# Quantitative stability
for overdetermined nonlocal problems
with parallel surfaces
and investigation of the stability exponents

Serena Dipierro  Serena Dipierro: Department of Mathematics and Statistics,
The University of Western Australia, 35 Stirling Highway, Crawley, Perth, WA
6009, Australia<EMAIL_ADDRESS>,  Giorgio Poggesi  Giorgio
Poggesi: Department of Mathematics and Statistics, The University of Western
Australia, 35 Stirling Highway, Crawley, Perth, WA 6009, Australia
<EMAIL_ADDRESS>,  Jack Thompson  Jack Thompson: Department of
Mathematics and Statistics, The University of Western Australia, 35 Stirling
Highway, Crawley, Perth, WA 6009, Australia<EMAIL_ADDRESS>and  Enrico Valdinoci  Enrico Valdinoci: Department of Mathematics and
Statistics, The University of Western Australia, 35 Stirling Highway, Crawley,
Perth, WA 6009, Australia<EMAIL_ADDRESS>
###### Abstract.

In this article, we analyze the stability of the parallel surface problem for
semilinear equations driven by the fractional Laplacian. We prove a
quantitative stability result that goes beyond that previously obtained in
[Cir+23].

Moreover, we discuss in detail several techniques and challenges in obtaining
the optimal exponent in this stability result. In particular, this includes an
upper bound on the exponent via an explicit computation involving a family of
ellipsoids. We also sharply investigate a technique that was proposed in
[Cir+18] to obtain the optimal stability exponent in the quantitative estimate
for the nonlocal Alexandrov’s soap bubble theorem, obtaining accurate
estimates to be compared with a new, explicit example.

###### 2020 Mathematics Subject Classification:

35R11, 47G20, 35N25, 35B50

##  1\. Introduction and main results

###  1.1. The long-standing tradition of overdetermined problems

Overdetermined problems are a broad class of partial differential equations
(PDEs) where ‘too many conditions’ are imposed on the solution. Since not
every region will admit a solution which satisfies all the conditions, the
objective in the study of overdetermined problems is to classify those regions
which do admit solutions.

Often, overdetermined problems arise naturally in applications as a
combination of a well-posed partial differential equation (PDE) which
describes the dynamics of a given physical system, as well as an extra
condition, often referred to as the overdetermined condition, which describes
a property or an optimal quality you would like the solution to possess. As
such, overdetermined problems have a close relationship with optimization,
free boundary problems and calculus of variations, particularly shape
optimization, as well as many other applications including fluid mechanics,
solid mechanics, thermodynamics, electrostatics, see [Ser71, FG08, HP18,
DPV21].

The study of overdetermined problems began in the early 1970’s with the
celebrated paper of Serrin [Ser71]. In this influential paper, Serrin proved
that, given a bounded domain $\Omega\subset\mathbb{R}^{n}$ with $C^{2}$
boundary and $f\in C^{0,1}_{\text{loc}}(\mathbb{R})$, if there exists a
positive solution $u\in C^{2}(\overline{\Omega})$ that satisfies the Dirichlet
boundary value problem

(1.3) $\displaystyle\left\\{\begin{array}[]{r@{ \ }l @{\quad\: \;} l}-\Delta
u&=f(u)\hfil\quad\>&\text{in }\Omega,\\\ u&=0\hfil\quad\>&\text{on
}\partial\Omega,\end{array}\right.$

as well as the overdetermined condition

(1.4) $\displaystyle\partial_{\nu}u$ $\displaystyle=c\qquad\text{on
}\partial\Omega$

for some constant $c\in\mathbb{R}\setminus\\{0\\}$, then $\Omega$ must be a
ball. Here $\nu$ is the unit outward pointing normal to $\partial\Omega$ and
$\partial_{\nu}u=\nabla u\cdot\nu$. The proof relies on a powerful technique,
now known as the _method of moving planes_ , a refinement of a reflection
principle conceived by Alexandrov in [Ale62] to prove the so-called _soap
bubble theorem_ , which states that the only connected closed hypersurfaces
embedded in a space form with everywhere constant mean curvature are spheres.

Since the work of Alexandrov and Serrin, the analysis of overdetermined
problems and the method of moving planes has seen intense research activity.
Some of the literature on the method of moving planes, overdetermined
problems, and symmetry in PDE include:

  * •

_Alternate/incomplete overdetermined conditions:_ One can consider elliptic
equations such (1.3) or otherwise with an alternate overdetermined condition
to (1.4), see [HP98, FG08, BHS14, DPV21]. The parallel surface problem fits
into this category, see below for references.

  * •

_Integral identities:_ Integral identities such as the Pohozaev identity are
often employed in the analysis of overdetermined problems, providing an
alternative approach to the method of moving planes. Such an alternative
approach was pioneered in [Wei71] and then further developed in, e.g., [PS89,
GL89, Bra+08, CS09, QX17, CV19, MP20, MP20a].

  * •

_Nonlinear elliptic equations:_ The original paper of Serrin [Ser71] dealt
with uniformly elliptic quasilinear equations including equations of mean
curvature type. Other (possibly degenerate) quasilinear and fully nonlinear
equations have been analyzed [GL89, DS07, FK08, Bra+08, BK11, BD13, SS15].

  * •

_Symmetry:_ The method of moving planes was famously used to prove symmetry
results for solutions to semilinear PDE in domains with symmetry in [GNN79,
GNN81]. For this type of results, an alternative approach which combines
Pohozaev identity and isoperimetric-type inequalities was pioneered in [Lio81]
and then further developed in [KP94, Ser13, DPV22].

  * •

_Nonlocal operators:_ For overdetermined problems and symmetry for nonlocal
equations, see [FJ15, Cab+18, Cir+18, SV19, Dip+22, Cir+23].

###  1.2. The nonlocal parallel surface problem and the main results of this
paper

In this paper, we are concerned with the nonlocal parallel surface problem.
The context of this problem is as follows: Let $n$ be a positive integer and
$s\in(0,1)$. Suppose that $G$ is an open bounded subset of $\mathbb{R}^{n}$
and let $\Omega=G+B_{R}$ for some $R>0$, where $A+B$ is the Minkowski sum of
sets defined by

$\displaystyle A+B=\\{a+b\text{ s.t. }a\in A,b\in B\\}.$

Moreover, let $(-\Delta)^{s}$ denote the fractional Laplacian defined by

$\displaystyle(-\Delta)^{s}u(x)$
$\displaystyle=c_{n,s}\mathrm{P.V.}\int_{\mathbb{R}^{n}}\frac{u(x)-u(y)}{|x-y|^{n+2s}}\mathop{}\\!dy$

where $c_{n,s}$ is a positive normalization constant and $\mathrm{P.V.}$
denotes the Cauchy principle value. Also, assume that
$f:\mathbb{R}\to\mathbb{R}$ is locally Lipschitz and satisfies $f(0)\geqslant
0$. Then, the nonlocal parallel surface problem asks: if there exists a
function $u$ that satisfies (in an appropriate sense) the equation

(1.8) $\displaystyle\left\\{\begin{array}[]{r@{ \ }l @{\quad\: \;}
l}(-\Delta)^{s}u&=f(u)\hfil\quad\>&\text{in }\Omega,\\\
u&=0\hfil\quad\>&\text{in }\mathbb{R}^{n}\setminus\Omega,\\\ u&\geqslant
0\hfil\quad\>&\text{in }\mathbb{R}^{n},\end{array}\right.$

as well as the overdetermined condition

(1.9) $\displaystyle u=\text{constant}\qquad\text{on }\partial G,$

then is $\Omega$ necessarily a ball?

This question is referred to as the rigidity problem for the parallel surface
problem. Furthermore, one can ask about the stability of this problem, that
is, heuristically, if $u$ satisfies (1.8) and ‘almost’ satisfies (1.9) then is
$\Omega$ ‘almost’ a ball? This is the subject of the current article. Of
course, one must be precise by what one means by ‘almost’—this will be made
clear in the proceeding paragraphs as we describe the literature and our main
results.

The local analogue of the parallel surface problem (i.e. the case $s=1$) was
introduced in [Sha12] as a discrete analogue of the original Serrin’s problem.
Moreover, it also appears in [MS10, CMS15] in the context of invariant
isothermic surfaces of a nonlinear non-degenerate fast diffusion equation.

To see why the parallel surface problem can be viewed as a discrete analogue
of Serrin’s problem, consider that $u\in C^{2}(\overline{\Omega})$ satisfies
(1.3) as well as $u=c_{k}>0$ on an countably infinite family of parallel
surfaces $\Gamma_{k}$ that are a distance $1/k$ from the boundary of $\Omega$.
Then $kc_{k}$ necessarily converges to some $c$ by regularity assumptions on
$u$ and $u$ satisfies (1.4). Consequently, $\Omega$ must be a ball.

To reiterate, in the parallel surface problem there is only a single parallel
surface (not a family as just described); regardless, this was enough to prove
in [Sha12, CMS15] the rigidity result: if there exists a solution to (1.3)
satisfying (1.9) then $\Omega$ must be a ball.

Subsequently, in [CMS16], the stability of (1.8) for the local problem was
addressed. In that article, the authors used the shape functional

$\displaystyle\rho(\Omega)=\inf\big{\\{}R-r\text{ s.t.
}B_{r}(x)\subset\Omega\subset B_{R}(x)\text{ for some }x\in\Omega\big{\\}}$

to quantify how close $\Omega$ is to being a ball and the semi-norm

$\displaystyle[u]_{\partial G}=\sup_{\begin{subarray}{c}x,y\in\partial G\\\
x\neq y\end{subarray}}\bigg{\\{}\frac{|u(x)-u(y)|}{|x-y|}\bigg{\\}}$

to quantify how close $u$ is to being constant on $\partial\Omega$. They
showed, under reasonable assumptions on $\Omega$, that

(1.10) $\displaystyle\rho(\Omega)\leqslant C[u]_{\partial G}.$

Moreover, for the nonlocal parallel surface problem, the rigidity problem was
answered affirmatively in [Cir+23] for the case $f=1$ and [Dip+22] for general
$f$. Moreover, [Cir+23] also addressed the stability problem (still for the
case $f=1$) and showed that

(1.11) $\displaystyle\rho(\Omega)\leqslant C[u]_{\partial G}^{\frac{1}{s+2}}.$

It is interesting to observe that (1.11) is sub-optimal in the sense that it
does not recover the estimate (1.10) when $s=1$. This is due to the
nonlocality of the fractional Laplacian which caused contributions from mass
‘far away’ to have a significant effect on the analysis.

This leads us to our main result.

