in many different
applications, such as image processing, curvature driven motion, deformations.
In this work, we focus on the problem of constant nonlocal curvature defined
via integrable kernel. Our results offer some extensions to the constant mean
curvature problem for nonintegrable kernels, where counterparts to
Alexandrov’s theorem in the nonlocal framework were established independently
by two separate groups: Ciraolo, Figalli, Maggi, Novaga, and respectively,
Cabré, Fall, Solà-Morales, Weth. Using the nonlocal Alexandrov’s theorem we
identify surfaces of constant mean nonlocal curvature for different integrable
kernels as unions of balls situated at distance $\delta$ apart, where $\delta$
measures the radius of nonlocal interactions.

###### Key words and phrases:

Nonlocal mean curvature, constant curvature, Alexandrov’s moving plane method,
integrable kernel of interaction, finite horizon.

###### 1991 Mathematics Subject Classification:

53A10, 45XX,

Research partially supported by,

##  1\. Introduction

In this paper, we study the constant nonlocal mean curvature problem, i.e.
find the surfaces for which the curvature is constant. After the seminal paper
[10] which introduced the concept of nonlocal curvature, there has been
increasing interest in problems involving this new concept. Given
$\Omega\subseteq\mathbb{R}^{n}$ with a sufficiently smooth boundary
$\partial\Omega$ and $0<s<1$, the authors Caffarelli, Roquejofre, and Savin of
[10] introduced

(1.1) $\displaystyle
H_{\Omega}^{s}(x)=\int_{\mathbb{R}^{n}}\frac{\chi_{\Omega^{c}}(y)-\chi_{\Omega}(y)}{|x-y|^{n+2s}}\,dy$

as a nonlocal mean curvature of $\Omega$ at $x\in\partial\Omega$. Here, we use
$\chi_{E}$ to denote the characteristic function of
$E\subseteq\mathbb{R}^{n}$. The kernel of interaction that appears in the
definition of $H^{s}_{\Omega}$ was chosen to be the strongly singular function
$J(x-y)$, with $J(z)=|z|^{-n-2s}$ for $z\in\mathbb{R}^{n}\setminus\\{0\\}$,
and $0<s<1$ (observe that $J\notin L^{1}(\mathbb{R}^{n})$). In [19], Mazón,
Rossi, and Toledo extended the concept of nonlocal curvature to include
integrable kernels, a case which is of interest to us as well. In particular,
our integrability assumption has a prototypical example given by

(1.2) $z\mapsto\frac{\chi_{B_{r}}(z)}{|z|^{n-\alpha}},$

for some $\alpha>0$ and $r>0$. Here $B_{r}\subseteq\mathbb{R}^{n}$ is the open
ball with radius $r$ centered at the origin. More generally, we assume the
integrable kernel $J\in L^{1}(\mathbb{R}^{n})$ satisfies the following

  * •

(rotational symmetry) there exists $\mu:\mathbb{R}\to\mathbb{R}$ such that
$J(z)=\mu(|z|)$ for all $z\neq 0$,

  * •

(compact support) there exists $r>0$ such that $\mu(\rho)=0$ for all $\rho\geq
r$,

  * •

(radially decreasing) if $0<r_{2}<r_{1}\leq r$, then $\mu(r_{1})<\mu(r_{2})$.

Given a measurable set $\Omega\subseteq\mathbb{R}^{n}$, the nonlocal curvature
at $x\in\mathbb{R}^{n}$, is defined as

(1.3)
$H^{J}_{\Omega}(x):=\int_{\mathbb{R}^{n}}J(x-y)(\chi_{\Omega}(y)-\chi_{\Omega^{c}}(y))=\int_{\mathbb{R}^{n}}J(x-y)\tau_{\Omega}(y)dy.$

For convenience, we will use $\tau_{\Omega}=\chi_{\Omega}-\chi_{\Omega^{c}}$.
Since $J$ is integrable and $\tau_{\Omega}\in L^{\infty}(\mathbb{R}^{n})$, we
see that $H_{\Omega}^{J}$ is well-defined and finite on all of
$\mathbb{R}^{n}$ without any regularity assumptions on $\partial\Omega$.

###  1.1. Motivation

In [10, 20], it was shown that the classical mean curvature problem appears in
the cellular automata problem in the following way. Fix $\varepsilon>0$, and
set $t_{0}=0$ and $\Omega_{0}=\Omega$. For each $k\in\mathcal{N}$, put
$t_{k}=k\varepsilon$. We iteratively define
$\Omega_{k}=\\{x\in\mathbb{R}^{n}:u_{k}(t_{k},x)\geq 1/2\\}$, where
$u_{k}:[t_{k-1},t_{k}]\times\mathbb{R}^{n}\to\mathbb{R}$ is the solution to
the heat equation $\partial_{t}u_{k}-\Delta u_{k}=0$ with initial condition
$u_{k}(t_{k-1},\cdot)=\chi_{\Omega_{k-1}}$. The sequence of surfaces
$\\{\partial\Omega_{k}\\}_{k=0}^{\infty}$ provide a discrete approximation to
the flow of $\Omega$ by mean curvature. A natural candidate for the flow by
nonlocal mean curvature uses the fractional heat equation
$u_{t}-(-\Delta)^{s}u=0$, where $0<s<1$ is fixed. Here the fractional
Laplacian operator, $(-\Delta)^{s}$ is defined by

$(-\Delta)^{s}u=\mathcal{F}^{-1}\left(|\xi|^{2s}\mathcal{F}u\right),$

where $\mathcal{F}$ is the Fourier transform and $\mathcal{F}^{-1}$ is its
inverse. The fractional Laplacian operator and in general fractional elliptic
and parabolic operators have been very well-studied, see for example [2, 3,
11, 12, 13].

In 1951 [9], Caccioppoli gave the following definition of perimeter. The
perimeter of $\Omega$ inside a measurable set $E\subseteq\mathbb{R}^{n}$ is
defined as

$Per(\Omega,E)=\sup\bigg{\\{}\int_{E}\chi_{\Omega}\operatorname{div}\phi\,dx:\phi\in
C^{\infty}_{0}(E,\mathbb{R}^{n}),|\phi|\leq 1\bigg{\\}},$

a formulation that is also known as the total variation of the function
$\chi_{\Omega}$. If $\partial\Omega$ is $C^{2}$, it is easy to show that
$Per(\Omega,\mathbb{R}^{n})=\mathcal{H}^{n-1}(\partial\Omega)$. In [10], the
authors defined the nonlocal energy functional as, when $s<1/2$,

$\left\|u\right\|^{2}_{H^{s}}=\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\frac{(u(x)-u(y))^{2}}{|x-y|^{n+2s}}dxdy.$

In this paper the authors showed that minimizing this energy functional of a
characteristic function $\chi_{\Omega}$ (which can be interpreted as a
nonlocal perimeter of $\Omega$), inside a bounded set $E$ under given nonlocal
boundary condition, one can obtain the minimal surfaces $S=\partial\Omega$
whose Euler-Lagrange equation is

$\int_{\mathbb{R}^{n}}(\chi_{\Omega}(y)-\chi_{\Omega^{c}}(y))|x-y|^{-n-2s}dy=0,\quad\text{for}\leavevmode\nobreak\
x\in S.$

The above-mentioned quantity,

$-\int_{\mathbb{R}^{n}}(\chi_{\Omega}(y)-\chi_{\Omega^{c}}(y))|x-y|^{-n-2s}dy$

is labeled as the nonlocal curvature. Introduction of the nonlocal minimal
surfaces and mean curvature motivated a rich body of literature in the last
ten years. In 2018, two different groups, [6, 17] studied the constant
nonlocal mean curvature problem, as defined above. Both groups used the idea
of the Alexandrov’s moving plane in the nonlocal setting and proved that a
ball is the only solution to the constant nonlocal mean curvature problem.
Similar type of constant nonlocal mean curvature problems in different
settings were studied in [7, 8]. Additionally, we mention some works on
nonlocal minimal graphs, [5] and nonlocal curvature flows [15]. For other
results concerning this particular type of nonlocal curvature, see [14, 18].

The motivation for the introduction and study of the concepts of perimeter and
curvature in the nonlocal framework with integrable kernel is born from
similar considerations as for other models, mainly, to eliminate restrictions
on the smoothness of the geometry (boundary) of the domain. The example from
image processing [16, 19] provides a simple illustration for considering new
geometrical measures for domains in the nonlocal setting.

For boundaries that are smooth the nonlocal curvature was shown to provide a
good approximation to the classical curvature [19]. However, note that the
nonlocal curvature has the advantage of being well-defined at corners or other
singular points of the boundary. Moreover, the definition makes sense for all
points of a set, not only on a boundary, a feature which may be useful for
irregular domains (such as cusps) as one may be able to capture the rate of
“narrowing” as the cusp, or other singular point, is approached. An important
application of the nonlocal curvature can be found for domains with “zig-zag”
boundaries where the classical curvature could be defined only piecewise on
segments; however, as the segments get smaller (and possibly approach a smooth
boundary), there is no measure for how curved or bent the boundary is.

Motivated by the significance of this version of nonlocal curvature, we are
interested in identifying surfaces with a prescribed nonlocal curvature, where
the nonlocal curvature is defined as in (1.3). In other words, for given $f$
we are interested in the solution $u$ of the equation

(1.4)
$\displaystyle\int_{\mathbb{R}^{n}}J(x-y)(\chi_{\Omega^{c}}(y)-\chi_{\Omega}(y))dy=f(x),\quad\text{where}\leavevmode\nobreak\
x\in\partial\Omega.$

Above, $\Omega$ is a bounded set, and we will further assume that $f$ is a
constant function. On the other hand, when $f$ is nonconstant we assume that
$\partial\Omega$ is the graph of a function. Although nonlocal curvature can
be defined for surfaces with no differentiability, for our results we will
have a standing assumption that the boundary is at least $C^{1}$. Indeed, this
is required by the main technique used for the constant curvature problem, the
Alexandrov’s moving plane method. Regarding this regularity assumption, we
mention that very recently, the authors in [4] were able to consider, for a
particular type of kernel, boundaries that are not differentiable. More
specifically, they considered the following problem. Let $\Omega$ a bounded
set which satisfies some non-degeneracy conditions. For any point
$x\in\partial^{*}\Omega$, where $\partial^{*}\Omega$ is the essential boundary
of $\Omega$, if

(1.5) $|B_{r}(x)\cap\Omega|=c,$

where $B_{r}(x)$ is the ball of radius $r$ centered at $x$, and $c>0$ and
$r>0$ are fixed, then the domain $\Omega$ must be a union of balls. Note that
(1.5) translates to constant mean curvature when the kernel is given by the
identity function for $B_{r}(x)$. Although the authors in [4] got the desired
solution in a more general setup, our methods provide an alternative approach
which may be valuable in handling different problems involving nonlocal
curvature.

###  1.2. Main contributions and significance

Our main interest in this paper is to study the constant nonlocal mean
curvature problem. For our first result, we will consider the case of an
infinite radius of interaction, i.e. $r=\infty$. In addition, assume that the
kernel of interaction $J$ is differentiable everywhere, except possibly at
$x=0$, and strictly decreasing. The proof of the following theorem is very
similar to the proofs given for the case of fractional Laplacian kernel given
in [6, 17] with the exception that $J$ is integrable. Since $J$ is integrable
near $x=0$, we assume some decay rate for $J$ and $\nabla J$ near $x=0$;
specifically, we assume

(1.6)
$\begin{cases}|J(x)|\leq\min\left\\{\frac{C}{|x|^{n-\alpha}},\frac{C}{|x|^{n+\alpha_{1}}}\right\\}\\\
|\nabla
J(x)|\leq\min\left\\{\frac{C}{|x|^{n+1-\alpha}},\frac{C}{|x|^{n+1+\alpha_{1}}}\right\\},\end{cases}$

for all $x\neq 0$ and for some $\alpha,\alpha_{1}>0$.

######  Theorem 1.1.

Let $J:\mathbb{R}^{n}\to\mathbb{R}$ be radially symmetric, strictly decreasing
nonnegative function which is $C^{\infty}(\mathbb{R}^{n}\setminus\\{0\\})$ and
satisfies the decay estimates (1.6). In addition, let $J(x)=\phi(|x|)$ for
some function $\phi:[0,\infty)\to\mathbb{R}_{+}$ with $\phi^{\prime}(t)<0$ for
all $t>0$. Let $\Omega$ be a nonempty bounded open set with $C^{1,\beta}$
boundary for some $\beta=\min\\{\alpha-1,0\\}\geq 0$ and with the property
that $H^{J}_{\Omega}$ is constant at every point on the boundary. Then
$\Omega$ is a ball.

Nest we consider the case when the kernel $J$ satisfies all the previous
conditions except it has compact support, precisely inside the ball $B_{r}$.
As $J$ is differentiable everywhere except possibly at $x=0$, $J=0$ on
$\partial B_{r}$. Also, as $J=0$ in $(B_{r})^{c}$, we don’t need to consider
the decay of $J$ near infinity. In this case, we assume

(1.7) $\begin{cases}|J(x)|\leq\frac{C}{|x|^{n-\alpha}}\\\ |\nabla
J(x)|\leq\frac{C}{|x|^{n+1-\alpha}}\end{cases}$

for all $x\neq 0$ for some $\alpha>0$.

Under these assumptions we obtain the following theorem:

######  Theorem 1.2.

Let $J:\mathbb{R}^{n}\to\mathbb{R}$ be radially symmetric, strictly decreasing
nonnegative function which is $C^{\infty}(\mathbb{R}^{n}\setminus\\{0\\})$ and
satisfies the estimates in (1.7). In addition to that if $J(x)=\phi(|x|)$ for
some function $\phi:[0,\infty)\to\mathbb{R}_{+}$ then $\phi^{\prime}(t)<0$ for
all $t>0$ and $J$ has a compact support in $B_{r}$. Let $\Omega$ be nonempty
bounded, open, set with $C^{1,\beta}$ boundary for some
$\beta=\min\\{\alpha-1,0\\}\geq 0$ and with the property that $H^{J}_{\Omega}$
is constant at every point on the boundary then $\Omega$ is a union of balls
which are at a distance at least $r$ from each other.