######  Theorem 1.1.

oAZAv7vy Let $G$ be an open bounded subset of $\mathbb{R}^{n}$ and
$\Omega=G+B_{R}$ for some $R>0$. Furthermore, let $\Omega$ and $G$ have
$C^{1}$ boundary, and let $f\in C^{0,1}_{\mathrm{loc}}(\mathbb{R})$ be such
that $f(0)\geqslant 0$. Suppose that $u$ satisfies (1.8) in the weak sense.

Then,

(1.12) $\displaystyle\rho(\Omega)$ $\displaystyle\leqslant C[u]_{\partial
G}^{\frac{1}{s+2}}$

with

$C:=C(n,s)\left[\frac{(\max\\{1,\operatorname{diam}\Omega\\})^{n+2s}\big{(}1+(\operatorname{diam}\Omega)^{2s}[f]_{C^{0,1}([0,\|u\|_{L^{\infty}(\Omega)}])}\big{)}}{R^{s}\big{(}f(0)+\|u\|_{L_{s}(\mathbb{R}^{n})}\big{)}}+(\operatorname{diam}\Omega)^{n-1}+\frac{|\Omega|}{R}\right]\\\
\times\frac{(\operatorname{diam}\Omega)^{n+2s+3}}{R^{2s}|\Omega|}\left(1+R^{2s}[f]_{C^{0,1}([0,\|u\|_{L^{\infty}(\Omega)}])}\right).$

For the precise definition of a solution satisfying (1.8) in the weak sense,
see Section 2.

oAZAv7vy is a direct extension of [Cir+23] to the case $f\neq 1$. Moreover,
oAZAv7vy relaxes some of the regularity assumptions on $\Omega$ appearing in
[Cir+23]. It is clear that the dependence on the volume $|\Omega|$ in the
constant of Theorem LABEL:oAZAv7vy can be removed by means of the following
bounds

$|B_{1}|R^{n}\leqslant|\Omega|\leqslant|B_{1}|(\operatorname{diam}\Omega)^{n},\text{
where }|B_{1}|\text{ is the volume of the unit ball in }\mathbb{R}^{n},$

which hold true in light of the monotonicity of the volume with respect to
inclusion.

Currently, an important open problem for the nonlocal parallel surface problem
is to understand the optimality of oAZAv7vy. Indeed, let $\overline{\beta}(s)$
be the optimal exponent in (1.12) defined as the supremum over
$\beta\in\mathbb{R}$ such that if $u=u_{f}$ is a weak solution of (1.8) for
some $f$ satisfying the assumptions of oAZAv7vy then $\rho(\Omega)\leqslant
C[u]_{\partial G}^{\beta}$. In this framework, an explicit expression for
$\overline{\beta}$ as a function of $s$ is still unknown. oAZAv7vy implies
that $\overline{\beta}(s)\geqslant\frac{1}{s+2}$ and [Sha12, CMS15] establish
that $\overline{\beta}(1)=1$, therefore we believe it is an interesting
problem to detect optimal stability exponents in the nonlocal setting, also to
recover, whenever possible, the classical exponent of the local cases in the
limit.

In Section 7, we explicitly construct a family of domains $G_{\varepsilon}$
that are small perturbations of a ball and corresponding solutions
$u_{\varepsilon}$ to (1.8) when $f=1$ which satisfy
$[u_{\varepsilon}]_{\partial G_{\varepsilon}}\simeq
C\rho(\Omega_{\varepsilon})$ as $\varepsilon\to 0^{+}$. This entails the
following result:

######  Theorem 1.2.

O4zCF33o Let $G$ be an open bounded subset of $\mathbb{R}^{n}$ and
$\Omega=G+B_{R}$ for some $R>0$. Furthermore, let $\Omega$ and $G$ have
$C^{1}$ boundary.

Then, we have that $\overline{\beta}(s)\leqslant 1$.

As far as the authors are aware, these are the only known estimates for
$\overline{\beta}(s)$. Furthermore, by considering the case in which
$\Omega=\Omega_{\varepsilon}$ is a small perturbation of a ball and exploting
interior regularity for the fractional Laplacian, one would expect that
$\rho(\Omega_{\varepsilon})\simeq[u_{\varepsilon}]_{\partial
G_{\varepsilon}}\simeq\varepsilon$ (up to constant) as $\varepsilon\to 0^{+}$
which suggests that $\overline{\beta}(s)=1$ for all $s\in(0,1]$. In Section 5,
we give a broad discussion on some of the challenges that the nonlocality of
the fractional Laplacian presents in obtaining this result. In particular, by
way of an example via the Poisson representation formula, we show that
estimates for a singular integral involving the reciprocal of the distance to
the boundary function play a key role in obtaining the anticipated optimal
result. This suggests, surprisingly, that fine geometric estimates for the
distance function close to the boundary are required to obtain the optimal
exponent.

###  1.3. Scrutiny of the stability exponent

In the recent literature, some inventive methods have been introduced to
improve the stability results obtained via the moving plane method. A
remarkable one was put forth in [Cir+18]. Roughly speaking, the setting
considered in [Cir+18, Proposition 3.1(b)] focused on the critical hyperplane
$\pi_{\lambda}=\\{x_{n}=\lambda\\}$ for the moving plane technique for a set
$\Omega$ which contains the ball $B_{r}$ and is contained in the ball $B_{R}$,
looked at the symmetric difference between $\Omega$ and its reflection
$\Omega^{\prime}$ and obtained a bound on the measure of the set

(1.13) $\big{\\{}x\in\Omega\triangle\Omega^{\prime}\text{ s.t.
}\operatorname{dist}(x,\pi_{\lambda})\leqslant\gamma\big{\\}}$

which was linear in both $\gamma$ and $R-r$.

This constituted a fundamental ingredient in [Cir+18] to achieve an optimal
stability exponent.

Unfortunately, we believe that, in the very broad generality in which the
result is stated in [Cir+18], this statement may not be true, and we present
an explicit counter-example.

Nevertheless, in our opinion, a weaker version of [Cir+18, Proposition 3.1(b)]
does hold true. We state and prove this new result and check its optimality
against the counter-example mentioned above. This construction plays a
decisive role in improving the stability exponent.

Though we refer the reader to Section 6 for full details of this strategy, let
us anticipate here some important details.

More specifically, in the forthcoming Theorem LABEL:m9q9X2hE, under the
additional assumption that the set $\Omega$ is of class $C^{\alpha}$, with
$\alpha>1$, we will bound the measure of the set in (1.13), up to constants,
by

(1.14) $\gamma(R-r)^{1-\frac{1}{\alpha}}.$

We stress that this bound formally recovers exactly the one stated in [Cir+18,
Proposition 3.1(b)] when $\alpha=+\infty$.

However, we believe that the bound in (1.14) is optimal and cannot be improved
(in particular, while the dependence in $\gamma$ of the estimate above is
linear, the dependence in $R-r$, surprisingly, is not!). This will be shown by
an explicit counter-example put forth in Theorem 6.6—while the analytical
details of this counter-example are very delicate, the foundational idea
behind it is sketched in Figure 2 (roughly speaking, the example is obtained
by a very small and localized modification of a ball to induce a critical
situation at the maximal location allowed by the regularity of the set; the
construction is technically demanding since the constraint for the set of
“being between two balls” induces two different scales, in the horizontal and
in the vertical directions, which in principle could provide different
contributions).

As a consequence of the bound (1.14) for the measure of the set in (1.13), as
given in m9q9X2hE, the stability exponent $1/(s+2)$ of oAZAv7vy can be
improved to $\alpha/(1+\alpha(s+1))$ provided that $\Omega$ is of class
$C^{\alpha}$ for $\alpha>1$: we refer to Section 6.2 and thm:improvement for
details.

###  1.4. Organization of paper

The paper is organized as follows. In Section 2, we summarize the notation and
basic definitions used throughout the article. In Section 3, we give several
quantitative maximum principles—in both the non-antisymmetric and
antisymmetric situations—that are required in the proof of oAZAv7vy and, in
Section 4, we give the proof of oAZAv7vy.

The remaining sections are broadly focused on the techniques and challenges in
obtaining the optimal stability exponent. In Section 5, we discuss the
surprising role that fine geometric estimates for the distance function close
to the boundary play in the attainment of the optimal exponent. In Section 6,
we accurately discuss the possibility of obtaining the optimal exponent and
comment about some criticalities in the existing literature, and, in Section
7, we construct an explicit family of solutions which implies O4zCF33o.

## Acknowledgements

All the authors are members of AustMS. GP is supported by the Australian
Research Council (ARC) Discovery Early Career Researcher Award (DECRA)
DE230100954 “Partial Differential Equations: geometric aspects and
applications” and is member of the Gruppo Nazionale Analisi Matematica
Probabilità e Applicazioni (GNAMPA) of the Istituto Nazionale di Alta
Matematica (INdAM). JT is supported by an Australian Government Research
Training Program Scholarship. EV is supported by the Australian Laureate
Fellowship FL190100081 “Minimal surfaces, free boundaries and partial
differential equations”.

##  2\. Preliminaries and notation

In this section we fix the notation that we will use throughout the article
and give some relevant definitions. Let $n\geqslant 1$ and $s\in(0,1)$. The
fractional Sobolev space $H^{s}(\mathbb{R}^{n})$ is defined as

$\displaystyle H^{s}(\mathbb{R}^{n})=\big{\\{}u\in L^{2}(\mathbb{R}^{n})\text{
such that }[u]_{H^{s}(\mathbb{R}^{n})}<+\infty\big{\\}}$

where $[\cdot]_{H^{s}(\mathbb{R}^{n})}$ is the Gagliardo semi-norm given by

$\displaystyle[u]_{H^{s}(\mathbb{R}^{n})}=c_{n,s}\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\frac{|u(x)-u(y)|^{2}}{|x-y|^{n+2s}}\mathop{}\\!dy\mathop{}\\!dx$

where $c_{n,s}$ is the same constant appearing in the definition of the
fractional Laplacian. Functions in $H^{s}(\mathbb{R}^{n})$ that are equal
almost everywhere are identified. Moreover, the bilinear form associated with
$[\cdot]_{H^{s}(\mathbb{R}^{n})}$ is denoted by
$\mathcal{E}:H^{s}(\mathbb{R}^{n})\times H^{s}(\mathbb{R}^{n})\to\mathbb{R}$
and is given by

$\displaystyle\mathcal{E}(u,v)=\frac{c_{n,s}}{2}\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\frac{(u(x)-u(y))(v(x)-y(y)}{|x-y|^{n+2s}}\mathop{}\\!dy\mathop{}\\!dx.$

We also define the following weighted $L^{1}$ norm via

$\displaystyle\|u\|_{L_{s}(\mathbb{R}^{n})}=\int_{\mathbb{R}^{n}}\frac{|u(x)|}{1+|x|^{n+2s}}\mathop{}\\!dx$

and the space $L_{s}(\mathbb{R}^{n})$ as

$\displaystyle
L_{s}(\mathbb{R}^{n})=\big{\\{}L^{1}_{\mathrm{loc}}(\mathbb{R}^{n})\text{ such
that }\|u\|_{L_{s}(\mathbb{R}^{n})}<+\infty\big{\\}}.$

Now, suppose that $\Omega$ is an open, bounded subset of $\mathbb{R}^{n}$ and
define the space $\mathcal{H}^{s}_{0}(\Omega)$ by

$\displaystyle\mathcal{H}^{s}_{0}(\Omega)$ $\displaystyle=\big{\\{}u\in
H^{s}(\mathbb{R}^{n})\text{ such that }u=0\text{ in
}\mathbb{R}^{n}\setminus\Omega\big{\\}}.$

Let $c:\Omega\to\mathbb{R}$ be a measurable function such that $c\in
L^{\infty}(\Omega)$ and $g\in L^{2}(\Omega)$. We say that a function $u\in
H^{s}(\mathbb{R}^{n})\cap L_{s}(\mathbb{R}^{n})$ satisfies
$(-\Delta)^{s}u+cu\geqslant g$ (respectively, $\leqslant$) in the _weak sense_
if

$\displaystyle\mathcal{E}(u,v)+\int_{\Omega}cuv\mathop{}\\!dx\geqslant\int_{\Omega}gv\mathop{}\\!dx\qquad(\text{respectively,
}\leqslant)$

for all $v\in\mathcal{H}^{s}_{0}(\Omega)$ with $v\geqslant 0$. In this case,
we also say that $u$ is a supersolution (respectively, subsolution) of
$(-\Delta)^{s}u+cu=g$ in the weak sense. Moreover, we say a function $u\in
H^{s}(\mathbb{R}^{n})\cap L_{s}(\mathbb{R}^{n})$ satisfies
$(-\Delta)^{s}u+cu=g$ in the _weak sense_ if

$\displaystyle\mathcal{E}(u,v)+\int_{\Omega}cuv\mathop{}\\!dx=\int_{\Omega}gv\mathop{}\\!dx$

for all $v\in\mathcal{H}^{s}_{0}(\Omega)$. This is equivalent to $u$ being
both a weak supersolution and a weak subsolution. Note that we are assuming _a
priori_ that weak solutions are in $L_{s}(\mathbb{R}^{n})$.

Next, we describe some notation regarding antisymmetric functions. This is
closely related to the notation of the method of moving planes; however, we
will defer our explanation of the method of moving planes until Section 4 in
the interests of simplicity.

A function $v:\mathbb{R}^{n}\to\mathbb{R}$ is said to be _antisymmetric_ with
respect to a plane $T$ if

$\displaystyle v(Q_{T}(x))=-v(x)\qquad\text{for all }x\in\mathbb{R}^{n}$

where $Q_{T}:\mathbb{R}^{n}\to\mathbb{R}^{n}$ is the function that reflects
$x$ across $T$. Often it will suffice to consider the case $T=\\{x_{1}=0\\}$,
in which case $Q_{T}(x)=x-2x_{1}e_{1}$.

For simplicity, we will refer to $v$ as _antisymmetric_ if it is antisymmetric
with respect to the plane $\\{x_{1}=0\\}$. Moreover, we sometimes write
$x^{\prime}$ to denote $Q_{T}(x)$ when it is clear from context what $T$ is.

Finally, some other notation we will employ through the article is: given a
set $A\subset\mathbb{R}^{n}$, the function
$\chi_{A}:\mathbb{R}^{n}\to\mathbb{R}$ denotes the characteristic function of
$A$, given by

$\displaystyle\chi_{A}(x)$ $\displaystyle=\begin{cases}1,&\text{if }x\in A,\\\
0,&\text{if }x\not\in A,\end{cases}$

and $\delta_{A}:\mathbb{R}^{n}\to[0,+\infty]$ denotes the distance function to
$A$, given by

$\displaystyle\delta_{A}(x)=\inf_{y\in A}|x-y|.$

We will denote by
$\mathbb{R}^{n}_{+}:=\\{x=(x_{1},\dots,x_{n})\in\mathbb{R}^{n}\;{\mbox{s.t.}}\;x_{1}>0\\}$
and, given $A\subseteq\mathbb{R}^{n}$, we will denote by
$A^{+}:=A\cap\mathbb{R}^{n}_{+}$.

Moreover, if $A$ is open and bounded with sufficiently regular boundary (we
only use the case in which $A$ has smooth boundary), we will denote by
$\psi_{A}$ the (unique) function in $C^{\infty}(A)\cap C^{s}(\mathbb{R}^{n})$
such that $\psi_{A}$ satisfies

(2.3) $\displaystyle\left\\{\begin{array}[]{r@{ \ }l @{\quad\: \;}
l}(-\Delta)^{s}\psi_{A}&=1\hfil\quad\>&\text{in }A,\\\
\psi_{A}&=0\hfil\quad\>&\text{in }\mathbb{R}^{n}\setminus
A.\end{array}\right.$

Also, we will use $\lambda_{1}(A)$ to denote the first Dirichlet eigenvalue of
the fractional Laplacian.

##  3\. Quantitative maximum principles up to the boundary

The basic idea of the proof of oAZAv7vy is to apply the method of moving
planes, as in the proof of the analogous rigidity result [Dip+22, Theorem
1.4], but replace ‘qualitative’ maximum principles, such as the strong maximum
principles, with ‘quantitative’ maximum principles, such as the Harnack
inequality.

The purpose of this section is to prove several such quantitative maximum
principles both in the non-antisymmetric and the antisymmetric setting. In
particular, we require that these maximum principles hold up to the boundary
of regions with very little boundary regularity. Moreover, we are careful to
keep track of precisely how constants depend on relevant quantities since we
believe that this may be useful in future analyses of the nonlocal parallel
surface problem.

The section is split into two parts: maximum principles for equations without
any antisymmetry assumptions and maximum principles for equations with
antisymmetry assumptions.

###  3.1. Equations without antisymmetry

In this subsection, we will give several maximum principles for linear
equations with zero-th order terms without any antisymmetry assumptions. This
will culminate in a quantitative analogue of the Hopf lemma for non-negative
supersolutions of general semilinear equations (see zAPw0npM below).

Our first result is as follows:

######  Proposition 3.1.

YCl8lL0I Let $\Omega$ be an bounded open subset of $\mathbb{R}^{n}$, $c\in
L^{\infty}(\Omega)$, and $g\in L^{2}(\Omega)$ with $g\geqslant 0$. Suppose
that $u$ satisfies

$\displaystyle\left\\{\begin{array}[]{r@{ \ }l @{\quad\: \;}
l}(-\Delta)^{s}u+cu&\geqslant g\hfil\quad\>&\text{in }\Omega,\\\ u&\geqslant
0\hfil\quad\>&\text{in }\mathbb{R}^{n}\end{array}\right.$

in the weak sense.

Then,

$\displaystyle u(x)$ $\displaystyle\geqslant
C\Big{(}\operatorname*{ess\,inf}_{\Omega}g+\|u\|_{L_{s}(\mathbb{R}^{n})}\Big{)}\delta_{\partial\Omega}^{2s}(x)\qquad\text{for
a.e. }x\in\Omega$

with
$C:=C(n,s)(\max\\{1,\operatorname{diam}\Omega\\})^{-n-2s}\big{(}1+(\operatorname{diam}\Omega)^{2s}\|c^{+}\|_{L^{\infty}(\Omega)}\big{)}^{-1}$.

The proof of YCl8lL0I essentially follows by, at each point $x\in\Omega$,
‘touching’ the solution $u$ from below by $\psi_{B}$ (recall the notation in
(2.3)) where $B$ is the largest ball contained in $\Omega$ and centred at $x$.
This is similar to the proof of [RS19, Theorem 2.2] where a nonlinear interior
analogue of YCl8lL0I was proven. To prove YCl8lL0I, however, some care must be
taken since the touching point may occur on the boundary of $\Omega$ (unlike
in the interior result of [RS19, Theorem 2.2]) where the PDE does not
necessarily hold. Moreover, YCl8lL0I is stated in the context of weak
solutions where pointwise techniques no longer make sense, so a simple
mollification argument needs to be made. We require the following lemma.

######  Lemma 3.2.

Voueel7o Let $\rho>0$, $c\in L^{\infty}(B_{\rho})$, and $g\in
C^{\infty}_{0}(\mathbb{R}^{n})$ be such that $g\geqslant 0$ in $B_{\rho}$.
Suppose that $u\in C^{\infty}_{0}(\mathbb{R}^{n})$ satisfies

$\displaystyle\left\\{\begin{array}[]{r@{ \ }l @{\quad\: \;}
l}(-\Delta)^{s}u+cu&\geqslant g\hfil\quad\>&\text{in }B_{\rho},\\\ u&\geqslant
0\hfil\quad\>&\text{in }\mathbb{R}^{n}.\end{array}\right.$

Then,

$\displaystyle u(x)\geqslant
C\Big{(}\inf_{B_{\rho}}g+\|u\|_{L_{s}(\mathbb{R}^{n})}\Big{)}\psi_{B_{\rho}}(x)\qquad\text{for
all }x\in B_{\rho}$

with
$C:=C(n,s)(\max\\{1,\rho\\})^{-n-2s}\big{(}1+\rho^{2s}\|c^{+}\|_{L^{\infty}(B_{\rho})}\big{)}^{-1}$.

We observe that Lemma LABEL:Voueel7o can be seen as a quantitative version of
a strong maximum principle (which can be proved directly): in particular, it
entails that if $u(x_{0})=0$ for some $x_{0}\in\Omega$, then $u$ vanishes
identically.