Our next theorem is for some $J$ which has compact support in $B_{r}$. $J$ is
radially symmetric, strictly decreasing function which is in
$C^{\infty}(B_{r}\setminus\\{0\\})$. Inside the ball $B_{r}$, $J$ satisfies
the estimate given in 1.7. But $J$ has a jump at the boundary $\partial
B_{r}$. In addition, if $J(x)=\phi(|x|)$ for some
$\phi:[0,\infty)\to\mathbb{R}_{+}$, then $\phi^{\prime}(t)<0$ in $(0,r)$. As
an example, one can consider the kernel given in (1.2). Next we impose the
following condition on the boundary of the set $\Omega$. We assume the
$C^{1}$-norm of the boundary is uniformly bounded and small so that
$\partial\Omega$ does not vary wildly inside $B_{r+\delta}(x)$ for any
$x\in\partial\Omega$ and for some $\delta>0$. Precisely we assume that for
every $x\in\partial\Omega$, $\partial\Omega$ is a graph of some $C^{1}$
function $f^{x}$ inside the ball $B_{r+\delta}(x)$ (for some $\delta>0$) such
that $\left\|\nabla f^{x}\right\|\leq M<1$.

######  Theorem 1.3.

Let $J:\mathbb{R}^{n}\to\mathbb{R}$ be radially symmetric, strictly decreasing
nonnegative function which is $C^{\infty}(B_{r}\setminus\\{0\\})$ and
satisfies the estimates in (1.7). In addition to that if $J(x)=\phi(|x|)$ for
some function $\phi:[0,\infty)\to\mathbb{R}_{+}$ then $\phi^{\prime}(t)<0$ in
$(0,r)$ and $\phi$ has a jump at $r$ from a positive value to $0$. Let
$\Omega$ be nonempty bounded open connected set with $C^{1,\beta}$ boundary
for some $\beta=\min\\{\alpha-1,0\\}\geq 0$. In addition to that we assume
that the boundary has uniformly bounded $C^{1}$-norm as mentioned above. Next
if $H^{J}_{\Omega}$ is constant at every point on the boundary then $\Omega$
is a union of balls which are at a distance at least $r$ from each other.

In all the theorems given above $J$ is strictly decreasing in its support set.
Now we present a result when $J$ is not necessarily strictly decreasing in its
support set. In particular, we assume that $J(x)=\chi_{B_{r}}$. But we don’t
have a sphere solution for all $r$. To see that, consider any set with
diameter $<r/2$. Then using the definition of nonlocal curvature at some point
$x_{1}\in\partial\Omega$ is

$H^{J}_{\Omega}(x_{1})=|B_{r}(x_{1})\cap\Omega^{c}|-|B_{r}(x_{1})\cap\Omega|=|B_{r}|-2|B_{r}(x_{1})\cap\Omega)|=|B_{r}|-2|\Omega|.$

The above expression is independent of $x_{1}$ and hence there is no unique
solution. As mentioned before, in [4], the authors studied this problem when
the boundary of the set is not smooth. They assumed a non-degeneracy condition
on the set, which says that if $\Omega$ is $r$-degenerate then

$\inf_{x_{1},x_{2}\in\partial\Omega^{*}}\frac{|\Omega\cap(B_{r}(x_{1})\Delta
B_{r}(x_{2}))|}{\left\|x_{1}-x_{2}\right\|}=0.$

Our aim is to present a different proof of the similar result under uniform
$C^{1}$-norm boundary condition stated before the Theorem 1.3.

######  Theorem 1.4.

Let $J(x)=\chi_{B_{r}}(x)$. Let $\Omega$ be nonempty connected bounded open
set which has $C^{1}$ boundary with similar condition as in Theorem 1.3 . It
also satisfies the constant mean curvature conditions, that is
$H^{J}_{\Omega}$ to be constant at every point on the boundary. Then $\Omega$
is a ball.

Before we present the proof of Theorems 1.1, 1.2, 1.3 and 1.4 let us first
describe the idea of the proof. In the case of a highly singular kernel, see
[6, 17], where the authors introduce and prove a nonlocal version of the
celebrated Alexandrov’s Theorem, [1]. Our proof will follow a similar
approach. For simplicity consider the kernel whose support set is
$\mathbb{R}^{n}$. Next we consider a hyperplane, $HP_{e}$, which is normal to
a unit vector $e\in\mathbb{S}^{n-1}$. One can see that $HP_{e}$ partitions
$\Omega$ into two disjoint sets, say $\Omega_{1},\Omega_{2}$ (either of them
can be empty). Let us start with a situation when one of them, suppose
$\Omega_{2}$, is empty. Let $R_{e}$ be reflection operator with respect to
$HP_{e}$ at that position. In that scenario, it is trivially true that
$R_{e}(\Omega_{2})\subset\Omega_{1}$. Then as we move $HP_{e}$ (translation in
the $e$ direction), $\Omega_{2}$ starts to become nonempty still satisfying
$R_{e}(\Omega_{2})\subset\Omega_{1}$. Here we are using same notation for the
quantities $\Omega_{1},\Omega_{2}$ and $R_{e}$ which, in principle, depend
also on the position of $HP_{e}$. Next after a particular translation,
$R_{e}(\Omega_{2})$ does not remain a subset of $\Omega_{1}$. Let us name that
translation of $HP_{e}$ as the critical hyperplane $HP^{*}_{e}$ and the
corresponding reflection operator as $R^{*}_{e}$. Next, we prove that
$\Omega\Delta R^{*}_{e}(\Omega)$ is an empty set, where

$\Omega\Delta R^{*}_{e}(\Omega)=(\Omega\setminus
R^{*}_{e}(\Omega))\cup(R^{*}_{e}(\Omega)\setminus\Omega)$

That implies $\Omega$ is symmetric with respect to the $e$-direction. Proof of
the above claim, in case of nonlocal curvature, depends on the strict
monotonicity, size conditions and support of $J$ along with the boundary
regularity of $\Omega$. Finally as this claim is true for any
$e\in\mathbb{S}^{n-1}$, we must have that $\Omega$ is a ball.

As mentioned before, our focus is on the study of the constant nonlocal mean
curvature problem when the boundary is at least $C^{1}$. Very recently, the
paper [4] solved the constant mean curvature problem for boundaries with less
regularity (even measurable), our work provides a different perspective and
approach to this problem. Specifically, for different integrable kernels we
investigate the differentiability of the curvature function on the boundary.
To this end, regularity properties of the boundary are required. Establishing
differentiablity of the curvature function is a critical step in solving the
constant mean curvature problem and is an independent problem deserving
attention on its own.

##  2\. Proof of Theorem 1.1

Proof of Theorem 1.1 is very similar to the proof given in [6, 17]. Since the
main idea of this proof will be used for the other related theorems, we give a
brief outline of the proof for the convenience of the reader. As mentioned
before, the principal idea of the proof is the Alexandrov’s method. We discuss
that with details in the proof. Another important step is to prove that
$H^{J}_{\Omega}\in C^{1}(\partial\Omega)$. This is in parallel to a result
given in [17, Lemma 2.1] and [6, Proposition 2.1]. As the proof of this result
is very similar except some minor modifications due to the regularity of the
boundary, we present the proof of that result in the Appendix. We provide only
the statement of that result in this section.

######  Lemma 2.1.

Suppose $\phi_{\varepsilon}\in C^{\infty}_{c}([0,\infty))$ is such that
$\phi_{\varepsilon}\geq 0,\phi^{\prime}_{\varepsilon}\leq 0$ and it satisfies
the following estimates, for all $t>0$

$\begin{cases}\max\\{t^{n-\alpha},t^{n+\alpha_{1}}\\}\phi_{\varepsilon}(t)+\max\\{t^{n-\alpha+1},t^{n+\alpha_{1}+1}\\}|\phi^{\prime}_{\varepsilon}(t)|\leq
C(n,\alpha,\alpha_{1})\\\
|\phi^{\prime}_{\varepsilon}(t)|\to|\phi^{\prime}(t)|\quad\text{as}\leavevmode\nobreak\
\varepsilon\to 0\\\
\phi_{\varepsilon}(t)=\phi(t)\quad\text{if}\leavevmode\nobreak\
t>\varepsilon\end{cases}$

where $\phi(t)$ has been defined in Theorem 1.1. Then we define

$H^{\varepsilon}_{\Omega}(x)=\int_{\mathbb{R}^{n}}\tau_{\Omega}(y)\phi_{\varepsilon}(|x-y|)dy,$

$x\in\mathbb{R}^{n}$. Next we assume that $\Omega$ is nonempty bounded open
set with $C^{1,\beta}$ boundary for some $\beta>0$ such that $\beta+\alpha>1$.
Then $H^{J}_{\Omega}\in C^{1}(\partial\Omega)$ and
$H^{\varepsilon}_{\Omega}\to H^{J}_{\Omega}$ in $C^{1}(\partial\Omega)$ as
$\varepsilon\to 0$.

Proof of the Lemma is given in the Appendix. Now we can write using Lemma 2.1

$\nabla H^{J}_{\Omega}(x)\cdot e=\lim_{\varepsilon\to 0}\nabla
H^{\varepsilon}_{\Omega}(x)\cdot e$

for any $x\in\partial\Omega$ and $e\in T_{x}(\partial\Omega)$. Next we prove
Theorem 1.1. But before we do, we want to introduce some notation which will
be useful in later sections also. For $e\in\mathbb{S}^{n-1}$,
$E\subset\mathbb{R}^{n}$, $\tau\in\mathbb{R}$, we define

$\displaystyle\pi_{\tau}=\\{x\in\mathbb{R}^{n}:x\cdot
e=\tau\\},\quad\hbox{plane perpendicular to $e$-vector}$
$\displaystyle(\pi_{\tau})_{+}=\\{x\in\mathbb{R}^{n}:x\cdot e>\tau\\},$
$\displaystyle E_{\tau}=E\cap(\pi_{\tau})_{+},$ $\displaystyle
R_{\tau}(x)=x-2(x\cdot e-\tau)e,$ $\displaystyle\quad\hbox{where $R_{\tau}$ is
the reflection operator with respect to $\pi_{\tau}$ plane},$ $\displaystyle
R_{\tau}(E)=\\{R_{\tau}(x):x\in E\\}.$

Next we define, for a bounded open set $\Omega$, $\mu=\sup\\{x\cdot
e:x\in\Omega\\}$. If the set $\Omega$ has a $C^{1}$ boundary, [17], then for
every $\tau<\mu$, sufficiently close to $\mu$,
$R_{\tau}(\Omega_{\tau})\subseteq\Omega$. Therefore we define

$\lambda:=\inf\\{\tau\in\mathbb{R}:R_{\tilde{\tau}}(\Omega_{\tilde{\tau}})\subset\Omega\quad\hbox{for
all}\leavevmode\nobreak\ \tilde{\tau}\in(\tau,\mu)\\}.$

We denote $\pi_{\lambda}$ to be the critical hyperplane for a fixed direction
$e$. For our convenience, we use $R$ in place of $R_{\lambda}$, in the
following discussion, to denote reflection with respect to the critical
hyperplane. Next from [1], we see that for any direction $e$, at least one of
the following two conditions will hold,

  1. (a)

interior touching: $\partial R(\Omega)$ is tangent to $\partial\Omega$ at some
point $x_{0}$ which is the refection of a point $x^{\prime}_{0}\in\partial
R(\Omega)\setminus\pi_{\lambda}$.

  2. (b)

non-transversal intersection: $\pi_{\lambda}$ is orthogonal to
$\partial\Omega$ at some point $x_{0}\in\partial\Omega\cap\pi_{\lambda}$.

###### Proof of Theorem 1.1.

As mentioned before, proof is similar as in [6, 17]. Hence we give a brief
outline where we mention the important steps only. We start by considering any
$e\in\mathbb{S}^{n-1}$. For a fixed $e\in\mathbb{S}^{n-1}$, we will have
either case $(a)$ or case (b). Then we want to prove, irrespective of case (a)
or (b), that $\Omega\setminus R(\Omega)=R(\Omega)\setminus\Omega=\varnothing$,
where $\varnothing$ is the null set. Since this is true for any $e$, $\Omega$
is symmetric in any direction and hence is a ball. Assume that, without the
loss of generality, $e=\langle 1,0,0,\cdots 0\rangle$. If $\pi_{\lambda}$ is
the critical hyperplane in this direction for some $\lambda$, then the
rotation operator with respect to $\pi_{\lambda}$, is given by
$R(x)=(-x_{1}+2\lambda,x_{2},\cdots,x_{n})$ for any
$x=(x_{1},x_{2},\cdots,x_{n})$.