###### Proof of Lemma LABEL:Voueel7o.

We will begin by proving the case $\rho=1$. Fix $\varepsilon>0$ small and let
$\tau\geqslant 0$ be the largest value such that
$u\geqslant\tau\psi_{B_{1-\varepsilon}}$ in $\mathbb{R}^{n}$. Now, by the
definition of $\tau$, there exists $x_{0}\in B_{1-\varepsilon}$ such that
$u(x_{0})=\tau\psi_{B_{1-\varepsilon}}(x_{0})$. On one hand,

$\displaystyle(-\Delta)^{s}(u-\tau\psi_{B_{1-\varepsilon}})(x_{0})+c(x_{0})(u-\tau\psi_{B_{1-\varepsilon}})(x_{0})$
$\displaystyle\geqslant\inf_{B_{1}}g-\tau\big{(}1+c(x_{0})\psi_{B_{1-\varepsilon}}(x_{0})\big{)}$
$\displaystyle\geqslant\inf_{B_{1}}g-C(n,s)\big{(}1+\|c^{+}\|_{L^{\infty}(B_{1})}\big{)}\tau.$

On the other hand, since $|x_{0}-y|^{-n-2s}\geqslant
C(n,s)(1+|y|^{n+2s})^{-1}$, we have that

$\displaystyle(-\Delta)^{s}(u-\tau\psi_{B_{1-\varepsilon}})(x_{0})+c(x_{0})(u-\tau\psi_{B_{1-\varepsilon}})(x_{0})$
$\displaystyle=-\int_{\mathbb{R}^{n}}\frac{(u-\tau\psi_{B_{1-\varepsilon}})(y)}{|x_{0}-y|^{n+2s}}\mathop{}\\!dy$
$\displaystyle\leqslant-C(n,s)\int_{\mathbb{R}^{n}}\frac{(u-\tau\psi_{B_{1-\varepsilon}})(y)}{1+|y|^{n+2s}}\mathop{}\\!dy$
$\displaystyle\leqslant-C(n,s)\big{(}\|u\|_{L_{s}(\mathbb{R}^{n})}-\tau\big{)}.$

Rearranging gives $\tau\geqslant
C(n,s)\big{(}1+\|c^{+}\|_{L^{\infty}(B_{1})}\big{)}^{-1}\big{(}\inf_{B_{1}}g+\|u\|_{L_{s}(\mathbb{R}^{n})}\big{)}$,
which implies that

$\displaystyle u(x)$ $\displaystyle\geqslant
C(n,s)\big{(}1+\|c^{+}\|_{L^{\infty}(B_{1})}\big{)}^{-1}\Big{(}\inf_{B_{1}}g+\|u\|_{L_{s}(\mathbb{R}^{n})}\Big{)}\psi_{B_{1-\varepsilon}}.$

Sending $\varepsilon\to 0^{+}$ gives the result when $\rho=1$.

Now, for the case that $\rho$ is not necessarily 1, let $u_{\rho}(x):=u(\rho
x)$. Then

$\displaystyle(-\Delta)^{s}u_{\rho}+\rho^{2s}c_{\rho}$
$\displaystyle\geqslant\rho^{2s}g_{\rho}\qquad\text{in }B_{1}$

where $c_{\rho}(x):=c(\rho x)$ and $g_{\rho}(x):=g(\rho x)$, so, for all $x\in
B_{\rho}$, we have that

$\displaystyle u(x)$ $\displaystyle=u_{\rho}(x/\rho)$ $\displaystyle\geqslant
C(n,s)\big{(}1+\rho^{2s}\|c^{+}_{\rho}\|_{L^{\infty}(B_{1})}\big{)}^{-1}\Big{(}\rho^{2s}\inf_{B_{1}}g_{\rho}+\|u_{\rho}\|_{L_{s}(\mathbb{R}^{n})}\Big{)}\psi_{B_{1}}(x/\rho)$
$\displaystyle=C(n,s)\rho^{-2s}\big{(}1+\rho^{2s}\|c^{+}\|_{L^{\infty}(B_{\rho})}\big{)}^{-1}\Big{(}\rho^{2s}\inf_{B_{\rho}}g+\|u_{\rho}\|_{L_{s}(\mathbb{R}^{n})}\Big{)}\psi_{B_{\rho}}(x).$

Finally,

$\displaystyle\|u_{\rho}\|_{L_{s}(\mathbb{R}^{n})}$
$\displaystyle=\int_{\mathbb{R}^{n}}\frac{|u(\rho
x)|}{1+|x|^{n+2s}}\mathop{}\\!dx=\rho^{2s}\int_{\mathbb{R}^{n}}\frac{|u(x)|}{\rho^{n+2s}+|x|^{n+2s}}\mathop{}\\!dx\geqslant\rho^{2s}(\max\\{1,\rho\\})^{-n-2s}\|u\|_{L_{s}(\mathbb{R}^{n})}$

which completes the proof. ∎

We can now give the proof of YCl8lL0I.

###### Proof of YCl8lL0I.

If $u,g\in C^{\infty}_{0}(\mathbb{R}^{n})$ then the proof follows immediately
from Voueel7o. Indeed, let $x\in\Omega$ be arbitrary. Then, applying Voueel7o
in the ball $B_{\rho}(x)$ with $\rho:=\delta_{\partial\Omega}(x)$, we obtain
that

$\displaystyle u(y)$ $\displaystyle\geqslant
C\Big{(}\operatorname*{ess\,inf}_{\Omega}g+\|u\|_{L_{s}(\mathbb{R}^{n})}\Big{)}\psi_{B_{\rho}}(y)\qquad\text{for
all }y\in B_{\rho}(x)$

with
$C=C(n,s)(\max\\{1,\operatorname{diam}\Omega\\})^{-n-2s}\big{(}1+(\operatorname{diam}\Omega)^{2s}\|c^{+}\|_{L^{\infty}(\Omega)}\big{)}^{-1}$.
Substituting $y=x$ completes the proof.

In the general case that $u\in H^{s}(\mathbb{R}^{n})$, if $u_{\varepsilon}$
and $g_{\varepsilon}$ are mollifications of $u$ and $g$ respectively then
$(-\Delta)^{s}u_{\varepsilon}+\|c^{+}\|_{L^{\infty}(\Omega)}u_{\varepsilon}\geqslant
g_{\varepsilon}$ in
$\Omega_{\varepsilon}:=\Omega\cap\\{\delta_{\partial\Omega}>\varepsilon\\}$.
Then, applying the result for smooth compactly supported functions, we have
that $u_{\varepsilon}\geqslant C\delta^{2s}_{\partial\Omega_{\varepsilon}}$
with $C$ as in the statement of Proposition LABEL:YCl8lL0I (and, in
particular, uniformly bounded in $\varepsilon$). Moreover, given
$x\in\Omega_{\varepsilon}$, if $y\in\partial\Omega_{\varepsilon}$ is the
closest point to $x$ on $\partial\Omega_{\varepsilon}$ and
$z\in\partial\Omega$ is the closest point to $y$ on $\partial\Omega$ then

$\displaystyle\delta_{\partial\Omega}(x)\leqslant|x-z|\leqslant|x-y|+|y-z|\leqslant\delta_{\partial\Omega_{\varepsilon}}(x)+\varepsilon,$

so $u_{\varepsilon}\geqslant C(\delta_{\partial\Omega}-\varepsilon)^{2s}$ in
$\Omega_{\varepsilon}$. Then $u_{\varepsilon}\to u$ a.e., so sending
$\varepsilon\to 0^{+}$ gives the result. ∎

Next, we obtain a refined version of YCl8lL0I when $\Omega$ satisfies the
uniform interior ball condition. As customary, we say that a bounded domain
$\Omega\subset\mathbb{R}^{n}$ satisfies the uniform interior ball condition
with radius $r_{\Omega}>0$ if for every point $x_{0}\in\partial\Omega$ there
exists a ball $B\subset\Omega$ of radius $r_{\Omega}$ such that its closure
intersects $\partial\Omega$ only at $x_{0}$.

######  Proposition 3.3.

1kapKUbs Let $\Omega$ be an open bounded subset of $\mathbb{R}^{n}$ with
$C^{1}$ boundary and satisfying the uniform interior ball condition with
radius $r_{\Omega}>0$, $c\in L^{\infty}(\Omega)$, and $g\in L^{2}(\Omega)$
with $g\geqslant 0$. Suppose that $u$ satisfies

$\displaystyle\left\\{\begin{array}[]{r@{ \ }l @{\quad\: \;}
l}(-\Delta)^{s}u+cu&\geqslant g\hfil\quad\>&\text{in }\Omega,\\\ u&\geqslant
0\hfil\quad\>&\text{in }\mathbb{R}^{n}\end{array}\right.$

in the weak sense.

Then,

$\displaystyle u(x)$ $\displaystyle\geqslant
C\Big{(}\operatorname*{ess\,inf}_{\Omega}g+\|u\|_{L_{s}(\mathbb{R}^{n})}\Big{)}\delta_{\partial\Omega}^{s}(x)\qquad\text{for
a.e. }x\in\Omega$

with
$C:=C(n,s)r_{\Omega}^{s}(\max\\{1,\operatorname{diam}\Omega\\})^{-n-2s}\big{(}1+(\operatorname{diam}\Omega)^{2s}\|c^{+}\|_{L^{\infty}(\Omega)}\big{)}^{-1}$.

###### Proof.

By an analogous argument to the one in the proof of YCl8lL0I, we may assume
that $u,g\in C^{\infty}_{0}(\mathbb{R}^{n})$. Let $x\in\Omega$ be arbitrary.
If $\delta_{\partial\Omega}(x)\geqslant r_{\Omega}$ then we are done by
YCl8lL0I, so let $\delta_{\partial\Omega}(x)<r_{\Omega}$. If $\bar{x}$ denotes
the closest point to $x$ on $\partial\Omega$ then there exists a ball
$B=B_{r_{\Omega}}(x_{0})$ such that $x\in B$, $B\subset\Omega$, and $B$
touches $\partial\Omega$ at $\bar{x}$. Then, applying Voueel7o in $B$, we have
that $u(y)\geqslant
C\big{(}\operatorname*{ess\,inf}_{\Omega}g+\|u\|_{L_{s}(\mathbb{R}^{n})}\big{)}\psi_{B_{r_{\Omega}}}(y)$
with
$C=C(n,s)(\max\\{1,\operatorname{diam}\Omega\\})^{-n-2s}\big{(}1+(\operatorname{diam}\Omega)^{2s}\|c^{+}\|_{L^{\infty}(\Omega)}\big{)}^{-1}$.
Substituting $y=x$, we obtain

$\displaystyle u(x)$ $\displaystyle\geqslant
C\Big{(}\operatorname*{ess\,inf}_{\Omega}g+\|u\|_{L_{s}(\mathbb{R}^{n})}\Big{)}(r_{\Omega}^{2}-|x-x_{0}|^{2})^{s}$
$\displaystyle=C\Big{(}\operatorname*{ess\,inf}_{\Omega}g+\|u\|_{L_{s}(\mathbb{R}^{n})}\Big{)}(r_{\Omega}+|x-x_{0}|)^{s}(r_{\Omega}-|x-x_{0}|)^{s}$
$\displaystyle\geqslant
Cr_{\Omega}^{s}\Big{(}\operatorname*{ess\,inf}_{\Omega}g+\|u\|_{L_{s}(\mathbb{R}^{n})}\Big{)}(r_{\Omega}-|x-x_{0}|)^{s}.$

Observing that $r_{\Omega}-|x-x_{0}|=\delta_{\partial\Omega}(x)$ completes the
proof. ∎

From 1kapKUbs, we immediately obtain the following corollary.