Case (a): We notice that
$x_{0},x^{\prime}_{0}=R(x_{0})\in\partial\Omega\cap\partial R(\Omega)$. Then

$H^{J}_{\Omega}(x_{0})-H^{J}_{\Omega}(R(x_{0}))=0.$

Using $\tau_{\Omega}(y)=(\chi_{\Omega^{c}}(y)-\chi_{\Omega}(y))$ for any
$y\in\mathbb{R}^{n}$,

$\displaystyle H^{J}_{\Omega}(R(x_{0}))$
$\displaystyle=-\int_{\mathbb{R}^{n}}\tau_{\Omega}(y)J(R(x_{0})-y)dy$
$\displaystyle=-\int_{\mathbb{R}^{n}}\tau_{\Omega}(y)J(-(x_{0})_{1}+2\lambda-
y_{1},(x_{0})_{2}-y_{2},\cdots,(x_{0})_{n}-y_{n})dy$
$\displaystyle=-\int_{\mathbb{R}^{n}}\tau_{\Omega}(y)J(-(x_{0})_{1}+(-y_{1}+2\lambda),(x_{0})_{2}-y_{2},\cdots,(x_{0})_{n}-y_{n})dy$
$\displaystyle=-\int_{\mathbb{R}^{n}}\tau_{\Omega}(y)J((x_{0})_{1}-(-y_{1}+2\lambda),(x_{0})_{2}-y_{2},\cdots,(x_{0})_{n}-y_{n})dy$
(using radial symmetry of $J$)
$\displaystyle=-\int_{\mathbb{R}^{n}}\tau_{\Omega}(y)J(x_{0}-R(y))dy$
$\displaystyle=-\int_{\mathbb{R}^{n}}\tau_{R(\Omega)}(y)J(x_{0}-y)dy\quad\hbox{(doing
a change of variable)}$ $\displaystyle=H^{J}_{R(\Omega)}(x_{0})$

Next we observe that $\tau_{\Omega}=-\tau_{R(\Omega)}=1$ in $\Omega\setminus
R(\Omega)$ and $\tau_{\Omega}=-\tau_{R(\Omega)}=-1$ in
$R(\Omega)\setminus\Omega$. Then we have,

(2.1) $\displaystyle 0$
$\displaystyle=H^{J}_{\Omega}(x_{0})-H^{J}_{\Omega}(R(x_{0}))$
$\displaystyle=H^{J}_{\Omega}(x_{0})-H^{J}_{R(\Omega)}(x_{0})$
$\displaystyle=\int_{\mathbb{R}^{n}}\bigg{(}\tau_{\Omega}(y)-\tau_{R(\Omega)}(y)\bigg{)}J(x_{0}-y)dy$
$\displaystyle=\int_{\Omega\setminus
R(\Omega)}J(x_{0}-y)dy-\int_{R(\Omega)\setminus\Omega}J(x_{0}-y)dy$
$\displaystyle=\int_{\Omega\setminus
R(\Omega)}\big{(}J(x_{0}-y)-J(x_{0}-R(y))dy$

From the construction, $x_{0}\in\Omega\setminus R(\Omega)$ and hence
$|x_{0}-y|\leq|x_{0}-R(y)|$ for all $y\in\Omega\setminus R(\Omega)$. Since $J$
is a strictly decreasing function, we have for all $y\in\Omega\setminus
R(\Omega)$, $\big{(}J(x_{0}-y)-J(x_{0}-R(y))$ is a positive quantity. That
implies $|\Omega\setminus R(\Omega)|=0$ and similarly
$|R(\Omega)\setminus\Omega|=0$.

Case (b): We see that $\pi_{\lambda}$ is orthogonal to $\partial\Omega$, which
implies that the vector $e=\langle 1,0,\cdots,0\rangle$ is tangent to
$\partial\Omega$ at the point $x_{0}$. Similarly, $e$ is tangent to $\partial
R(\Omega)$ at $x_{0}$. Since $H^{J}_{\Omega}$ is a constant function on
$\partial\Omega$, $H^{J}_{R(\Omega)}$ is constant on $\partial R(\Omega)$. So
the tangential derivative of $H^{J}_{\Omega}$, at the point $x_{0}$ will be
zero, i.e $\partial_{e}H^{J}_{\Omega}(x_{0})=\nabla H_{\Omega}(x_{0})\cdot
e=0$. Using similar reasons, $\partial_{e}H^{J}_{R(\Omega)}(x_{0})=0$. From
Lemma 2.1, we know that
$\partial_{e}H^{J}_{\Omega}(x_{0})=\lim_{\varepsilon\to
0}\partial_{e}H^{\varepsilon}_{\Omega}(x_{0})$. As
$H^{\varepsilon}_{J}(x_{0})=\int_{\mathbb{R}^{n}}\tau_{\Omega}(y)\phi_{\varepsilon}(|x_{0}-y|)dy,$
denoting $J_{\varepsilon}(x)=\phi_{\varepsilon}(|x|),$ we can write

$\partial_{e}H^{J}_{\varepsilon}(x_{0})=\int_{\mathbb{R}^{n}}\tau_{\Omega}(y)\nabla
J_{\varepsilon}(|x_{0}-y|)\cdot
e\,dy=\int_{\mathbb{R}^{n}}\tau_{\Omega}(y)\phi^{\prime}_{\varepsilon}(|x_{0}-y|)\frac{x_{0}-y}{|x_{0}-y|}\cdot
e\,dy$

Since $e=\langle 1,0,\cdots,0\rangle$,

$\partial_{e}H^{J}_{\varepsilon}(x_{0})=\int_{\mathbb{R}^{n}}\tau_{\Omega}(y)\phi^{\prime}_{\varepsilon}(|x_{0}-y|)\frac{(x_{0})_{1}-y_{1}}{|x-y|}\,dy$

Again using the fact
$\partial_{e}H^{J}_{\Omega}(x_{0})-\partial_{e}H^{J}_{R(\Omega)}(x_{0})=0$, we
have

$\displaystyle 0$ $\displaystyle=\lim_{\varepsilon\to
0}\int_{\mathbb{R}^{n}}\big{(}\tau_{\Omega}(y)-\tau_{R(\Omega)}(y))\phi^{\prime}_{\varepsilon}(|x_{0}-y|)\frac{(x_{0})_{1}-y_{1}}{|x_{0}-y|}\,dy$
$\displaystyle=\lim_{\varepsilon\to 0}\bigg{[}\int_{\Omega\setminus
R(\Omega)}\phi^{\prime}_{\varepsilon}(|x_{0}-y|)\frac{(x_{0})_{1}-y_{1}}{|x_{0}-y|}\,dy-\int_{R(\Omega)\setminus\Omega}\phi^{\prime}_{\varepsilon}(|x_{0}-y|)\frac{(x_{0})_{1}-y_{1}}{|x_{0}-y|}\,dy\bigg{]}$

Next we observe that $(x_{0})_{1}=\lambda$ and if $y\in\Omega\setminus
R(\Omega)$ then $y_{1}<\lambda$. On the other hand, if $y\in
R(\Omega)\setminus\Omega$ then $y_{1}>\lambda$. Hence

(2.2) $0=\lim_{\varepsilon\to 0}\bigg{[}\int_{\Omega\setminus
R(\Omega)}\phi^{\prime}_{\varepsilon}(|x_{0}-y|)\frac{|\lambda-
y_{1}|}{|x_{0}-y|}\,dy+\int_{R(\Omega)\setminus\Omega}\phi^{\prime}_{\varepsilon}(|x_{0}-y|)\frac{|\lambda-
y_{1}|}{|x_{0}-y|}\,dy\bigg{]}$

Since $\phi^{\prime}_{\varepsilon}(t)\leq 0$ for all $t$ and
$\phi_{\varepsilon}(t)=\phi(t)$ when $t>\varepsilon$, then
$\phi^{\prime}_{\varepsilon}(|x_{0}-y|)<0$ when $|x_{0}-y|>\varepsilon$ for
any $y\in\mathbb{R}^{n}$. After defining,

$E^{1}_{\varepsilon}=\\{y\in\Omega\setminus
R(\Omega):|x_{0}-y|>\varepsilon\\},E^{2}_{\varepsilon}=\\{y\in
R(\Omega)\setminus\Omega:|x_{0}-y|>\varepsilon\\},$

we see that $|E^{1}_{\varepsilon}|=|E^{2}_{\varepsilon}|=0$ for every
$\varepsilon>0$. That gives us $|\Omega\setminus
R(\Omega)|=|R(\Omega)\setminus\Omega|=0$. Then we see that, in both cases
$|\Omega\setminus R(\Omega)|=|R(\Omega)\setminus\Omega|=0$. But $\Omega$ has
$C^{1,\beta}$ boundary and hence $\Omega=R(\Omega)$. ∎

##  3\. Proof of Theorem 1.2

Since $J$ is supported on a finite set, precisely on $B_{r}$, the solution is
not always a ball. If we impose a condition that $\Omega$ is a connected set,
then the solution is a ball otherwise in general it will be a union of balls
of same size, where the balls are at least $r$ distance away from each other.
In the previous case, as the kernel is supported on $\mathbb{R}^{n}$, the
point $x_{0}$ (interior touching point or point of non-transversal
intersection) ‘can see’ the whole $\Omega\setminus R(\Omega)$. But here, it
‘can see’ only the part of $\Omega\setminus R(\Omega)$ which is inside
$B_{r}(x_{0})$. So here we may have different critical hyperplanes, for the
same direction, with respect to which different connected components of
$\Omega$ will be symmetric. We also need to use a covering type argument since
$\Omega$ is bounded.

###### Proof.

Since $J$ is $C^{\infty}$ everywhere in $\mathbb{R}^{n}$ except at $x=0$, and
is radially strictly decreasing to $0$ in $B_{r}$, we can repeat the
computation given in Lemma 2.1 and get a similar lemma which proves that
$H^{J}_{\Omega}$ is in $C^{1}(\partial\Omega)$. For that, we need to define a
function $\phi:[0,\infty)\to\mathbb{R}_{+}$ such that $J(x)=\phi(|x|)$. Then
we know that $\phi^{\prime}(t)<0$ in $(t,r)$ for all $0<t<r$. Suppose
$\phi_{\varepsilon}\in C^{\infty}_{c}([0,\infty))$ is such that
$\phi_{\varepsilon}\geq 0,\phi^{\prime}_{\varepsilon}\leq 0$ and it satisfies
the following estimates, for all $t>0$

$\begin{cases}t^{n-\alpha}\phi_{\varepsilon}(t)+t^{n-\alpha+1}|\phi^{\prime}_{\varepsilon}(t)|\leq
C(n,\alpha,\alpha_{1})\\\
|\phi^{\prime}_{\varepsilon}(t)|\to|\phi^{\prime}(t)|\quad\text{as}\leavevmode\nobreak\
\varepsilon\to 0\\\
\phi_{\varepsilon}(t)=\phi(t)\quad\text{if}\leavevmode\nobreak\
t>\varepsilon\end{cases}$

If we define

$H^{\varepsilon}_{\Omega}(x)=\int_{\mathbb{R}^{n}}\tau_{\Omega}(y)\phi_{\varepsilon}(|x-y|)dy,\quad
x\in\mathbb{R}^{n}$

then $H^{J}_{\Omega}\in C^{1}(\partial\Omega)$ and
$H^{\varepsilon}_{\Omega}\to H^{J}_{\Omega}$ in $C^{1}(\partial\Omega)$ as
$\varepsilon\to 0$ using the very similar steps as in the proof of Lemma 2.1.
Next we fix a direction vector $e$. Without loss of generality, we assume that
$e=\langle 1,0,\cdots,0\rangle$. Then Alexandrov’s moving plane method
provides the two situations with either interior touching or non-transversal
intersection. Let $x_{0}$ be one such point in $\partial\Omega$ and
$\pi_{\lambda}$ be the critical hyperplane corresponding to the direction $e$.
We will denote such type of point as ‘point of contact’. Let $R_{\lambda}$ be
the reflection operator with respect to $\pi_{\lambda}$, which will be denoted
as operator $R$ for simplicity. Next we define,

${\pi_{\lambda}}_{+}=\\{x:x_{1}>\lambda\\}\quad\text{and}\leavevmode\nobreak\
\quad{\pi_{\lambda}}_{-}=\\{x:x_{1}<\lambda\\}$

We also define

$\begin{cases}\Omega_{+}=\Omega\cap{\pi_{\lambda}}_{+},\quad\text{and}\leavevmode\nobreak\
\quad\Omega_{-}=\Omega\cap{\pi_{\lambda}}_{-}\\\
\partial\Omega_{+}=\partial\Omega\cap{\pi_{\lambda}}_{+},\quad\text{and}\leavevmode\nobreak\
\quad\partial\Omega_{-}=\partial\Omega\cap{\pi_{\lambda}}_{-}\end{cases}$

We know that if $x_{0}$ is an interior touching point then
$x_{0}\in\partial\Omega_{-}$. On the other hand, if $x_{0}$ is a non-
transversal intersection then $x_{0}\in\partial\Omega\cap\pi_{\lambda}$. Next
if we repeat all the computations as in the proof of Theorem 1.1, we get the
following two equations similar to (2.1) and (2.2),

$0=\int_{\Omega\setminus R(\Omega)}\big{(}J(x_{0}-y)-J(x_{0}-R(y))dy$

and,

$0=\lim_{\varepsilon\to 0}\bigg{[}\int_{\Omega\setminus
R(\Omega)}\phi^{\prime}_{\varepsilon}(|x_{0}-y|)\frac{|\lambda-
y_{1}|}{|x_{0}-y|}\,dy+\int_{R(\Omega)\setminus\Omega}\phi^{\prime}_{\varepsilon}(|x_{0}-y|)\frac{|\lambda-
y_{1}|}{|x_{0}-y|}\,dy\bigg{]}$

where $y_{1}=y\cdot e$. Now, in the interior case,
$x_{0}\in\partial\Omega_{-}$. Then for any $y\in\Omega\setminus R(\Omega)$,
$|x_{0}-y|<|x_{0}-R(y)|$. As $J$ is strictly decreasing and $J$ has support in
$B_{r}$, we have,

(3.1) $\displaystyle|(\Omega\Delta R(\Omega))\cap B_{r}(x_{0})|=0$

On the other hand, in the non-transversal case,
$\phi^{\prime}_{\varepsilon}\leq 0$ for all $t\geq 0$. Since
$\phi_{\varepsilon}(t)=\phi(t)$ when $t>\varepsilon$ then
$\phi^{\prime}_{\varepsilon}<0$ in the interval $(\varepsilon,r)$. Again
defining,

$E^{1}_{\varepsilon}=\\{y\in\Omega\setminus
R(\Omega):|x_{0}-y|>\varepsilon\\},E^{2}_{\varepsilon}=\\{y\in
R(\Omega)\setminus\Omega:|x_{0}-y|>\varepsilon\\},$

we see that

$|E^{1}_{\varepsilon}\cap B_{r}(x_{0})|=|E^{2}_{\varepsilon}\cap
B_{r}(x_{0})|=0$

for every $\varepsilon>0$. That gives us $|\Omega\setminus
R(\Omega)|=|R(\Omega)\setminus\Omega|=0$ inside $B_{r}(x_{0})$ and hence (3.1)
is true. From (3.1), we have

(3.2) $|(\Omega\setminus R(\Omega))\cap
B_{r}(x_{0})|=|(R(\Omega)\setminus\Omega)\cap B_{r}(x_{0})|=0$