######  Corollary 3.4.

zAPw0npM Let $\Omega$ be an open bounded subset of $\mathbb{R}^{n}$ with
$C^{1}$ boundary and satisfying the uniform interior ball condition with
radius $r_{\Omega}>0$. Let $f\in
C^{0,1}_{\mathrm{loc}}(\overline{\Omega}\times\mathbb{R})$ satisfy
$f_{0}:=\inf_{x\in\Omega}f(x,0)\geqslant 0$. Suppose that $u$ satisfies

$\displaystyle\left\\{\begin{array}[]{r@{ \ }l @{\quad\: \;}
l}(-\Delta)^{s}u&\geqslant f(x,u)\hfil\quad\>&\text{in }\Omega,\\\ u&\geqslant
0\hfil\quad\>&\text{in }\mathbb{R}^{n}\end{array}\right.$

in the weak sense.

Then,

$\displaystyle u(x)$ $\displaystyle\geqslant
C_{\ast}\big{(}f_{0}+\|u\|_{L_{s}(\mathbb{R}^{n})}\big{)}\delta_{\partial\Omega}^{s}(x)\qquad\text{for
a.e. }x\in\Omega$

with
$C_{\ast}:=C(n,s)r_{\Omega}^{s}(\max\\{1,\operatorname{diam}\Omega\\})^{-n-2s}\big{(}1+(\operatorname{diam}\Omega)^{2s}[f]_{C^{0,1}(\overline{\Omega}\times[0,\|u\|_{L^{\infty}(\Omega)}])}\big{)}^{-1}$.

###### Proof.

Since $f=f(x,z)$ is locally lipschitz, $\partial_{z}f$ exists a.e., so we may
define

$\displaystyle c(x):=-\int_{0}^{1}\partial_{z}f(x,tu(x))\mathop{}\\!dt.$

Observe that
$\|c\|_{L^{\infty}(\Omega)}\leqslant[f]_{C^{0,1}(\overline{\Omega}\times[0,\|u\|_{L^{\infty}(\Omega)}])}$.
Then, $(-\Delta)^{s}u+cu\geqslant f(\cdot,0)$ in $\Omega$ in the weak sense,
so the result follows by 1kapKUbs. ∎

######  Remark 3.5.

One could also obtain an analogous result to zAPw0npM without assuming that
$\Omega$ has $C^{1}$ boundary and satisfies the uniform interior ball
condition; this can be achieved by applying YCl8lL0I instead of 1kapKUbs. ∎

###  3.2. Equations with antisymmetry

The purpose of this subsection is to prove the following proposition.

######  Proposition 3.6.

TV1cTSyn Let $H\subset\mathbb{R}^{n}$ be a halfspace, $U$ be an open subset of
$H$, and $a\in H$ and $\rho>0$ such that $B_{\rho}(a)\cap H\subset U$.
Moreover, let $c\in L^{\infty}(U)$ be such that

(3.1) $\|c^{+}\|_{L^{\infty}(U)}<\lambda_{1}(B_{\rho}(a)\cap H).$

Suppose that $v$ is antisymmetric with respect to $\partial H$ and satisfies

$\displaystyle\left\\{\begin{array}[]{r@{ \ }l @{\quad\: \;}
l}(-\Delta)^{s}v+cv&\geqslant 0\hfil\quad\>&\text{in }U,\\\ v&\geqslant
0\hfil\quad\>&\text{in }H\end{array}\right.$

in the weak sense.

If $K\subset H$ is a non-empty open set that is disjoint from $B_{\rho}(a)$
and

$\inf_{{x\in K}\atop{y\in B_{\rho}(a)\cap H}}|Q_{\partial
H}(x)-y|^{-1}\geqslant M\geqslant 0$

then

(3.2) $\displaystyle v(x)$ $\displaystyle\geqslant C\|\delta_{\partial
H}v\|_{L^{1}(K)}\delta_{\partial H}(x)\qquad\text{for a.e. }x\in
B_{\rho/2}(a)\cap H$

with

$\displaystyle
C:=C(n,s)\rho^{2s}M^{n+2s+2}\Big{(}1+\rho^{2s}\|c^{+}\|_{L^{\infty}(U)}\Big{)}^{-1}.$

TV1cTSyn can be viewed as a quantitative version of Proposition 3.3 in [FJ15].
A similar result was also obtained in [Cir+23] in the case $c=0$. One
advantage that TV1cTSyn has over both of the results of [FJ15] and [Cir+23] is
that it allows the ball $B_{\rho}(a)$ to go right up to and, indeed, overlap
the plane of symmetry $\partial H$. To allow for this possibility required the
construction of an antisymmetric barrier given in gmjWA2gy below.

However, a disadvantage of TV1cTSyn is that it is not a boundary estimate, in
the sense that (3.2) holds in $B_{\rho/2}(a)\cap H$ and not $B_{\rho}(a)\cap
H$, so it does not give any information up to $\partial U\setminus\partial H$.
This is also a by-product of the barrier given in gmjWA2gy. In theory, one
should be able to fix this issue by adjusting the barrier to behave like
distance to the power $s$ close to the boundary of its support, but this was
unnecessary for the purposes of our results.

To prove TV1cTSyn, we require two lemmata.

######  Lemma 3.7.

xwrqefNE Let $v\in C^{\infty}_{0}(\mathbb{R}^{n})$ be antisymmetric with
respect to $\\{x_{1}=0\\}$.

Then, $(-\Delta)^{s}v$ is a smooth function in $\mathbb{R}^{n}$ that is
antisymmetric with respect to $\\{x_{1}=0\\}$. Furthermore, if
$w(x):=v(x)/x_{1}$ then

(3.3) $\displaystyle|(-\Delta)^{s}v(x)|$ $\displaystyle\leqslant
C\Big{(}\|w\|_{L^{\infty}(\mathbb{R}^{n})}+\|x_{1}^{-1}\partial_{1}w\|_{L^{\infty}(\mathbb{R}^{n})}+\|D^{2}w\|_{L^{\infty}(\mathbb{R}^{n})}\Big{)}x_{1}$

for all $x\in\mathbb{R}^{n}_{+}$. The constant $C$ depends only on $n$ and
$s$.

###### Proof.

The proof that $(-\Delta)^{s}v$ is a smooth function and antisymmetric follows
from standard properties of the fractional Laplacian, so we will omit it and
focus on the proof of (3.3). For all $\tilde{x}\in\mathbb{R}^{n+2}$, define
$\tilde{w}(\tilde{x}):=w\big{(}\sqrt{\tilde{x}_{1}^{2}+\tilde{x}_{2}^{2}+\tilde{x}_{3}^{2}},\tilde{x}_{4},\dots,\tilde{x}_{n+2}\big{)}$.
By Bochner’s relation, we have that

(3.4) $\displaystyle(-\Delta)^{s}v(x)$
$\displaystyle=x_{1}(-\Delta)^{s}_{\mathbb{R}^{n+2}}\tilde{w}(x_{1},0,0,x_{2},\dots,x_{n})$

where $(-\Delta)^{s}_{\mathbb{R}^{n+2}}$ is the fractional Laplacian in
$\mathbb{R}^{n+2}$ (and $(-\Delta)^{s}$ still refers to the fractional
Laplacian in $\mathbb{R}^{n}$). For more details regarding Bochner’s relation
and a proof of (3.4), we refer the interested reader to the upcoming note
[Dip+23].

Thus, applying standard estimates for the fractional Laplacian, we have that,

$\displaystyle|(-\Delta)^{s}v(x)|$ $\displaystyle\leqslant
C\Big{(}\|D^{2}\tilde{w}\|_{L^{\infty}(\mathbb{R}^{n+2})}+\|\tilde{w}\|_{L^{\infty}(\mathbb{R}^{n+2})}\Big{)}x_{1}$

for all $x\in\mathbb{R}^{n}_{+}$ and $C>0$ depending on $n$ and $s$. Clearly,
$\|\tilde{w}\|_{L^{\infty}(\mathbb{R}^{n+2})}\leqslant\|w\|_{L^{\infty}(\mathbb{R}^{n})}$.
Moreover, by a direct computation, one can check that

$\displaystyle\partial_{ij}\tilde{w}(\tilde{x})$ $\displaystyle=$
$\displaystyle\partial_{11}w\Big{(}\sqrt{\tilde{x}_{1}^{2}+\tilde{x}_{2}^{2}+\tilde{x}_{3}^{2}},\tilde{x}_{4},\dots,\tilde{x}_{n+2}\Big{)}\frac{\tilde{x}_{i}\tilde{x}_{j}}{\tilde{x}_{1}^{2}+\tilde{x}_{2}^{2}+\tilde{x}_{3}^{2}}$
$\displaystyle+\partial_{1}w\Big{(}\sqrt{\tilde{x}_{1}^{2}+\tilde{x}_{2}^{2}+\tilde{x}_{3}^{2}},\tilde{x}_{4},\dots,\tilde{x}_{n+2}\Big{)}\left[\frac{\delta_{ij}}{\sqrt{\tilde{x}_{1}^{2}+\tilde{x}_{2}^{2}+\tilde{x}_{3}^{2}}}-\frac{\tilde{x}_{i}\tilde{x}_{j}}{(\tilde{x}_{1}^{2}+\tilde{x}_{2}^{2}+\tilde{x}_{3}^{2})^{3/2}}\right]$
$\displaystyle\qquad{\mbox{ if }}i,j\in\\{1,2,3\\},$
$\displaystyle\partial_{ij}\tilde{w}(\tilde{x})$ $\displaystyle=$
$\displaystyle\partial_{1j-2}w\Big{(}\sqrt{\tilde{x}_{1}^{2}+\tilde{x}_{2}^{2}+\tilde{x}_{3}^{2}},\tilde{x}_{4},\dots,\tilde{x}_{n+2}\Big{)}\frac{\tilde{x}_{i}}{\sqrt{\tilde{x}_{1}^{2}+\tilde{x}_{2}^{2}+\tilde{x}_{3}^{2}}}$
$\displaystyle\qquad{\mbox{ if }}i\in\\{1,2,3\\}\;{\mbox{ and
}}\;j\in\\{4,\dots,n+2\\},$ $\displaystyle\partial_{ij}\tilde{w}(\tilde{x})$
$\displaystyle=$
$\displaystyle\partial_{i-2j-2}w\Big{(}\sqrt{\tilde{x}_{1}^{2}+\tilde{x}_{2}^{2}+\tilde{x}_{3}^{2}},\tilde{x}_{4},\dots,\tilde{x}_{n+2}\Big{)}\qquad{\mbox{
if }}i,j\in\\{4,\dots,n+2\\},$

and therefore

$\displaystyle\|D^{2}\tilde{w}\|_{L^{\infty}(\mathbb{R}^{n+2})}$
$\displaystyle\leqslant
C\Big{(}\|x_{1}^{-1}\partial_{1}w\|_{L^{\infty}(\mathbb{R}^{n})}+\|D^{2}w\|_{L^{\infty}(\mathbb{R}^{n})}\Big{)}$

for some universal constant $C>0$, which implies the desired result. ∎

We now construct the barrier that will be essential to allow $B_{\rho}(a)$ in
TV1cTSyn to come up to $\partial H$.