As $\partial\Omega$ is $C^{1,\beta}$ for $\beta>0$ then for any
$x_{0}\in\partial\Omega$, $|B_{\varepsilon}(x_{0})\cap\Omega|>0$ for every
$\varepsilon>0$. Then (3.2) can only happen, when $\Omega=R(\Omega)$ inside
$B_{r}(x_{0})$. Or we can say that their boundaries coincide inside
$B_{r}(x_{0})$. Let $\Omega$ be a union of disjoint connected components. Then
$x_{0}$ lies on the boundary of one such component, say $\mathcal{N}_{x}$.
Again if $x_{0}$ is interior touching point then
$x_{0}\in\partial{\mathcal{N}_{x}}_{-}$. On the other hand, if it is a non-
transversal intersection point, then
$x_{0}\in\partial\mathcal{N}_{x}\cap\pi_{\lambda}$. We already showed that,
since $x_{0}$ is a point of contact, $\Omega=R(\Omega)$ in $B_{r}(x_{0})$ and
hence the boundaries coincide. If $\partial{\mathcal{N}_{x}}_{-}\subseteq
B_{r}(x_{0})$ then we can say that

(3.3) $\Omega=R(\Omega),\quad\text{everywhere in }\leavevmode\nobreak\
{\mathcal{N}_{x}}.$

Otherwise, we choose a point $z_{0}\in\partial{\mathcal{N}_{x}}_{-}\cap
B_{r}(x_{0}))$ such that $|x_{0}-z_{0}|\geq r/2$. Notice that $z_{0}$ is a
point of interior touching with respect to the hyperplane $\pi_{\lambda}$.
Using similar analysis as in interior touching, we have

$|(\Omega\setminus R(\Omega))\cap
B_{r}(z_{0})|=|(R(\Omega)\setminus\Omega)\cap B_{r}(z_{0})|=0$

that implies $\Omega=R(\Omega)$ inside $B_{r}(z_{0})$. Again if
$\partial{\mathcal{N}_{x}}_{-}\subseteq B_{r}(x_{0})\cup B_{r}(z_{0})$ then we
have (3.3). Otherwise, we choose a point
$z_{1}\in\partial{\mathcal{N}_{x}}_{-}\cap B_{r}(z_{0})$ such that
$|x_{0}-z_{1}|\geq r/2$ and $|z_{1}-z_{0}|\geq r/2$. Likewise, in the next
iteration, we choose a point $z_{2}\in\partial{\mathcal{N}_{x}}_{-}\cap
B_{r}(z_{1})$ such that $|z_{2}-x_{0}|>r/2,|z_{2}-z_{0}|>r/2$ and
$|z_{2}-z_{1}|>r/2$ if $\partial{\mathcal{N}_{x}}_{-}$ is not contained in the
union of those balls, i.e, $B_{r}(x_{0})\cup B_{r}(z_{0})\cup B_{r}(z_{1})$.
This construction allows only finitely many iterations since $\partial\Omega$
and hence $\partial\mathcal{N}_{x}$ is compact. Hence, we should get (3.3).
That further implies that ${\mathcal{N}_{x}}_{+}=R({\mathcal{N}_{x}}_{-})$ and
$\mathcal{N}_{x}$ is symmetric about $\pi_{\lambda}$.

Next we consider any two disjoint connected components of $\Omega$. Let they
be $\mathcal{P}$ and $\mathcal{Q}$. Then we define,

$d(\mathcal{P},\mathcal{Q})=\inf\\{\operatorname{dist}(x,y):x\in\partial\mathcal{P},y\in\partial\mathcal{Q}\\}.$

Similarly we can define,

$d(\mathcal{P}_{-},\mathcal{Q}_{-})=\inf\\{\operatorname{dist}(x,y):x\in\partial\mathcal{P}_{-},y\in\partial\mathcal{Q}_{-}\\},$

and

$d(\mathcal{P}_{+},\mathcal{Q}_{+})=\inf\\{\operatorname{dist}(x,y):x\in\partial\mathcal{P}_{+},y\in\partial\mathcal{Q}_{+}\\}.$

In all these definitions above, if any set, say $\mathcal{P}_{+}$, is empty
then the distance would be $+\infty$. Now suppose there is a component
$\mathcal{P}$ of $\Omega$ such that $d(\mathcal{P},\mathcal{N}_{x})<r$ then
one can see that $d(\mathcal{P}_{-},{\mathcal{N}_{x}}_{-})<r$. From the
definition of infimum, there exit
$a\in\partial{\mathcal{N}_{x}}_{-},b\in\mathcal{P}_{-}$ such that $d(a,b)<r$.
Since $\Omega$ has $C^{1,\beta}$ boundary, we must have
$|B_{r}(a)\cap\mathcal{P}_{-}|>0$. Now we already know that $a$ is an interior
point and $\Omega=R(\Omega)$ everywhere in $B_{r}(a)$. Hence
$\Omega=R(\Omega)$ inside $B_{r}(a)\cap\mathcal{P}_{-}$ and hence $b$ is an
interior point. Now we extend the analysis on $b$ to the whole
$\mathcal{P}_{-}$ and get

$\Omega=R(\Omega),\quad\hbox{inside}\leavevmode\nobreak\
\mathcal{N}_{x}\cup\mathcal{P}.$

For a component $\mathcal{Q}$, if $d(\mathcal{Q},\mathcal{N}_{x})\geq r$ then
we can not extend analysis on $\mathcal{Q}$ from $\mathcal{N}_{x}$ directly.
On the other hand, if $d(\mathcal{Q},\mathcal{P})<r$ then we can extend the
analysis from $\mathcal{N}_{x}$ to $\mathcal{Q}$ via $\mathcal{P}$. In that
way, we say that $\mathcal{N}_{x}$ can ‘influence’ $\mathcal{Q}$. After doing
analysis on each components, we get $\Omega$ to be union of sets of the
following two types, which are at least at a distance $r$ from each other,

  1. (i)

A set, $S$ which is symmetric about $\pi_{\lambda}$. The set $S$ may not be
connected itself, rather it is a union of connected symmetric components of
the form ${\mathcal{N}_{x}}$ described as before, where one component can
‘influence’ other components.

  2. (ii)

A set $\mathcal{NSP}$ which is not symmetric about $\pi_{\lambda}$. Again,
this set $\mathcal{NSP}$ may not be connected itself, rather it is union of
connected components which have no contact point. But again, one component can
‘influence’ the other components.

In the above construction, any such two sets (either symmetric or non-
symmetric) have mutual distance at least $r$. One can see that there can be
only finitely many such sets as $\Omega$ is bounded. Now we denote the
corresponding critical hyperplane $\pi_{\lambda}$ to be
$\pi^{e}_{\lambda^{(1)}}$. With respect to this hyperplane, we get finitley
many symmetric sets of the form $S$ and finitely many non-symmetric sets of
the form $\mathcal{NSP}$. We write union of symmetric sets, in the first
iteration as $\Omega_{s^{(e,1)}}=\cup_{i=1}^{m_{1}}S^{(e,1)}_{i}$ and union of
non-symmetric sets as
$\Omega_{n^{(e,1)}}=\cup_{i=1}^{l_{1}}\mathcal{NSP}^{(e,1)}_{i}$. We denote
$\Omega^{(e,1)}=\Omega=\Omega_{s^{(e,1)}}\cup\Omega_{n^{(e,1)}}$. In the next
iteration, we only consider $\Omega_{n^{(e,1)}}$ and denote
$\Omega^{(e,2)}=\Omega_{n^{(e,1)}}$. We already know each set in
$\Omega^{e,2}$ does not have any contact point with respect to
$\pi^{e}_{\lambda^{(1)}}$. Therefore we move the plane again until we get a
new critical hyperplane $\pi^{e}_{\lambda^{(2)}}$. With respect to this
hyperplane, we again have
$\Omega^{(e,2)}=\Omega_{s^{(e,2)}}\cup\Omega_{n^{(e,2)}}$. We continue this
process for finitely many times as $\Omega$ is bounded and get $\Omega$ to be
a symmetric in the direction of $e$ but for different critical hyperplanes. If
the number of iteration or critical hyeprplane is $\Lambda^{e}$ then,

$\Omega=\cup^{\Lambda^{e}}_{j=1}\cup^{m_{j}}_{i=1}S^{(e,j)}_{i}$

where $S^{(e,j)}_{i}$ is one of the $m_{j}$ number of symmetric sets (in
spirit of definition given in (i)) with respect to the critical hyeprplane
$\pi^{e}_{\lambda^{(j)}}$.

Next we change direction to any $v\in\mathbb{S}^{n-1}$ such that $v\neq e$. We
use the Alexandrov’s principle and get a critical hyperplane in the first
iteration for this vector $v$. Let us denote that hyperplane as
$\pi^{v}_{\lambda^{(1)}}$. Consider a subset, $S^{e}$ of $\Omega$, which is
symmetric with respect to the direction $e$ (in sense of definition (i)) and
boundary of that set contains a point of contact with respect to
$\pi^{v}_{\lambda^{(1)}}$. Hence the component (of $S^{e}$) that contains the
point of contact can be proved to be symmetric with respect to
$\pi^{v}_{\lambda^{(1)}}$ using the earlier analysis. Now remember that
$S^{e}$ contains different connected components where one component can
‘influence’ another component. Then it evident that $S^{e}$ is symmetric with
respect to $\pi^{v}_{\lambda^{(1)}}$. Similarly all other sets of the form
$S^{(e,j)}_{i}$ which are symmetric in the direction $e$, can be proved to be
symmetric in the direction $v$ with respect to some critical hyperplane
$\pi^{v}_{\lambda^{(k)}}$, where $k=1,2,\cdots,\Lambda^{v}$ for some
$0<\Lambda^{v}<\infty$. As $v$ is any vector, each set of the form $S^{e}$ is
a ball. Then $\Omega$ is a union of balls of same radii, which are at least
$r$ distance away from each other. ∎

##  4\. Proof of Theorem 1.3

As $J$ has jump at $\partial\Omega$ we can’t directly use the computations as
in the proof of Theorem 1.1. We can think $J(x)=\tilde{J}(x)\chi_{B_{r}}(x)$,
where $\tilde{J}$ satisfies all the conditions of the kernel given in Theorem
1.1. Let us define $\phi,\phi_{\varepsilon}:[0,\infty)\to\mathbb{R}_{+}$
exactly as in the proof of Theorem 1.1 so that $\tilde{J}(x)=\phi(|x|)$ and
$\phi_{\varepsilon},\phi^{\prime}_{\varepsilon}$ converges to
$\phi,\phi^{\prime}$ respectively. We also introduce another sequence of
functions. First consider $\mathcal{T}:(-\infty,\infty)\to\mathbb{R}_{-}$ be
such that $\mathcal{T}$ is $C^{\infty}_{c}(-\infty,\infty)$ with support in
$(-r/2,r/2)$ and $\int_{\mathbb{R}}\mathcal{T}(t)dt=-1$. Now define the
function $g:[0,\infty)\to\mathbb{R}_{+}$ such that $g(t)=1$ when $0\leq t\leq
r$. Then it is evident that $\chi_{B_{r}}(x)=g(|x|)$. Next we define the
sequence $g_{\varepsilon}:[0,\infty)\to\mathbb{R}_{+}$ such that
$g_{\varepsilon}(0)=1$ and
$g^{\prime}_{\varepsilon}(t)=\frac{1}{\varepsilon}\mathcal{T}\bigg{(}\frac{t-r}{\varepsilon}\bigg{)}$.
Now we define,
$\Lambda_{\varepsilon}:[0,\infty)\to\mathbb{R}_{+}:t\mapsto\phi_{\varepsilon}(t)g_{\varepsilon}(t)$.

Before we introduce the main result of this section, we want to prove a
supporting lemma. For this result to be true we want the boundary of $\Omega$
to be $C^{1}$ with uniformly bounded $C^{1}$-norm as mentioned in Theorem 1.3.

######  Lemma 4.1.

Let $\Omega$ be a bounded set with $C^{1}$-boundary as mentioned in Theorem
1.3. Consider the following function,

$H^{c}_{\Omega}(x)=\int_{\mathbb{R}^{n}}\chi_{\Omega}(y)\chi_{B_{r}(x)}(y)dy$

for all $x\in\mathbb{R}^{n}$. Then $H^{c}_{\Omega}(x)$ has directional
derivatives at every $x\in\mathbb{R}^{n}$.