######  Lemma 3.8.

gmjWA2gy Let $a\in\overline{\mathbb{R}^{n}_{+}}$ and $\rho>0$. There exists a
function $\varphi\in C^{\infty}_{0}(B_{\rho}(a)\cup B_{\rho}(a^{\prime}))$
such that $\varphi$ is antisymmetric with respect to
$\partial\mathbb{R}^{n}_{+}$,
$\rho^{2s}\chi_{B_{\rho/2}(a)}(x)x_{1}\leqslant\varphi(x)\leqslant
C\rho^{2s}x_{1}$ in $\mathbb{R}^{n}_{+}$, and

(3.5) $|(-\Delta)^{s}\varphi(x)|\leqslant Cx_{1}\qquad\text{in
}B_{\rho}^{+}(a).$

The constant $C$ depends only on $n$ and $s$.

Recall that $B_{\rho}^{+}(a)=B_{\rho}(a)\cap\mathbb{R}^{n}_{+}$, and notice
that $B_{\rho}^{+}(a)$ in Lemma LABEL:gmjWA2gy coincides with $B_{\rho}(a)$ if
$\rho\in(0,a_{1})$ with $a=(a_{1},\dots,a_{n})$.

###### Proof of Lemma LABEL:gmjWA2gy.

Let $\eta\in C^{\infty}_{0}(B_{1})$ be a radial function such that $\eta=1$ in
$B_{1/2}$ and $0\leqslant\eta\leqslant 1$ in $\mathbb{R}^{n}$. Define

$\displaystyle\varphi(x)$
$\displaystyle:=\rho^{2s}x_{1}\bigg{(}\eta\bigg{(}\frac{x-a}{\rho}\bigg{)}+\eta\bigg{(}\frac{x-a^{\prime}}{\rho}\bigg{)}\bigg{)}.$

Then $\varphi\in C^{\infty}_{0}(B_{\rho}(a)\cup B_{\rho}(a^{\prime}))$, it is
antisymmetric, and satisfies
$\rho^{2s}\chi_{B_{\rho/2}(a)}(x)x_{1}\leqslant\varphi\leqslant
2\rho^{2s}x_{1}$ in $\mathbb{R}^{n}_{+}$.

Moreover, let

$\bar{\varphi}(x):=x_{1}\left(\eta\left(x-\frac{a}{\rho}\right)+\eta\left(x-\frac{a^{\prime}}{\rho}\right)\right)$

so that $\varphi(x)=\rho^{2s+1}\bar{\varphi}(x/\rho)$. From xwrqefNE, it
follows that $|(-\Delta)^{s}\bar{\varphi}|\leqslant Cx_{1}$ for some $C>0$
depending on $n$ and $s$, which implies that

$\displaystyle|(-\Delta)^{s}\varphi(x)|$
$\displaystyle=\rho|(-\Delta)^{s}\bar{\varphi}(x/\rho)|\leqslant Cx_{1}.\qed$

Now, we will give the proof of TV1cTSyn.

###### Proof of TV1cTSyn.

Without loss of generality, we may assume that $H=\mathbb{R}^{n}_{+}$. We also
denote by $B=B_{\rho}(a)$. Recall that, given $A\subset\mathbb{R}^{n}$, we use
the notation $A^{+}=A\cap\mathbb{R}^{n}_{+}$. Let $\tau\geqslant 0$ be a
constant to be chosen later and $w:=\tau\varphi+(\chi_{K}+\chi_{K^{\prime}})v$
where $\varphi$ is as in gmjWA2gy and
$K^{\prime}:=Q_{\partial\mathbb{R}^{n}_{+}}(K)$. Furthermore, let
$\xi\in\mathcal{H}^{s}_{0}(B^{+})$ with $\xi\geqslant 0$. By formula (3.5) in
gmjWA2gy, we have that

$\displaystyle\mathcal{E}(w,\xi)$
$\displaystyle=\tau\mathcal{E}(\varphi,\xi)+\mathcal{E}(\chi_{K}v,\xi)+\mathcal{E}(\chi_{K^{\prime}}v,\xi)$
$\displaystyle\leqslant C\tau\int_{B^{+}}x_{1}\xi(x)\mathop{}\\!dx-
c_{n,s}\int_{B^{+}}\int_{K}\frac{\xi(x)v(y)}{|x-y|^{n+2s}}\mathop{}\\!dy\mathop{}\\!dx-
c_{n,s}\int_{B^{+}}\int_{K^{\prime}}\frac{\xi(x)v(y)}{|x-y|^{n+2s}}\mathop{}\\!dy\mathop{}\\!dx$
$\displaystyle=\int_{B^{+}}\bigg{[}C\tau
x_{1}-c_{n,s}\int_{K}\bigg{(}\frac{1}{|x-y|^{n+2s}}-\frac{1}{|x^{\prime}-y|^{n+2s}}\bigg{)}v(y)\mathop{}\\!dy\bigg{]}\xi(x)\mathop{}\\!dx$

where $x^{\prime}:=Q_{\partial H}(x)$.

Since, for all $x\in B^{+}$ and $y\in K$, we have that

$\displaystyle\frac{1}{|x-y|^{n+2s}}-\frac{1}{|x^{\prime}-y|^{n+2s}}$
$\displaystyle=\frac{n+2s}{2}\int_{|x-y|^{2}}^{|x^{\prime}-y|^{2}}t^{-\frac{n+2s+2}{2}}\mathop{}\\!dt$
$\displaystyle\geqslant\frac{n+2s}{2}\;\frac{|x^{\prime}-y|^{2}-|x-y|^{2}}{|x^{\prime}-y|^{n+2s+2}}$
$\displaystyle=2(n+2s)\frac{x_{1}y_{1}}{|x^{\prime}-y|^{n+2s+2}}$
$\displaystyle\geqslant 2(n+2s)M^{n+2s+2}x_{1}y_{1},$

it follows that

$\displaystyle\mathcal{E}(w,\xi)$ $\displaystyle\leqslant
C\Big{(}\tau-\tilde{C}M^{n+2s+2}\|y_{1}v\|_{L^{1}(K)}\Big{)}\int_{B^{+}}x_{1}\xi(x)\mathop{}\\!dx$

with $C$ and $\tilde{C}$ depending only on $n$ and $s$.

Hence, using that $w=\tau\varphi\leqslant C\rho^{2s}\tau x_{1}$ in $B^{+}$ by
gmjWA2gy, we have that

$\displaystyle\mathcal{E}(w,\xi)+\int_{B^{+}}c(x)w(x)\xi(x)\mathop{}\\!dx$
$\displaystyle\leqslant
C\Big{[}\tau\big{(}1+\rho^{2s}\|c^{+}\|_{L^{\infty}(U)}\big{)}-\tilde{C}M^{n+2s+2}\|y_{1}v\|_{L^{1}(K)}\Big{]}\int_{B^{+}}x_{1}\xi(x)\mathop{}\\!dx$

(after possibly relabeling $C$ and $\tilde{C}$, but still depending only on
$n$ and $s$). Choosing

$\displaystyle\tau$
$\displaystyle:=\frac{1}{2}\tilde{C}M^{n+2s+2}\|y_{1}v\|_{L^{1}(K)}\big{(}1+\rho^{2s}\|c^{+}\|_{L^{\infty}(U)}\big{)}^{-1}$

we obtain that

$\displaystyle(-\Delta)^{s}w+cw$ $\displaystyle\leqslant 0\qquad\text{in
}B^{+}.$

Moreover, in $\mathbb{R}^{n}_{+}\setminus B^{+}$, we have that
$w=v\chi_{K}\leqslant v$, so, recalling also (3.1), [FJ15, Proposition 3.1]
implies that, in $B^{+}$,

$\displaystyle v(x)$ $\displaystyle\geqslant
w(x)=\frac{1}{2}\tilde{C}M^{n+2s+2}\big{(}1+\rho^{2s}\|c^{+}\|_{L^{\infty}(U)}\big{)}^{-1}\|y_{1}v\|_{L^{1}(K)}\varphi(x).$

Recalling that $\varphi\geqslant\rho^{2s}x_{1}\chi_{B_{\rho/2}^{+}(a)}$ (by
Lemma LABEL:gmjWA2gy), we obtain the final result. ∎

##  4\. The stability estimate and proof of Theorem LABEL:oAZAv7vy

The proof of oAZAv7vy makes use of the method of moving planes. Before we
begin our discussion of this technique and give the proof of oAZAv7vy, we must
fix some notation. Let $\mu\in\mathbb{R}$, $e\in\mathbb{S}^{n-1}$, and
$A\subset\mathbb{R}^{n}$. Then we have the following standard definitions:

$\displaystyle\pi_{\mu}$ $\displaystyle=\\{x\in\mathbb{R}^{n}\text{ s.t.
}x\cdot e=\mu\\}$ $\displaystyle\text{a hyperplane orthogonal to }e$
$\displaystyle H_{\mu}$ $\displaystyle=\\{x\in\mathbb{R}^{n}\text{ s.t.
}x\cdot e>\mu\\}$ $\displaystyle\text{the right-hand half space with respect
to }\pi_{\mu}$ $\displaystyle H_{\mu}^{\prime}$
$\displaystyle=\\{x\in\mathbb{R}^{n}\text{ s.t. }x\cdot e<\mu\\}$
$\displaystyle\text{the left-hand half space with respect to }\pi_{\mu}$
$\displaystyle A_{\mu}$ $\displaystyle=A\cap H_{\mu}$ $\displaystyle\text{the
portion of }A\text{ on the right-hand side of }\pi_{\mu}$ $\displaystyle
x_{\mu}^{\prime}$ $\displaystyle=x-2(x\cdot e-\mu)e$ $\displaystyle\text{the
reflection of }x\text{ across }\pi_{\mu}$ $\displaystyle A_{\mu}^{\prime}$
$\displaystyle=\\{x\in\mathbb{R}^{n}\text{ s.t. }x^{\prime}_{\mu}\in
A_{\mu}\\}$ $\displaystyle\text{the reflection of }A_{\mu}\text{ across
}\pi_{\mu}$

Note that in some articles such as [Cir+18] $A^{\prime}_{\mu}$ is used to
denote the reflection of $A$ (instead of $A_{\mu}$) across $\pi_{\mu}$.