###### Proof.

First we define,

$H^{c}_{\varepsilon}(x)=\int_{\Omega}g_{\varepsilon}(|x-y|)dy$

then, for $e\in T_{x}(\partial\Omega)$, since $g_{\varepsilon}$ is
$C^{\infty}$, we can write,

$\nabla H^{c}_{\varepsilon}(x)\cdot e=\int_{\Omega}\nabla
g_{\varepsilon}(|x-y|)\cdot edy.$

Consider a point $x\in\partial\Omega$, $e$ is any tangent vector of $x$. Next
we define,

$F(x,e,\lambda)=\int_{\Omega\cap\partial B_{\lambda}(x)}(e\cdot
e_{y})d\sigma(y),$

where $e_{y}$ is the unit normal vector at each $y\in\Omega\cap\partial
B_{\lambda}(x)$. Then we have,

$\displaystyle\nabla H^{c}_{\varepsilon}(x)\cdot e$
$\displaystyle=\int_{\Omega}\nabla g_{\varepsilon}(|x-y|)\cdot edy$
$\displaystyle=\int_{\Omega}g^{\prime}_{\varepsilon}(|x-y|)\frac{(x-y)}{|x-y|}\cdot
edy$ $\displaystyle=\int^{\infty}_{0}\int_{\Omega\cap\partial
B_{\lambda}(x)}g^{\prime}_{\varepsilon}(\lambda)(e_{y}\cdot
e)d\sigma(y)d\lambda\quad[\hbox{where}\leavevmode\nobreak\ x-y=\lambda e_{y}]$
$\displaystyle=\int^{\infty}_{0}g^{\prime}_{\varepsilon}(\lambda)F(x,e,\lambda)d\lambda$
$\displaystyle=\int^{\infty}_{0}\frac{1}{\varepsilon}\mathcal{T}\bigg{(}\frac{\lambda-r}{\varepsilon}\bigg{)}F(x,e,\lambda)d\lambda$

Now we want to prove that $F(x,e,\lambda)$ is a continuous function of
$\lambda$ when $0\leq\lambda<r+\delta$. After rotation and translation, we can
assume $x=0$, $\nabla f^{x}(0)=0$. For simplicity, we will now denote
$f^{x}=f$. Next we define
$B^{\prime}_{\rho}=\\{y^{\prime}:|y^{\prime}|<\rho\\}$,
$L_{\rho}=(-\rho,\rho)\times B^{\prime}_{\rho}$,

$\partial\Omega\cap L_{r+\delta}=\\{(y^{\prime},f(y^{\prime})):y^{\prime}\in
B^{\prime}_{r+\delta}\\}=(Id\times
f)\big{(}B^{\prime}_{r+\delta}\big{)}=G_{r+\delta},$

and

$\Omega\cap L_{r+\delta}=\\{(y^{\prime},y_{n}):y_{n}>f(y^{\prime})\\}.$

We have, for some $\lambda_{1}<\lambda_{2}<r+\delta$,

$F(0,e,\lambda_{2})-F(0,e,\lambda_{1})=\int_{\Omega\cap\partial
B_{\lambda_{2}}}(e\cdot e_{y})d\sigma(y)-\int_{\Omega\cap\partial
B_{\lambda_{1}}}(e\cdot e_{y})d\sigma(y)$

If we define $h(y)=e\cdot y$ for any $y\in\mathbb{R}^{n}$ then $h(y)$ is a
harmonic function and $\nabla_{y}h=e$. Next we observe that since
$\left\|\nabla f\right\|<1$, then for any tangent direction $\nu$ at the point
$0$, $f$ intersects $\partial B_{\lambda}$ at only one point. That further
implies

$\mathcal{H}^{n-1}(\partial\Omega\cap\partial
B_{\lambda}(x))=0\quad\hbox{when}\leavevmode\nobreak\ \lambda<r+\delta.$

Next using the Gauss divergence theorem and proper orientation of the normal
vector, we see that

$\int_{\Omega\cap\partial B_{\lambda_{2}}}(e\cdot
n_{y})d\sigma(y)+\int_{\Omega\cap\partial B_{\lambda_{1}}}(e\cdot
n_{y})d\sigma(y)+\int_{\partial\Omega\cap(B_{\lambda_{2}}\setminus
B_{\lambda_{1}})}(e\cdot n_{y})d\sigma(y)=\int_{\omega}\Delta hdy=0,$

where $\omega$ is the region bounded by the surfaces $\Omega\cap\partial
B_{\lambda_{2}},\Omega\cap\partial
B_{\lambda_{1}},\Omega\cap(B_{\lambda_{2}}\setminus B_{\lambda_{1}})$ and
$n_{y}$ is the normal vector with the correct orientation. Therefore,

$\displaystyle F(0,e,\lambda_{2})-F(0,e,\lambda_{1})$
$\displaystyle=-\int_{\partial\Omega\cap(B_{\lambda_{2}}\setminus
B_{\lambda_{1}})}(e\cdot n_{y})d\sigma(y)$
$\displaystyle=-\int_{\partial\Omega\cap B_{\lambda_{2}}}(e\cdot
n_{y})d\sigma(y)+\int_{\partial\Omega\cap B_{\lambda_{1}}}(e\cdot
n_{y})d\sigma(y).$

Next we assume that $P_{\lambda}$ is the projection of $\partial\Omega\cap
B_{\lambda}$ on the $x_{n}=0$ hyperplane. Then $P_{\lambda_{1}}\subseteq
P_{\lambda_{2}}\subseteq B^{\prime}_{r+\delta}$. Using $n_{y}=\frac{(-\nabla
f(y^{\prime}),1)}{\sqrt{1+|\nabla f(y^{\prime})|^{2}}}$ and $e\cdot e_{n}=0$,
we have $e\cdot n_{y}=\frac{-\nabla f(y^{\prime})}{\sqrt{1+|\nabla
f(y^{\prime})|^{2}}}$ and $d\sigma(y)=\sqrt{1+|\nabla
f(y^{\prime})|^{2}}dy^{\prime}$

$\displaystyle F(0,e,\lambda_{2})-F(0,e,\lambda_{1})$
$\displaystyle=-\int_{P_{\lambda_{2}}}(-\nabla
f(y^{\prime}))dy^{\prime}+\int_{P_{\lambda_{1}}}(-\nabla
f(y^{\prime}))dy^{\prime}$ $\displaystyle=\int_{P_{\lambda_{2}}\setminus
P_{\lambda_{1}}}\nabla f(y^{\prime})dy^{\prime}$

That implies,

$\displaystyle\big{|}F(0,e,\lambda_{2})-F(0,e,\lambda_{1})\big{|}$
$\displaystyle\leq\int_{P_{\lambda_{2}}\setminus P_{\lambda_{1}}}|\nabla
f(y^{\prime})|dy^{\prime}$
$\displaystyle\leq\mathcal{H}^{n-1}\big{(}P_{\lambda_{2}}\setminus
P_{\lambda_{1}}\big{)}=\mathcal{H}^{n-1}(P_{\lambda_{2}})-\mathcal{H}^{n-1}(P_{\lambda_{1}})$

Next we fix a unit tangent vector $\nu$ of point $x=0$. We want to find
$\mathcal{H}^{1}(P_{\lambda_{2}}\cap\nu)-\mathcal{H}^{1}(P_{\lambda_{1}}\cap\nu)$.

Let $Q$ be the intersection point of $f$ with $\partial B_{\lambda_{1}}$ and
$P$ be the intersection of $f$ with $\partial B_{\lambda_{2}}$ along the
direction $\nu$. Next we define
$G(x^{\prime})=\sqrt{(x^{\prime})^{2}+f(x^{\prime})^{2}}$ and along that
tangent direction $\nu$, we use $\rho$ to denote the distance (from $0$) of a
point in $x_{n}=0$ hyperplane. Then $x^{\prime}=\rho\nu$ and $Q$ corresponds
to $G(x^{\prime}_{1})=\lambda_{1}$, $P$ corresponds to
$G(x^{\prime}_{2})=\lambda_{2}$. Or we can also say that

$G(\rho_{1};\nu)=\sqrt{\rho^{2}_{1}+f(\rho_{1}\nu)^{2}}=\lambda_{1},\quad
G(\rho_{2};\nu)=\sqrt{\rho^{2}_{2}+f(\rho_{2}\nu)^{2}}=\lambda_{2}$

Since for a fixed $\nu$, there is a unique $\rho$ for each $\lambda$ and hence
we we can write $\rho=Gn(\lambda;\nu)$. Then

$G(Gn(\lambda;\nu);\nu)=\lambda.$

Next differentiating w.r.t $\lambda$,

$\frac{dG}{d\rho}\frac{d\rho}{d\lambda}=1.$

Now

$\frac{dG}{d\rho}=\nabla_{x^{\prime}}G\cdot\frac{x^{\prime}}{\rho}$

where $\nabla_{x^{\prime}}G=\frac{x^{\prime}+f(x^{\prime})\nabla
f(x^{\prime})}{G(x^{\prime})}$. Using the fact that $|\nabla
f(x^{\prime})|\leq M$, we have $|f(x^{\prime})\nabla f(x^{\prime})\cdot
x^{\prime}|\leq M^{2}|x^{\prime}|^{2}$. That further implies
$\frac{dG}{d\rho}\geq\frac{(1-M^{2})|x^{\prime}|^{2}}{|x^{\prime}|G(x^{\prime})}=\frac{(1-M^{2})\rho}{G(x^{\prime})}=\frac{(1-M^{2})\rho}{\lambda}$.
Then we have

$\frac{d\rho}{d\lambda}\leq\frac{\lambda}{(1-M^{2})\rho}$

Next, we notice that $\rho\geq\frac{\lambda}{\sqrt{1+M^{2}}}$ which implies

$\frac{d\rho}{d\lambda}\leq\frac{\sqrt{1+M^{2}}}{(1-M^{2})}.$

Hence we have, if $(\lambda_{2}-\lambda_{1})$ is small,

$\mathcal{H}^{1}(P_{\lambda_{2}}\cap\nu)-\mathcal{H}^{1}(P_{\lambda_{2}}\cap\nu)=Gn(\lambda_{2};\nu)-Gn(\lambda_{1};\nu)\leq
C(\lambda_{2}-\lambda_{1})$

This is true for any tangent vector $\nu$ which further implies that the
distance between the $\partial P_{\lambda_{2}}$ and $\partial P_{\lambda_{1}}$
is always bounded by $C(\lambda_{2}-\lambda_{1})$ in any direction, when
$(\lambda_{2}-\lambda_{1})$ is small. Furthermore, we have

$\displaystyle\mathcal{H}^{n-1}(P_{\lambda_{2}})-\mathcal{H}^{n-1}(P_{\lambda_{2}})$
$\displaystyle=\int_{B^{n-2}}\int^{Gn(\lambda_{2};\nu)}_{Gn(\lambda_{1};\nu)}\rho^{n-2}d\rho\,d\Theta(\nu)$
$\displaystyle=\int_{B^{n-2}}\frac{1}{n-1}\big{[}Gn(\lambda_{2};\nu)^{n-1}-Gn(\lambda_{1};\nu)^{n-1}]\,d\Theta(\nu)$
$\displaystyle\leq C_{n}(\lambda_{2}-\lambda_{1})$

As a consequence, $F(0,e,\lambda)$ is continuous in $\lambda$ and the modulus
of continuity does not depend on $x,e$. Then we can write,

$\displaystyle\bigg{|}\int^{\infty}_{0}\frac{1}{\varepsilon}\mathcal{T}\bigg{(}\frac{\lambda-r}{\varepsilon}\bigg{)}F(x,e,\lambda)\,d\lambda+F(x,e,r)\bigg{|}$
$\displaystyle=\bigg{|}\int^{\infty}_{0}\frac{1}{\varepsilon}\mathcal{T}\bigg{(}\frac{\lambda-r}{\varepsilon}\bigg{)}F(x,e,\lambda)\,d\lambda-\int^{\infty}_{0}\frac{1}{\varepsilon}\mathcal{T}\bigg{(}\frac{\lambda-r}{\varepsilon}\bigg{)}F(x,e,r)d\lambda\bigg{|}$
$\displaystyle=\bigg{|}\int^{\infty}_{0}\frac{1}{\varepsilon}\mathcal{T}\bigg{(}\frac{\lambda-r}{\varepsilon}\bigg{)}\big{(}F(x,e,\lambda)-F(x,e,r)\big{)}\,d\lambda\bigg{|}$
$\displaystyle\leq\int_{|\lambda-r|\leq
r\varepsilon/2}\bigg{|}\frac{1}{\varepsilon}\mathcal{T}\bigg{(}\frac{\lambda-r}{\varepsilon}\bigg{)}\big{(}F(x,e,\lambda)-F(x,e,r)\big{)}\bigg{|}\,d\lambda$
$\displaystyle\quad+\int_{|\lambda-r|>r\varepsilon/2}\bigg{|}\frac{1}{\varepsilon}\mathcal{T}\bigg{(}\frac{\lambda-r}{\varepsilon}\bigg{)}\big{(}F(x,e,\lambda)-F(x,e,r)\big{)}\bigg{|}\,d\lambda$

Since $F(x,e,\lambda)$ is uniformly continuous in the interval
$[0,r+\delta/2]$ then for small $\varepsilon_{0}$, as $\varepsilon\to 0$,

$\int^{\infty}_{0}\frac{1}{\varepsilon}\mathcal{T}\bigg{(}\frac{\lambda-r}{\varepsilon}\bigg{)}F(x,e,\lambda)\,d\lambda\to
F(x,e,r)=-\int_{\Omega\cap\partial B_{r}(x)}(e_{y}\cdot e)\,d\sigma(y).$

Then $H^{c}_{\varepsilon}$ converges in $C^{1}(\partial\Omega)$ norm. ∎

######  Remark 4.2.

Without the given boundary conditions, as described in Theorem 1.3, curvature
function may not be differentiable. Let us consider the following example.
Suppose $0\in\partial\Omega$ and the boundary function $f$ follows some part
of the sphere $\partial B_{r}$. Then we can see that $F$ is not continuous at
the point $\lambda=r$.

######  Lemma 4.3.

Let $\Omega$ be a bounded set with $C^{1}$-boundary as mentioned in Theorem
1.3. Consider the following function,

$H^{c}_{\Omega}(x)=\int_{\mathbb{R}^{n}}\tau_{\Omega}(y)\chi_{B_{r}(x)}(y)dy$

for all $x\in\mathbb{R}^{n}$. Then $H^{c}_{\Omega}$ is in
$C^{1}(\partial\Omega)$.

###### Proof.

This proof is a consequence of Lemma 4.1 after defining

$F(x,e,\lambda)=\int_{\Omega^{c}\cap\partial B_{\lambda}(x)}(e\cdot
e_{y})d\sigma(y)-\int_{\Omega\cap\partial B_{\lambda}(x)}(e\cdot
e_{y})d\sigma(y),$

where $e_{y}$ is the unit normal vector at each $y\in\Omega\cap\partial
B_{\lambda}(x)$.

∎

######  Lemma 4.4.

We consider $\Lambda_{\varepsilon}$ which is defined as above. Then we define

$H^{\varepsilon}_{\Omega}(x)=\int_{\mathbb{R}^{n}}\tau_{\Omega}(y)\Lambda_{\varepsilon}(|x-y|)dy,$

$x\in\mathbb{R}^{n}$. Next we assume that $\Omega$ is nonempty bounded open
set with $C^{1}$-boundary as mentioned in Theorem 1.3. In addition to that,
$\partial\Omega$ is $C^{1,\beta}$ for some $\beta=\min\\{\alpha-1,0\\}$. Then
$H^{J}_{\Omega}\in C^{1}(\partial\Omega)$ and $H^{\varepsilon}_{\Omega}\to
H^{J}_{\Omega}$ in $C^{1}(\partial\Omega)$ as $\varepsilon\to 0$.

###### Proof.

Using the definition of $H^{\varepsilon}_{\Omega}$, we compute, for some $e\in
T_{x}(\partial\Omega)$,

$\nabla H^{\varepsilon}_{\Omega}(x)\cdot
e=\int_{\mathbb{R}^{n}}\tau_{\Omega}(y){\phi_{\varepsilon}}(|x-y|)\nabla
g_{\varepsilon}(|x-y|)\cdot
e\,dy+\int_{\mathbb{R}^{n}}\tau_{\Omega}(y)(\nabla{\phi_{\varepsilon}}(|x-y|)\cdot
e)g_{\varepsilon}(|x-y|)\,dy$

We notice that

$\int_{\mathbb{R}^{n}}\tau_{\Omega}(y)(\nabla{\phi_{\varepsilon}}(|x-y|)\cdot
e)dy\to\int_{\mathbb{R}^{n}}\tau_{\Omega}(y)(\nabla{\phi}(|x-y|)\cdot e)dy$

uniformly in $C^{0}(\partial\Omega)$ using Lemma 2.1. Then we have,

$\displaystyle\bigg{|}\int_{\mathbb{R}^{n}}\tau_{\Omega}(y)(\nabla{\phi_{\varepsilon}}(|x-y|)\cdot
e)dy-\int_{\mathbb{R}^{n}}\tau_{\Omega}(y)(\nabla{\phi_{\eta}}(|x-y|)\cdot
e)dy\bigg{|}$
$\displaystyle=\bigg{|}\int_{\mathbb{R}^{n}}\tau_{\Omega}(y)[(\nabla{\phi_{\varepsilon}}(|x-y|)-\nabla{\phi_{\eta}}(|x-y|))\cdot
e]dy\bigg{|}$
$\displaystyle=\bigg{|}\int_{\mathbb{R}^{n}}\tau_{\Omega}(z+x)[(\nabla{\phi_{\varepsilon}}(|z|)-\nabla{\phi_{\eta}}(|z|))\cdot
e]dz\bigg{|}$
$\displaystyle=\bigg{|}\int_{\mathbb{R}^{n}}\tau_{\Omega}(z+x)(\phi_{\varepsilon}^{\prime}(|z|)-\phi^{\prime}_{\eta}(|z|))\bigg{(}\frac{z\cdot
e}{|z|}\bigg{)}dz\bigg{|}$
$\displaystyle=\bigg{|}\int_{B_{\eta}}\tau_{\Omega}(z+x)(\phi_{\varepsilon}^{\prime}(|z|)-\phi^{\prime}_{\eta}(|z|))\bigg{(}\frac{z\cdot
e}{|z|}\bigg{)}dz\bigg{|}\to 0,\quad\hbox{when}\leavevmode\nobreak\
\varepsilon,\eta\to 0.$

In the last line, $B_{\eta}$ denotes the ball of radius $\eta$ centered at $0$
and we assume $\eta<\varepsilon$. Then we use the fact that
$\phi_{\varepsilon}(t)=\phi_{\eta}(t)$ when $t>\varepsilon$. Next from the
uniform convergence of we have

$\bigg{|}\int_{B_{\eta}}\tau_{\Omega}(z+x)(\phi_{\varepsilon}^{\prime}(|z|)-\phi^{\prime}_{\eta}(|z|))\bigg{(}\frac{z\cdot
e}{|z|}\bigg{)}dz\bigg{|}\to 0$

Next we define
$A_{\varepsilon}=\\{z:0<g_{\varepsilon}(|z|)<1\\}=B_{r(1+\eta/2)}\setminus
B_{r(1-\eta/2)}$. Since $g_{\varepsilon}\to g$, then $|A_{\varepsilon}|\to 0$.
Then we have, for very small $\varepsilon,\eta>0$ and $\eta>\varepsilon$,

$\displaystyle\bigg{|}\int_{\mathbb{R}^{n}}\tau_{\Omega}(y)(\nabla{\phi_{\varepsilon}}(|x-y|)\cdot
e)g_{\varepsilon}(|x-y|)\,dy-\int_{\mathbb{R}^{n}}\tau_{\Omega}(y)(\nabla{\phi_{\eta}}(|x-y|)\cdot
e)g_{\eta}(|x-y|)\,dy\bigg{|}$
$\displaystyle=\bigg{|}\int_{\mathbb{R}^{n}}\tau_{\Omega}(x+z)(\phi_{\varepsilon}^{\prime}(|z|)g_{\varepsilon}(|z|)-\phi^{\prime}_{\eta}(|z|)g_{\eta}(|z|))\bigg{(}\frac{z\cdot
e}{|z|}\bigg{)}dz\bigg{|}$
$\displaystyle\leq\bigg{|}\int_{\mathbb{R}^{n}\setminus(B_{\eta}\cup
A_{\eta})}\tau_{\Omega}(x+z)(\phi_{\varepsilon}^{\prime}(|z|)g_{\varepsilon}(|z|)-\phi^{\prime}_{\eta}(|z|)g_{\eta}(|z|))\bigg{(}\frac{z\cdot
e}{|z|}\bigg{)}dz\bigg{|}$
$\displaystyle\qquad+\bigg{|}\int_{B_{\eta}}\tau_{\Omega}(x+z)(\phi_{\varepsilon}^{\prime}(|z|)g_{\varepsilon}(|z|)-\phi^{\prime}_{\eta}(|z|)g_{\eta}(|z|))\bigg{(}\frac{z\cdot
e}{|z|}\bigg{)}dz\bigg{|}$
$\displaystyle\qquad+\bigg{|}\int_{A_{\eta}}\tau_{\Omega}(x+z)(\phi_{\varepsilon}^{\prime}(|z|)g_{\varepsilon}(|z|)-\phi^{\prime}_{\eta}(|z|)g_{\eta}(|z|))\bigg{(}\frac{z\cdot
e}{|z|}\bigg{)}dz\bigg{|}$

From the construction of $\phi_{\varepsilon}$ and $g_{\varepsilon}$, we notice
that $\phi_{\varepsilon}=\phi_{\eta}$ in $(\eta,\infty)$ and
$g_{\varepsilon}=g_{\eta}$ in $[0,r(1-\eta/2))\cup(r(1+\eta/2),\infty)$. Since
$\varepsilon,\eta$ are very small,, $B_{\eta}\cap A_{\eta}=\emptyset$. Using
that we get,

$\displaystyle\bigg{|}\int_{\mathbb{R}^{n}}\tau_{\Omega}(y)(\nabla{\phi_{\varepsilon}}(|x-y|)\cdot
e)g_{\varepsilon}(|x-y|)\,dy-\int_{\mathbb{R}^{n}}\tau_{\Omega}(y)(\nabla{\phi_{\eta}}(|x-y|)\cdot
e)g_{\eta}(|x-y|)\,dy\bigg{|}$
$\displaystyle\leq\bigg{|}\int_{B_{\eta}}\tau_{\Omega}(x+z)(\phi_{\varepsilon}^{\prime}(|z|)-\phi^{\prime}_{\eta}(|z|))\bigg{(}\frac{z\cdot
e}{|z|}\bigg{)}dz\bigg{|}$
$\displaystyle\qquad+\bigg{|}\int_{A_{\eta}}\tau_{\Omega}(x+z)(g_{\varepsilon}(|z|)-g_{\eta}(|z|))\bigg{(}\frac{z\cdot
e}{|z|}\bigg{)}dz\bigg{|}$

We see that the first term converges to $0$ as $\eta,\varepsilon\to 0$. On the
other hand, the second term is bounded by $4|A_{\eta}|$ which converges to $0$
when $\eta\to 0$. Hence we have

$\int_{\mathbb{R}^{n}}\tau_{\Omega}(y)(\nabla{\phi_{\varepsilon}}(|x-y|)\cdot
e)g_{\varepsilon}(|x-y|)\,dy\to\int_{\mathbb{R}^{n}}\tau_{\Omega}(y)(\nabla{\phi}(|x-y|)\cdot
e)g(|x-y|)\,dy$

uniformly for every $x\in\partial\Omega$.

Next we consider the following term,

$\int_{\mathbb{R}^{n}}\tau_{\Omega}(y){\phi_{\varepsilon}}(|x-y|)\nabla
g_{\varepsilon}(|x-y|)\cdot e\,dy.$

We can write

$\displaystyle\int_{\mathbb{R}^{n}}\tau_{\Omega}(y)\phi_{\varepsilon}(|x-y|)\nabla
g_{\varepsilon}(|x-y|)\cdot e\,dy$
$\displaystyle=\int_{\Omega^{c}}\phi_{\varepsilon}(|x-y|)g^{\prime}_{\varepsilon}(|x-y|)\frac{(x-y)}{|x-y|}\cdot
e\,dy$
$\displaystyle\quad-\int_{\Omega}\phi_{\varepsilon}(|x-y|)g^{\prime}_{\varepsilon}(|x-y|)\frac{(x-y)}{|x-y|}\cdot
e\,dy$ $\displaystyle=\int^{\infty}_{0}\int_{\Omega^{c}\cap\partial
B_{\gamma}(x)}\phi_{\varepsilon}(\gamma)g^{\prime}_{\varepsilon}(\gamma)(e_{y}\cdot
e)\,d\sigma(y)d\gamma$
$\displaystyle\quad-\int^{\infty}_{0}\int_{\Omega\cap\partial
B_{\gamma}(x)}\phi_{\varepsilon}(\gamma)g^{\prime}_{\varepsilon}(\gamma)(e_{y}\cdot
e)\,d\sigma(y)d\gamma,\quad[\text{where}\leavevmode\nobreak\ (x-y)=\gamma
e_{y}]$

If we define

$F(x,e,\lambda)=\int_{\Omega^{c}\cap\partial B_{\lambda}(x)}(e\cdot
e_{y})d\sigma(y)-\int_{\Omega\cap\partial B_{\lambda}(x)}(e\cdot
e_{y})d\sigma(y),$

as before, then we notice that $F(x,e,\lambda)$ is continuous function in
$[0,r+\delta/2]$. Next, using the explicit expression of
$g^{\prime}_{\varepsilon}$ we get,

$\int^{\infty}_{0}\phi_{\varepsilon}(\gamma)g^{\prime}_{\varepsilon}(\gamma)F(x,e,\gamma)\,d\gamma=\int^{\infty}_{0}\phi_{\varepsilon}(\gamma)\frac{1}{\varepsilon}\mathcal{T}\bigg{(}\frac{\gamma-r}{\varepsilon}\bigg{)}F(x,e,\gamma)\,d\gamma$

Then

$\displaystyle\bigg{|}\int^{\infty}_{0}\phi_{\varepsilon}(\gamma)\frac{1}{\varepsilon}\mathcal{T}\bigg{(}\frac{\gamma-r}{\varepsilon}\bigg{)}F(x,e,\gamma)\,d\gamma+\phi(r)F(x,e,r)\bigg{|}$
$\displaystyle=\bigg{|}\int^{\infty}_{0}\phi_{\varepsilon}(\gamma)\frac{1}{\varepsilon}\mathcal{T}\bigg{(}\frac{\gamma-r}{\varepsilon}\bigg{)}F(x,e,\gamma)\,d\gamma-\int^{\infty}_{0}\frac{1}{\varepsilon}\mathcal{T}\bigg{(}\frac{\gamma-r}{\varepsilon}\bigg{)}\phi(r)F(x,e,r)d\gamma\bigg{|}$
$\displaystyle=\bigg{|}\int^{\infty}_{0}\frac{1}{\varepsilon}\mathcal{T}\bigg{(}\frac{\gamma-r}{\varepsilon}\bigg{)}\big{(}\phi_{\varepsilon}(\gamma)F(x,e,\gamma)-\phi(r)F(x,e,r)\big{)}\,d\gamma\bigg{|}$
$\displaystyle\leq\int_{|\gamma-r|\leq
r\varepsilon/2}\bigg{|}\frac{1}{\varepsilon}\mathcal{T}\bigg{(}\frac{\gamma-r}{\varepsilon}\bigg{)}\big{(}\phi_{\varepsilon}(\gamma)F(x,e,\gamma)-\phi(r)F(x,e,r)\big{)}\bigg{|}\,d\gamma$
$\displaystyle\quad+\int_{|\gamma-r|>r\varepsilon/2}\bigg{|}\frac{1}{\varepsilon}\mathcal{T}\bigg{(}\frac{\gamma-r}{\varepsilon}\bigg{)}\big{(}\phi_{\varepsilon}(\gamma)F(x,e,\gamma)-\phi(r)F(x,e,r)\big{)}\bigg{|}\,d\gamma$

Using the continuity of $F$, as $\varepsilon\to 0$,

$\displaystyle\int_{\mathbb{R}^{n}}\tau_{\Omega}(y)\phi_{\varepsilon}(|x-y|)\nabla
g_{\varepsilon}(|x-y|)\cdot e\,dy$ $\displaystyle\xrightarrow[\varepsilon\to
0]{}-\phi(r)\bigg{[}\int_{\Omega^{c}\cap\partial B_{r}(x)}(e_{y}\cdot
e)\,d\sigma(x,y)-\int_{\Omega\cap\partial B_{r}(x)}(e_{y}\cdot
e)\,d\sigma(y)\bigg{]}$ $\displaystyle\quad=-\phi(r)\int_{\partial
B_{r}(x)}\tau_{\Omega}(y)(e_{y}\cdot e)\,d\sigma(x,y)$

Then $H^{\varepsilon}_{\Omega}$ converges in $C^{1}(\partial\Omega)$ norm. ∎

Now we will prove the Theorem 1.3. Again we move hyperplane and do the
reflection for a vector $e\in\mathbb{S}^{n-1}$. Suppose $\pi_{\lambda}$ is the
critical hyperplane in this case and for simplicity $e=\langle
1,0,\cdots,0\rangle$.

###### Proof of Theorem 1.3.

As before, we consider two different cases.