The method of moving planes works as follows. Fix a direction
$e\in\mathbb{S}^{n-1}$ and suppose that $\Omega$ is a bounded open subset of
$\mathbb{R}^{n}$ with $C^{1}$ boundary. Since $\Omega$ is bounded, for $\mu$
sufficiently large the hyperplane $\pi_{\mu}$ does not intersect $\Omega$.
Furthermore, by decreasing the value of $\mu$, at some point $\pi_{\mu}$ will
intersect $\overline{\Omega}$. We denote the value of $\mu$ at this point by

$\displaystyle\Lambda=\Lambda_{e}:=\sup\\{x\cdot e\text{ s.t. }x\in\Omega\\}.$

From here, we continue to decrease the value of $\mu$. Initially, since
$\partial\Omega$ is $C^{1}$, the reflection of $\Omega_{\mu}$ across
$\pi_{\mu}$ will be contained within $\Omega$, that is
$\Omega_{\mu}^{\prime}\subset\Omega$ for $\mu<\Lambda$ but with $\mu$
sufficiently close to $\Lambda$. Eventually, as we continue to make $\mu$
smaller, there will come a point when this is no longer the case. More
precisely, there exists $\lambda=\lambda_{e}\in\mathbb{R}$ such that for all
$\mu\in[\lambda,\Lambda)$, it occurs that
$\Omega^{\prime}_{\mu}\subset\Omega$, but
$\Omega^{\prime}_{\mu}\not\subset\Omega$ for $\mu<\lambda$. We may write
$\lambda$ more explicitly as

(4.1) $\displaystyle\lambda:=\inf\Big{\\{}\tilde{\mu}\in\mathbb{R}\text{ s.t.
}\Omega^{\prime}_{\mu}\subset\Omega\text{ for all
}\mu\in(\tilde{\mu},\Lambda)\Big{\\}}.$

When $\mu=\lambda$, geometrically speaking, there are two possibilities that
can occur:

_Case 1:_ The boundary of $\Omega^{\prime}_{\lambda}$ is internally tangent to
$\partial\Omega$ at some point not on $\pi_{\lambda}$, that is, there exists
$p\in(\partial\Omega\cap\partial\Omega^{\prime}_{\lambda})\setminus\pi_{\lambda}$;
or

_Case 2:_ The critical hyperplane $\pi_{\lambda}$ is orthogonal to the
$\partial\Omega$ at some point, that is, there exists
$p\in\partial\Omega\cap\pi_{\mu}$ such that the normal of $\partial\Omega$ at
$p$ is contained in the plane $\pi_{\mu}$.

At this stage, let us introduce the function

(4.2) $\displaystyle v_{\mu}(x):=u(x)-u(x_{\mu}^{\prime}),\qquad\text{for all
}\mu\in\mathbb{R}\text{ and }x\in\mathbb{R}^{n}.$

It follows that, in $\Omega_{\mu}^{\prime}$,

$\displaystyle(-\Delta)^{s}v_{\mu}(x)$
$\displaystyle=f(u(x))-f(u(x_{\mu}^{\prime}))=-c_{\mu}(x)v_{\mu}(x)$

where

$\displaystyle c_{\mu}(x)$
$\displaystyle=\int_{0}^{1}f^{\prime}((1-t)u(x)+tu(x_{\mu}^{\prime}))\mathop{}\\!dt.$

Note that

$\displaystyle\|c_{\mu}\|_{L^{\infty}(\Omega_{\mu}^{\prime})}\leqslant[f]_{C^{0,1}([0,\|u\|_{L^{\infty}(\Omega)}])}.$

Hence, $v_{\mu}$ is an antisymmetric function that satisfies

$\displaystyle\left\\{\begin{array}[]{r@{ \ }l @{\quad\: \;}
l}(-\Delta)^{s}v_{\mu}+c_{\mu}v_{\mu}&=0\hfil\quad\>&\text{in
}\Omega_{\mu}^{\prime},\\\ v_{\mu}&=u\hfil\quad\>&\text{in }(\Omega\cap
H_{\mu}^{\prime})\setminus\Omega_{\mu}^{\prime},\\\
v_{\mu}&=0\hfil\quad\>&\text{in
}H_{\mu}^{\prime}\setminus\Omega,\end{array}\right.$

with $c_{\mu}\in L^{\infty}(\Omega_{\mu}^{\prime})$.

In the situations where one expects that $\Omega$ should be a ball, the goal
of the method of moving planes is to prove that $v_{\lambda}\equiv 0$ i.e. $u$
is even with respect to reflections across the critical hyperplane
$\pi_{\lambda}$. Since the direction $e$ was arbitrary, one can then deduce
that $u$ must be radial with respect to some point. The proof that
$v_{\lambda}\equiv 0$ is achieved through repeated applications of the maximum
principle.

######  Remark 4.1.

VbKq6OVB From the preceding exposition, it is clear that in several instances
we will need to evaluate $u$ and $v_{\mu}$ at a single point. This is
technically an issue since, in oAZAv7vy, $u$ is only assumed to be in
$H^{s}(\mathbb{R}^{n})\cap L^{\infty}(\mathbb{R}^{n})$. However, by standard
regularity theory, we have that111Here we are using the notation that for
$\alpha>0$ with $\alpha$ not an integer,
$C^{\alpha}(\Omega):=C^{k,\beta}(\Omega)$ where $k$ is the integer part of
$\alpha$ and $\beta=\alpha-k\in(0,1)$. $u\in C^{2s+1-\varepsilon}(\Omega)$ for
all $\varepsilon>0$ such that $2s+1-\varepsilon$ is not an integer. In
particular, this implies that $u\in C^{1}(\Omega)$ which will be essential for
the proof of the theorem. Indeed, using that $f$ is locally Lipschitz, we have
that $(-\Delta)^{s}u=-cu+f(0)$ with
$c(x)=-\int_{0}^{1}f^{\prime}(tu(x))\mathop{}\\!dt$, so $(-\Delta)^{s}u\in
L^{\infty}(\Omega)$. Hence, it follows that $u\in C^{2s-\varepsilon}(\Omega)$
for all $\varepsilon\in(0,2s)$, $2s-\varepsilon\not\in\mathbb{Z}$, see [RS14,
Proposition 2.3]. Then, it follows that $(-\Delta)^{s}u=f(u)\in
C^{\min\\{1-\varepsilon,2s-\varepsilon\\}}(\Omega)$, so by [RS14, Proposition
2.2] and a bootstrapping argument (if necessary), we obtain that $u\in
C^{2s+1-\varepsilon}(\Omega)$. ∎

###  4.1. Uniform stability in each direction

In this subsection, we will use the maximum principle of Section 3 to prove
uniform stability for each direction $e\in\mathbb{S}^{n-1}$, that is, we will
show, for each $e\in\mathbb{S}^{n-1}$, that $\Omega$ is almost symmetric with
respect to $e$. This is stated precisely in 7TQmUHhl. We will repeatedly use
that fact that $v_{\lambda}$ is $C^{1}$ which follows from VbKq6OVB. Before
proving 7TQmUHhl, we have two lemmata.

######  Lemma 4.2.

JFhLgBx4 Let $\Omega$ be a bounded open set with $C^{1}$ boundary,
$e\in\mathbb{S}^{n-1}$, and $v_{\mu}$ as in (4.2).

Then, for all $\mu\in[\lambda,\Lambda]$, we have that $v_{\mu}\geqslant 0$ in
$\Omega_{\mu}^{\prime}$.

The proof of JFhLgBx4 is given as part of the proof of Theorem 1.4 in
[Dip+22], so we will not include it again here. However, we would like to
emphasise that, even though [Dip+22, Theorem 1.4] assumes the solution $u$ of
(1.8) is constant on $\partial G$ (i.e. $[u]_{\partial G}=0$), this assumption
was unnecessary to obtain the much weaker result of JFhLgBx4.

We now give the second lemma.

######  Lemma 4.3.

goNy8kYt Let $\Omega$ and $G$ be open bounded sets with $C^{1}$ boundary such
that $\Omega=G+B_{R}$ for some $R>0$. Let $f\in
C^{0,1}_{\mathrm{loc}}(\mathbb{R})$ be such that $f(0)\geqslant 0$ and $u\in
H^{s}(\mathbb{R}^{n})$ be a solution of (1.8).

Then, for each $e\in\mathbb{S}^{n-1}$, we have that

(4.3) $\displaystyle\int_{(\Omega\cap
H^{\prime}_{\lambda})\setminus\Omega_{\lambda}^{\prime}}\delta_{\pi_{\lambda}}(x)u(x)\mathop{}\\!dx\leqslant
C[u]_{\partial G},$

where

(4.4) $\displaystyle
C:=C(n,s)R^{-2s}(\operatorname{diam}\Omega)^{n+2s+2}\Big{(}1+R^{2s}[f]_{C^{0,1}([0,\|u\|_{L^{\infty}(\Omega)}])}\Big{)}.$

###### Proof.

Without loss of generality, take $e=-e_{1}$ and $\lambda=0$. We now apply the
method of moving planes to $G$. First, suppose that we are in the first case,
namely, the boundary of $G^{\prime}_{\lambda}$ is internally tangent to
$\partial G$ at some point not on $\\{x_{1}=0\\}$, and let $p\in(\partial
G\cap\partial G_{\lambda}^{\prime})\setminus\\{x_{1}=0\\}$.