  1. (a)

Interior touching at the point $x_{0}$. In this case, if we follow similar
computation as in the proof of Theorem 1.1, we get

$|(\Omega\setminus R(\Omega))\cap
B_{r}(x_{0})|=|(R(\Omega)\setminus\Omega)\cap B_{r}(x_{0})|=0$

  2. (b)

Non-transversal intersection at the point $x_{0}$. Again we follow similar
type of computation as in the proof of Theorem 1.1. That gives us
$\partial_{e}H^{J}_{\Omega}(x_{0})-\partial_{e}H^{J}_{R(\Omega)}(x_{0})=0$.
Now we know that

$\partial_{e}H^{J}_{\Omega}(x_{0})=\lim_{\varepsilon\to
0}\int_{\mathbb{R}^{n}}\tau_{\Omega}\phi^{\prime}_{\varepsilon}(|x_{0}-y|)\frac{(x_{0})_{1}-y_{1}}{|x_{0}-y|}g_{\varepsilon}(|x_{0}-y|)dy-\phi(r)\int_{\partial
B_{r}(x)}\tau_{\Omega}(y)\frac{(x_{0})_{1}-y_{1}}{|x_{0}-y|}\,d\sigma(x,y)$

Then

$\displaystyle
0=\partial_{e}H^{J}_{\Omega}(x_{0})-\partial_{e}H^{J}_{R(\Omega)}(x_{0})$
$\displaystyle=\lim_{\varepsilon\to
0}\int_{\mathbb{R}^{n}}(\tau_{\Omega}-\tau_{R(\Omega)})\phi^{\prime}_{\varepsilon}(|x_{0}-y|)\frac{(x_{0})_{1}-y_{1}}{|x_{0}-y|}g_{\varepsilon}(|x_{0}-y|)dy$
$\displaystyle\quad-\phi(r)\int_{\partial
B_{r}(x)}(\tau_{\Omega}(y)-\tau_{R(\Omega)}(y))\frac{(x_{0})_{1}-y_{1}}{|x_{0}-y|}\,d\sigma(x,y)$
$\displaystyle=\lim_{\varepsilon\to 0}\bigg{[}\int_{\Omega\setminus
R(\Omega)}\phi^{\prime}_{\varepsilon}(|x_{0}-y|)g_{\varepsilon}(|x_{0}-y|)\frac{(x_{0})_{1}-y_{1}}{|x_{0}-y|}\,dy$
$\displaystyle\qquad-\int_{R(\Omega)\setminus\Omega}\phi^{\prime}_{\varepsilon}(|x_{0}-y|)g_{\varepsilon}(|x_{0}-y|)\frac{(x_{0})_{1}-y_{1}}{|x_{0}-y|}\,dy\bigg{]}$
$\displaystyle\qquad-\phi(r)\bigg{[}\int_{(\Omega\setminus
R(\Omega))\cap\partial
B_{r}(x)}\frac{(x_{0})_{1}-y_{1}}{|x_{0}-y|}\,dy-\int_{(R(\Omega)\setminus\Omega)\cap\partial
B_{r}(x)}\frac{(x_{0})_{1}-y_{1}}{|x_{0}-y|}\,dy\bigg{]}$

Next we observe that $(x_{0})_{1}=\lambda$ and if $y\in\Omega\setminus
R(\Omega)$ then $y_{1}<\lambda$. On the other hand, if $y\in
R(\Omega)\setminus\Omega$ then $y_{1}>\lambda$. Hence

$\displaystyle\lim_{\varepsilon\to 0}\bigg{[}\int_{\Omega\setminus
R(\Omega)}\phi^{\prime}_{\varepsilon}(|x_{0}-y|)g_{\varepsilon}(|x_{0}-y|)\frac{|\lambda-
y_{1}|}{|x_{0}-y|}\,dy+\int_{R(\Omega)\setminus\Omega}\phi^{\prime}_{\varepsilon}(|x_{0}-y|)g_{\varepsilon}(|x_{0}-y|)\frac{|\lambda-
y_{1}|}{|x_{0}-y|}\,dy\bigg{]}$
$\displaystyle\quad-\phi(r)\bigg{[}\int_{(\Omega\setminus
R(\Omega))\cap\partial B_{r}(x)}\frac{|\lambda-
y_{1}|}{|x_{0}-y|}\,dy+\int_{(R(\Omega)\setminus\Omega)\cap\partial
B_{r}(x)}\frac{|\lambda-y_{1}|}{|x_{0}-y|}\,dy\bigg{]}=0$

We know $g_{\varepsilon}\geq 0$ in $B_{r}$,
$\phi(r)>0,\phi^{\prime}_{\varepsilon}\leq 0$. From this we can say that

$|(\Omega\setminus R(\Omega))\cap
B_{r}(x_{0})|=|(R(\Omega)\setminus\Omega)\cap B_{r}(x_{0})|=0$

and

$\mathcal{H}^{n-1}((\Omega\setminus R(\Omega))\cap\partial
B_{r}(x))=\mathcal{H}^{n-1}((R(\Omega)\setminus\Omega)\cap\partial
B_{r}(x))=0$

Hence in both cases, we are getting that,

$|(\Omega\setminus R(\Omega))\cap
B_{r}(x_{0})|=|(R(\Omega)\setminus\Omega)\cap B_{r}(x_{0})|=0$

Now using the arguments as in the proof of Theorem 1.2, we get the desired
result. ∎

##  5\. Constant Curvature problem with constant kernel

For this case, the prototypical kernel is

$J(x)=\chi_{B_{r}}(x),$

where $B_{r}$ is the ball of radius $r$ centered at $0$. Then we see that the
nonlocal curvature at a point $x$ is

$\displaystyle H^{J}_{\Omega}(x)$
$\displaystyle=\int_{\mathbb{R}^{n}}(\chi_{\Omega^{c}}(y)-\chi_{\Omega}(y))J(x-y)dy$
$\displaystyle=\int_{\mathbb{R}^{n}}(\chi_{\Omega^{c}}(y)-\chi_{\Omega}(y))\chi_{B_{r}(x)}(y)dy$
$\displaystyle=|\Omega^{c}\cap B_{r}(x)|-|\Omega\cap B_{r}(x)|$
$\displaystyle=|B_{r}|-2|\Omega\cap B_{r}(x)|$

As before, we assume $e=\langle 1,0,\cdots,0\rangle$ with $\pi_{\lambda}$ to
be the critical hyperplane for the direction $e$. $R_{\lambda}$ is the
reflection operator with respect to $\pi_{\lambda}$, which will be denoted as
operator $R$ for simplicity. Here again, we define,

${\pi_{\lambda}}_{+}=\\{x:x_{1}>\lambda\\}\quad\text{and}\leavevmode\nobreak\
\quad{\pi_{\lambda}}_{-}=\\{x:x_{1}<\lambda\\}$

We also define

$\begin{cases}\Omega_{+}=\Omega\cap{\pi_{\lambda}}_{+},\quad\text{and}\leavevmode\nobreak\
\quad\Omega_{-}=\Omega\cap{\pi_{\lambda}}_{-}\\\
\partial\Omega_{+}=\partial\Omega\cap{\pi_{\lambda}}_{+},\quad\text{and}\leavevmode\nobreak\
\quad\partial\Omega_{-}=\partial\Omega\cap{\pi_{\lambda}}_{-}\end{cases}$

###### Proof of Theorem 1.4.

As before, we fix the direction $e$ and apply the Alexandrov’s moving plane
method. Next we consider two different cases.

Case (a): Interior touching at the point $x_{0}$. In this case, if we follow
similar computation as in the proof of Theorem 1.1, we get,

(5.1) $\displaystyle 0$
$\displaystyle=H^{J}_{\Omega}(x_{0})-H^{J}_{\Omega}(R(x_{0}))$
$\displaystyle=H^{J}_{\Omega}(x_{0})-H^{J}_{R(\Omega)}(x_{0})$
$\displaystyle=\int_{\mathbb{R}^{n}}\bigg{(}\tau_{\Omega}(y)-\tau_{R(\Omega)}(y)\bigg{)}J(x_{0}-y)dy$
$\displaystyle=\int_{\Omega\setminus
R(\Omega)}J(x_{0}-y)dy-\int_{R(\Omega)\setminus\Omega}J(x_{0}-y)dy$
$\displaystyle=\int_{\Omega\setminus
R(\Omega)}\big{(}J(x_{0}-y)-J(x_{0}-R(y))dy$ $\displaystyle=|(\Omega\setminus
R(\Omega))\cap(B_{r}(x_{0})\setminus B_{r}(R(x_{0}))|$

$\Omega$ is a connected set with $C^{1}$ boundary and hence for any
$x_{1},x_{2}\in\partial\Omega$, if $x_{1}\neq x_{2}$, we have

$|\Omega\cap(B_{r}(x_{1})\Delta B_{r}(x_{2}))|>0.$

Hence from (5.1), we have $\Omega=R(\Omega)$ inside the set
$B_{r}(x_{0})\setminus B_{r}(R(x_{0}))$ or the boundaries of $\Omega$ and
$R(\Omega)$ coincide inside the set $B_{r}(x_{0})\setminus B_{r}(R(x_{0}))$.
In fact, boundaries of $\Omega$ and $R(\Omega)$ coincide inside the set
$B_{r}(x_{0})\setminus\overline{B_{r}(R(x_{0}))}$. Let $y\in
B_{r}(x_{0})\setminus\overline{B_{r}(R(x_{0}))}$ be such a point on
$\partial\Omega_{-}$. Then $y$ is a point of interior touching. Since
$B_{r}(x_{0})\setminus\overline{B_{r}(R(x_{0}))}$ is open, there exists an
open neighborhood $V\subseteq B_{r}(x_{0})\setminus\overline{B_{r}(R(x_{0}))}$
such that on the set, $V\cap\partial\Omega_{-}$, $\Omega$ and $R(\Omega)$
coincide and $y\in V\cap\partial\Omega_{-}$. Since $y\in
B_{r}(x_{0})\setminus\overline{B_{r}(R(x_{0}))}$, it is evident that $x_{0}\in
B_{r}(y)\setminus\overline{B_{r}(R(y))}$ and similarly there exits an open
neighborhood $U\subseteq B_{r}(y)\setminus\overline{B_{r}(R(y))}$ such that
$\Omega=R(\Omega)$ inside $U$ and $x_{0}\in U\cap\partial\Omega_{-}$. Let
$\partial^{i}\Omega$ be the set of all interior points. Then, it is an open
set subset of $\partial\Omega_{-}$. Next consider a point
$x_{nc}\in\partial\Omega_{-}\setminus\partial^{i}\Omega$. That point is not a
point of contact, i.e neither an interior point nor a point of non-transversal
intersection. Next we consider the set
$\partial\Omega_{-}\cap(B_{r}(x_{nc})\setminus\overline{B_{r}(R(x_{nc}))})$.
Every point in that set would be a noncontact point. Otherwise, if there is
one contact point, say $z_{c}$, that would be an interior point and
$x_{nc}\in\partial\Omega_{-}\cap(B_{r}(z_{c})\setminus\overline{B_{R}(R(z_{c}))})$,
which makes $x_{nc}$ to be an interior point. That is a contradiction. Next,
using the similar argument as before, we see that the set of non-contact point
of $\partial\Omega_{-}$ is also an open subset of $\partial\Omega_{-}$. But
that is not possible since $\partial\Omega_{-}$ is a connected set. That
implies $\partial^{i}\Omega=\partial\Omega_{-}$ and hence all points on
$\partial\Omega_{-}$ are interior touching points.

Case (b): Non-transversal intersection at the point $x_{0}$. Using Lemma 4.4,
we see that

$\partial_{e}H^{J}_{\Omega}(x_{0})=-\int_{\partial
B_{r}(x)}\tau_{\Omega}(y)\frac{(x_{0})_{1}-y_{1}}{|x_{0}-y|}\,d\sigma(x,y)$

Similar analysis as in the proof of Theorem 1.3, provides,

(5.2) $\displaystyle\mathcal{H}^{n-1}((\Omega\setminus R(\Omega))\cap\partial
B_{r}(x_{0}))=\mathcal{H}^{n-1}((R(\Omega)\setminus\Omega)\cap\partial
B_{r}(x_{0}))=0.$

Since $\Omega$ has $C^{1}$-boundary, for every point $x_{0}\in\partial\Omega$,

$\mathcal{H}^{n-1}(\Omega\cap\partial B_{r}(x_{0}))>0.$

In particular,

$\displaystyle\mathcal{H}^{n-1}(\Omega_{-}\cap\partial B_{r}(x_{0}))>0.$

Then (5.2) implies $\Omega=R(\Omega)$ on $\partial
B_{r}(x_{0})\cap\Omega_{-}$. From the given property of the boundary, there
will be a point $z\in\partial\Omega_{-}\cap\partial B_{r}(x_{0})$. But then
that point would be an interior touching point. Next we do the analysis as in
Case (a) to conclude all the points on $\partial\Omega_{-}$ are interior
touching points.

Then in both cases, we proved that $\Omega$ is symmetric with respect to the
direction $e$. Since $e$ is arbitrary, $\Omega$ is a ball. ∎

##  6\. Conclusions and future directions

The results of this paper showcases the extent to which previous methods (such
as the Alexandrov’s moving plane, establishing differentiability of the
nonlocal curvature) can handle problems of constant nonlocal mean curvature,
while allowing kernels which are only integrable. In fact, the kernel’s
integrability becomes an advantage as it enables our analysis to eliminate
some of the smoothness restrictions on the boundary, which were present in
previous works; indeed, we are able to consider boundaries with $C^{1}$
regularity whenever the kernel is constant. So far it remains unclear if
continuous boundaries could be handled with a modified version of this
approach. For this to work, one would need a generalized normal vector, for
which the authors are investigating some possible candidates.

In conclusion, the main problem arising for future studies is considering sets
with little regularity of the boundary. Note that the nonlocal curvature
function is not even $C^{1}$ in the absence of the same regularity for the
boundary. We are also interested in the nonconstant prescribed curvature
problem. More precisely, for the prescribed curvature problem, we aim to
establish wellposedness and regularity results for a function $u$, whose graph
$\partial E$ satisfies the equation $H^{J}_{\partial E}(x)=f(x)$, when $f$ is
nonconstant. The difficulty of this problem begins with identifying
appropriate boundary conditions, expected to be imposed on nonzero measure
sets. Regularity results for the graph of $u$ would be dependent on the
assumptions imposed on the prescribed nonlocal curvature $f$. To the authors’
knowledge this problem is open when the kernel of nonlocal interaction is
integrable or not.

##  7\. Appendix

###### Proof of Lemma 2.1.

Proof of this Lemma is very similar as in [17]. Here we mention some important
steps of the computation using the growth conditions of our kernel and
regularity of the boundary. Consider a point
$x=(x^{\prime},x_{n})\in\partial\Omega$. As the boundary is $C^{1,\beta}$,
there exists a small neighborhood of $x$ where the boundary can be written as
the graph of some function $f:\mathbb{R}^{n-1}\to\mathbb{R}$. Without loss of
generality, we can assume that $x=0$, $\nabla f(0)=0$. Then we define

$B^{\prime}_{\rho}=\\{x^{\prime}:|x^{\prime}|<\rho\\},L_{\rho}=(-\rho,\rho)\times
B^{\prime}_{\rho},$

and

$\partial\Omega\cap L_{\rho}=\\{(x^{\prime},f(x^{\prime}):x^{\prime}\in
B^{\prime}_{\rho}\\}=(Id\times f)(B^{\prime}_{\rho}).$

For the $C^{1}$ convergence, first we define

$\psi_{\varepsilon}(t)=-\frac{1}{t^{n}}\int^{\infty}_{t}\phi_{\varepsilon}(\tau)\tau^{n-1}d\tau$

Then

$\operatorname{div}(x\psi_{\varepsilon})=\phi_{\varepsilon}(|x|).$

Using divergence theorem,

$H^{\varepsilon}_{\Omega}(x)=-2\int_{\partial\Omega}\psi_{\varepsilon}(|x-y|)(x-y)\cdot\nu_{y}d\sigma$

Taking derivative in $x$-variable and for some $e\in
T_{x}(\partial\Omega)\cap\mathbb{S}^{n-1}$, we have

$\nabla H^{\varepsilon}_{\Omega}(x)\cdot
e=2\int_{\partial\Omega}\bigg{(}\psi_{\varepsilon}(|x-y|)\nu_{y}\cdot
e+\frac{\psi^{\prime}_{\varepsilon}(|x-y|)}{|x-y|}((x-y)\cdot\nu_{y})((x-y)\cdot
e)\bigg{)}d\sigma$

Next we want to evaluate the following quantity, $\left\|\nabla
H^{\varepsilon}_{\Omega}(x)\cdot e-\nabla H^{\eta}_{\Omega}(x)\cdot
e\right\|_{C^{0}}$ when $\varepsilon,\eta\to 0$. As mentioned before, we can
assume that $x=0,\nabla f(0)=0$. Then we have,

$\nabla H^{\varepsilon}_{\Omega}(0)\cdot
e=2\int_{\partial\Omega}\bigg{(}\psi_{\varepsilon}(|y|)\nu_{y}\cdot
e+\frac{\psi^{\prime}_{\varepsilon}(|y|)}{|y|}(y\cdot\nu_{y})(y\cdot
e)\bigg{)}d\sigma$

Next we choose a $0<\gamma<\rho$ and define the set $G_{\gamma}=(Id\times
f)(B^{\prime}_{\gamma})$. By the construction of
$\psi_{\varepsilon},\psi^{\prime}_{\varepsilon}$, they converge uniformly in
$[t,\infty)$ for any $t>0$. Then we have

$\displaystyle\bigg{|}\int_{\partial\Omega\setminus
G_{\gamma}}\bigg{(}\psi_{\varepsilon}(|y|)\nu_{y}\cdot
e+\frac{\psi^{\prime}_{\varepsilon}(|y|)}{|y|}(y\cdot\nu_{y})(y\cdot
e)\bigg{)}d\sigma$ $\displaystyle\quad-\int_{\partial\Omega\setminus
G_{\gamma}}\bigg{(}\psi_{\eta}(|y|)\nu_{y}\cdot
e+\frac{\psi^{\prime}_{\eta}(|y|)}{|y|}(y\cdot\nu_{y})(y\cdot
e)\bigg{)}d\sigma\bigg{|}\to 0$

as $\varepsilon,\eta\to 0$. Next inside the set $G_{\gamma}$, we can write
$\nu_{y}=\frac{(-\nabla f(y^{\prime}),1)}{\sqrt{1+|\nabla
f(y^{\prime})|^{2}}}$ where $y=(y^{\prime},f(y^{\prime}))$ in $G_{\gamma}$.
One can also notice that, as $e\in T_{0}(\partial\Omega)$ then $e$ does not
have any component in $x_{n}$ direction. Then we write,

(7.1) $\displaystyle
2\int_{G_{\gamma}}\bigg{(}\psi_{\varepsilon}(|y|)\nu_{y}\cdot
e+\frac{\psi^{\prime}_{\varepsilon}(|y|)}{|y|}(y\cdot\nu_{y})(y\cdot
e)\bigg{)}d\sigma$
$\displaystyle=2\int_{B^{\prime}_{\gamma}}\bigg{(}-\psi_{\varepsilon}(\sqrt{|y^{\prime}|^{2}+f^{2}})\frac{\nabla
f\cdot e}{\sqrt{1+|\nabla f(y^{\prime})|^{2}}}$
$\displaystyle\qquad+\frac{\psi^{\prime}_{\varepsilon}(\sqrt{|y^{\prime}|^{2}+f^{2}})}{\sqrt{|y^{\prime}|^{2}+f^{2}}}\frac{f-\nabla
f\cdot y^{\prime}}{\sqrt{1+|\nabla f(y^{\prime})|^{2}}}(y^{\prime}\cdot
e)\bigg{)}\sqrt{1+|\nabla f(y^{\prime})|^{2}}\,dy^{\prime}$
$\displaystyle=2\int_{B^{\prime}_{\gamma}}\bigg{(}-\psi_{\varepsilon}(\sqrt{|y^{\prime}|^{2}+f^{2}})(\nabla
f\cdot
e)+\frac{\psi^{\prime}_{\varepsilon}(\sqrt{|y^{\prime}|^{2}+f^{2}})}{\sqrt{|y^{\prime}|^{2}+f^{2}}}(f-\nabla
f\cdot y^{\prime})(y^{\prime}\cdot e)\bigg{)}\,dy^{\prime}$

By changing $y^{\prime}\mapsto-y^{\prime}$ and adding the two expressions, we
get the following, same as in [17],

(7.2)
$\displaystyle-\int_{B^{\prime}_{\gamma}}\psi_{\varepsilon}(\sqrt{|y^{\prime}|^{2}+f^{2}})(\nabla
f(y^{\prime})\cdot e+\nabla f(-y^{\prime})\cdot e)\,dy^{\prime}$
$\displaystyle+\int_{B^{\prime}_{\gamma}}\big{(}\psi_{\varepsilon}(\sqrt{|y^{\prime}|^{2}+f(y^{\prime})^{2}})-\psi_{\varepsilon}(\sqrt{|y^{\prime}|^{2}+f(-y^{\prime})^{2}})\big{)}\nabla
f(y^{\prime})\cdot e\,dy^{\prime}$
$\displaystyle+\int_{B^{\prime}_{\gamma}}\frac{\psi^{\prime}_{\varepsilon}(\sqrt{|y^{\prime}|^{2}+f(y^{\prime})^{2}})}{\sqrt{|y^{\prime}|^{2}+f(y^{\prime})^{2}}}\big{(}[f(y^{\prime})-f(-y^{\prime})](y^{\prime}\cdot
e)-[\nabla f(y^{\prime})\cdot y^{\prime}+\nabla f(-y^{\prime})\cdot
y^{\prime}](y^{\prime}\cdot e)\big{)}\,dy^{\prime}$
$\displaystyle+\int_{B^{\prime}_{\gamma}}\bigg{(}\frac{\psi^{\prime}_{\varepsilon}(\sqrt{|y^{\prime}|^{2}+f(y^{\prime})^{2}})}{\sqrt{|y^{\prime}|^{2}+f(y^{\prime})^{2}}}-\frac{\psi^{\prime}_{\varepsilon}(\sqrt{|y^{\prime}|^{2}+f(-y^{\prime})^{2}})}{\sqrt{|y^{\prime}|^{2}+f(-y^{\prime})^{2}}}\bigg{)}[f(-y^{\prime})+\nabla
f(-y^{\prime})\cdot y^{\prime}](y^{\prime}\cdot e)\,dy^{\prime}$

Using the regularity of $f$, we have the following estimates,

$|\nabla f(y^{\prime})\cdot e+\nabla f(-y^{\prime})\cdot e|\leq
C|y^{\prime}|^{\beta},\quad|\nabla f(-y^{\prime})|\leq
C|y^{\prime}|^{\beta},\quad|f(y^{\prime})|\leq C|y^{\prime}|^{1+\beta},$

which yield

$|f(y^{\prime})-f(-y^{\prime})|\leq C|y^{\prime}|^{1+\beta},\quad|\nabla
f(y^{\prime})\cdot y^{\prime}+\nabla f(-y^{\prime})\cdot y^{\prime}|\leq
C|y^{\prime}|^{1+\beta}.$

Moreover,

$\displaystyle\bigg{|}\psi_{\varepsilon}(\sqrt{|y^{\prime}|^{2}+f(y^{\prime})^{2}})-\psi_{\varepsilon}(\sqrt{|y^{\prime}|^{2}+f(-y^{\prime})^{2}})$
$\displaystyle\bigg{|}\leq
C\frac{|f(y^{\prime})^{2}-f(-y^{\prime})^{2}|}{\max\\{|y^{\prime}|^{n-\alpha+2},|y^{\prime}|^{n+\alpha_{1}+2}\\}}$
$\displaystyle\leq\frac{C}{\max\\{|y^{\prime}|^{n-\alpha-\beta},|y^{\prime}|^{n+\alpha_{1}-\beta}\\}}$

and also

$\displaystyle\bigg{|}\frac{\psi^{\prime}_{\varepsilon}(\sqrt{|y^{\prime}|^{2}+f(y^{\prime})^{2}})}{\sqrt{|y^{\prime}|^{2}+f(y^{\prime})^{2}}}-\frac{\psi^{\prime}_{\varepsilon}(\sqrt{|y^{\prime}|^{2}+f(-y^{\prime})^{2}})}{\sqrt{|y^{\prime}|^{2}+f(-y^{\prime})^{2}}}\bigg{|}$
$\displaystyle\leq
C\frac{|f(y^{\prime})^{2}-f(-y^{\prime})^{2}|}{\max\\{|y^{\prime}|^{n-\alpha+4},|y^{\prime}|^{n+\alpha_{1}+4}\\}}$
$\displaystyle\leq\frac{C}{\max\\{|y^{\prime}|^{n-\alpha-\beta+2},|y^{\prime}|^{n+\alpha_{1}-\beta+2}\\}}.$

As $\gamma$ is very small, we can assume that

$\displaystyle\bigg{|}\psi_{\varepsilon}(\sqrt{|y^{\prime}|^{2}+f(y^{\prime})^{2}})-\psi_{\varepsilon}(\sqrt{|y^{\prime}|^{2}+f(-y^{\prime})^{2}})\bigg{|}$
$\displaystyle\leq\frac{C}{|y^{\prime}|^{n-\alpha-\beta}}$
$\displaystyle\bigg{|}\frac{\psi^{\prime}_{\varepsilon}(\sqrt{|y^{\prime}|^{2}+f(y^{\prime})^{2}})}{\sqrt{|y^{\prime}|^{2}+f(y^{\prime})^{2}}}-\frac{\psi^{\prime}_{\varepsilon}(\sqrt{|y^{\prime}|^{2}+f(-y^{\prime})^{2}})}{\sqrt{|y^{\prime}|^{2}+f(-y^{\prime})^{2}}}\bigg{|}$
$\displaystyle\leq\frac{C}{|y^{\prime}|^{n+2-\alpha-\beta}}.$

Thus if we estimate each term in (7.2), we get

$\bigg{|}\int_{B^{\prime}_{\gamma}}\psi_{\varepsilon}(\sqrt{|y^{\prime}|^{2}+f^{2}})(\nabla
f(y^{\prime})\cdot e+\nabla f(-y^{\prime})\cdot e)\,dy^{\prime}\bigg{|}\leq
C\int^{\gamma}_{0}\frac{1}{\kappa^{n-\alpha}}\kappa^{\beta}\kappa^{n-2}\,d\kappa=C\gamma^{\alpha+\beta-1}$

and

$\displaystyle\bigg{|}\int_{B^{\prime}_{\gamma}}\big{(}\psi_{\varepsilon}(\sqrt{|y^{\prime}|^{2}+f(y^{\prime})^{2}})$
$\displaystyle-\psi_{\varepsilon}(\sqrt{|y^{\prime}|^{2}+f(-y^{\prime})^{2}})\big{)}\nabla
f(y^{\prime})\cdot e\,dy^{\prime}\bigg{|}$ $\displaystyle\leq
C\int^{\gamma}_{0}\frac{1}{\kappa^{n-\alpha-\beta}}\kappa^{\beta}\kappa^{n-2}\,d\kappa$
$\displaystyle\leq C\gamma^{\alpha+2\beta-1}.$

Similarly the third and fourth term will also be bounded by
$C\gamma^{\alpha+\beta-1}$. Hence

$\displaystyle\limsup_{\varepsilon,\eta\to
0}\bigg{|}\int_{G_{\gamma}}\bigg{(}\psi_{\varepsilon}(|y|)\nu_{y}\cdot
e+\frac{\psi^{\prime}_{\varepsilon}(|y|)}{|y|}(y\cdot\nu_{y})(y\cdot
e)\bigg{)}d\sigma$
$\displaystyle\quad-\int_{G_{\gamma}}\bigg{(}\psi_{\eta}(|y|)\nu_{y}\cdot
e+\frac{\psi^{\prime}_{\eta}(|y|)}{|y|}(y\cdot\nu_{y})(y\cdot
e)\bigg{)}d\sigma\bigg{|}\leq C\gamma^{\alpha+\beta-1}.$

By taking the limit $\gamma\to 0$ we obtain the desired convergence. ∎

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