By TV1cTSyn with $U:=\Omega^{\prime}_{\lambda}$, $K:=(\Omega\cap
H_{\lambda}^{\prime})\setminus\Omega_{\lambda}^{\prime}$ and
$B_{\rho}(a):=B_{R/2}(p)$ (notice that condition (3.1) is satisfied, possibly
taking a smaller ball centered at $p$), we have that

(4.5) $\displaystyle\frac{v_{\lambda}(p)}{p_{1}}$ $\displaystyle\geqslant
C\int_{(\Omega\cap
H_{\lambda}^{\prime})\setminus\Omega_{\lambda}^{\prime}}y_{1}v_{\lambda}(y)\mathop{}\\!dy=C\int_{(\Omega\cap
H_{\lambda}^{\prime})\setminus\Omega_{\lambda}^{\prime}}y_{1}u(y)\mathop{}\\!dy.$

Note that, since $x_{\lambda}^{\prime}$ belongs to the reflection of $\Omega$
across $\\{x_{1}=0\\}$ for each $x\in(\Omega\cap
H_{\lambda}^{\prime})\setminus\Omega_{\lambda}^{\prime}$, we have that

$\displaystyle\inf_{{x\in(\Omega\cap
H^{\prime}_{\lambda})\setminus\Omega_{\lambda}^{\prime}}\atop{y\in
B_{R/2}^{+}(p)}}|x_{\lambda}^{\prime}-y|^{-1}$
$\displaystyle\geqslant(\operatorname{diam}\Omega)^{-1},$

so TV1cTSyn implies that the constant in (4.5) is given by

$\displaystyle
C=C(n,s)R^{2s}(\operatorname{diam}\Omega)^{-n-2s-2}\big{(}1+R^{2s}\|c^{+}\|_{L^{\infty}(\Omega_{\lambda}^{\prime})}\big{)}^{-1}.$

Moreover, we have that

$\displaystyle\frac{v_{\lambda}(p)}{p_{1}}=\frac{u(p)-u(p_{\lambda}^{\prime})}{p_{1}}=\frac{2(u(p)-u(p_{\lambda}^{\prime}))}{|p_{1}-(p_{\lambda}^{\prime})_{1}|}\leqslant
2[u]_{\partial G},$

which, along with (4.5), gives (4.3).

Now, let us suppose that we are in the second case and let $p\in\partial
G\cap\\{x_{1}=0\\}$ be such that the normal of $\partial G$ at $p$ is
contained in $\\{x_{1}=0\\}$. Proceeding in a similar fashion as the first
case, we apply TV1cTSyn with $U:=\Omega^{\prime}_{\lambda}$, $K:=(\Omega\cap
H_{\lambda}^{\prime})\setminus\Omega_{\lambda}^{\prime}$ and
$B_{\rho}(a):=B_{R/2}(p+he_{1})$ with $h>0$ very small, to obtain that

$\displaystyle\int_{(\Omega\cap
H_{\lambda}^{\prime})\setminus\Omega_{\lambda}^{\prime}}x_{1}u(x)\mathop{}\\!dx\leqslant
C\frac{v_{\lambda}(p+he_{1})}{h}$

with $C$ in the same form as in (4.4) and, in particular, independent of $h$.
Sending $h\to 0^{+}$, we obtain that

$\displaystyle\int_{(\Omega\cap
H_{\lambda}^{\prime})\setminus\Omega_{\lambda}^{\prime}}x_{1}u(x)\mathop{}\\!dx\leqslant
C\partial_{1}v_{\lambda}(p)\leqslant C[u]_{\partial G}$

which gives (4.3) in this case as well. ∎

We are now able to obtain uniform stability for each direction in 7TQmUHhl
below.

######  Proposition 4.4.

7TQmUHhl Let $\Omega$ be an open bounded set with $C^{1}$ boundary and
satisfying the uniform interior ball condition with radius $r_{\Omega}>0$ and
$G$ be an open bounded set with $C^{1}$ boundary such that $\Omega=G+B_{R}$
for some $R>0$. Let $f\in C^{0,1}_{\mathrm{loc}}(\mathbb{R})$ be such that
$f(0)\geqslant 0$ and $u\in H^{s}(\mathbb{R}^{n})$ be a solution of (1.8).

For $e\in\mathbb{S}^{n-1}$, let $\Omega^{\prime}$ denote the reflection of
$\Omega$ with respect to the critical hyperplane $\pi_{\lambda}$.

Then,

(4.6) $\displaystyle|\Omega\triangle\Omega^{\prime}|\leqslant
C_{\star}[u]_{\partial G}^{\frac{1}{s+2}},$

where

$\displaystyle C_{\star}$ $\displaystyle:=$ $\displaystyle
C(n,s)\Big{(}C_{\ast}^{-1}\big{(}f(0)+\|u\|_{L_{s}(\mathbb{R}^{n})}\big{)}^{-1}+(\operatorname{diam}\Omega)^{n-1}+r_{\Omega}^{-1}|\Omega|\Big{)}$
$\displaystyle\qquad\times
R^{-2s}(\operatorname{diam}\Omega)^{n+2s+2}\Big{(}1+R^{2s}[f]_{C^{0,1}([0,\|u\|_{L^{\infty}(\Omega)}])}\Big{)}$

and $C_{\ast}$ is as in zAPw0npM.

###### Proof.

Without loss of generality, take $e=-e_{1}$ and $\lambda=0$. By zAPw0npM, we
have that

(4.7) $\displaystyle\int_{(\Omega\cap
H_{\lambda}^{\prime})\setminus\Omega_{\lambda}^{\prime}}x_{1}u(x)\mathop{}\\!dx$
$\displaystyle\geqslant
C_{\ast}\big{(}f(0)+\|u\|_{L_{s}(\mathbb{R}^{n})}\big{)}\int_{(\Omega\cap
H_{\lambda}^{\prime})\setminus\Omega_{\lambda}^{\prime}}x_{1}\delta_{\partial\Omega}^{s}(x)\mathop{}\\!dx.$

Fix $\gamma>0$. By Chebyshev’s inequality, (4.7) and goNy8kYt, we have that

(4.8) $\begin{split}\Big{|}\big{\\{}x\in(\Omega\cap
H_{\lambda}^{\prime})\setminus\Omega_{\lambda}^{\prime}\text{ s.t.
}x_{1}\delta_{\partial\Omega}^{s}(x)>\gamma\big{\\}}\Big{|}\leqslant\;&\frac{1}{\gamma}\int_{(\Omega\cap
H_{\lambda}^{\prime})\setminus\Omega_{\lambda}^{\prime}}x_{1}\delta_{\partial\Omega}^{s}(x)\mathop{}\\!dx\\\
\leqslant\;&\frac{CC_{\ast}^{-1}\big{(}f(0)+\|u\|_{L_{s}(\mathbb{R}^{n})}\big{)}^{-1}}{\gamma}[u]_{\partial
G}.\end{split}$

Moreover,

$\displaystyle\Big{|}\big{\\{}x\in(\Omega\cap
H_{\lambda}^{\prime})\setminus\Omega_{\lambda}^{\prime}\text{ s.t.
}x_{1}\delta_{\partial\Omega}^{s}(x)\leqslant\gamma\big{\\}}\Big{|}$
$\displaystyle=$ $\displaystyle\Big{|}\big{\\{}x\in(\Omega\cap
H_{\lambda}^{\prime})\setminus\Omega_{\lambda}^{\prime}\text{ s.t.
}x_{1}\delta_{\partial\Omega}^{s}(x)\leqslant\gamma,x_{1}<\gamma^{\frac{1}{s+1}}\big{\\}}\Big{|}$
$\displaystyle\qquad+\Big{|}\big{\\{}x\in(\Omega\cap
H_{\lambda}^{\prime})\setminus\Omega_{\lambda}^{\prime}\text{ s.t.
}x_{1}\delta_{\partial\Omega}^{s}(x)\leqslant\gamma,x_{1}\geqslant\gamma^{\frac{1}{s+1}}\big{\\}}\Big{|}$
$\displaystyle\leqslant$ $\displaystyle\Big{|}\big{\\{}x\in\Omega^{+}\text{
s.t.
}x_{1}<\gamma^{\frac{1}{s+1}}\big{\\}}\Big{|}+\Big{|}\big{\\{}x\in\Omega\text{
s.t.
}\delta_{\partial\Omega}(x)\leqslant\gamma^{\frac{1}{s+1}}\big{\\}}\Big{|}.$

Furthermore, we have the estimate

$\displaystyle\Big{|}\big{\\{}x\in\Omega^{+}\text{ s.t.
}x_{1}<\gamma^{\frac{1}{s+1}}\big{\\}}\Big{|}$
$\displaystyle\leqslant(\operatorname{diam}\Omega)^{n-1}\gamma^{\frac{1}{s+1}}$

and, by [Cir+23, Lemma 5.2] in the case that $\partial\Omega$ is $C^{2}$ and
more generally in [Tam76], we have that

$\displaystyle\Big{|}\big{\\{}x\in\Omega\text{ s.t.
}\delta_{\partial\Omega}(x)\leqslant\gamma^{\frac{1}{s+1}}\big{\\}}\Big{|}$
$\displaystyle\leqslant\frac{2n|\Omega|}{r_{\Omega}}\gamma^{\frac{1}{s+1}}.$

Thus,

(4.9) $\displaystyle\Big{|}\big{\\{}x\in(\Omega\cap
H_{\lambda}^{\prime})\setminus\Omega_{\lambda}^{\prime}\text{ s.t.
}x_{1}\delta_{\partial\Omega}^{s}(x)\leqslant\gamma\big{\\}}\Big{|}\leqslant\left[(\operatorname{diam}\Omega)^{n-1}+\frac{2n|\Omega|}{r_{\Omega}}\right]\gamma^{\frac{1}{s+1}}.$

From this and (4.8), we deduce that

$\displaystyle|(\Omega\cap
H_{\lambda}^{\prime})\setminus\Omega^{\prime}_{\lambda}|\leqslant\frac{CC_{\ast}^{-1}\big{(}f(0)+\|u\|_{L_{s}(\mathbb{R}^{n})}\big{)}^{-1}}{\gamma}[u]_{\partial
G}+\left[(\operatorname{diam}\Omega)^{n-1}+\frac{2n|\Omega|}{r_{\Omega}}\right]\gamma^{\frac{1}{s+1}}.$

Hence,

$\displaystyle|\Omega\triangle\Omega^{\prime}|$ $\displaystyle=2|(\Omega\cap
H_{\lambda}^{\prime})\setminus\Omega^{\prime}_{\lambda}|\leqslant\frac{2CC_{\ast}^{-1}\big{(}f(0)+\|u\|_{L_{s}(\mathbb{R}^{n})}\big{)}^{-1}}{\gamma}[u]_{\partial
G}+2\left[(\operatorname{diam}\Omega)^{n-1}+\frac{2n|\Omega|}{r_{\Omega}}\right]\gamma^{\frac{1}{s+1}}$

for all $\gamma>0$. Choosing $\gamma=[u]_{\partial G}^{\frac{s+1}{s+2}}$, we
obtain that

$|\Omega\triangle\Omega^{\prime}|\leqslant\left[2CC_{\ast}^{-1}\big{(}f(0)+\|u\|_{L_{s}(\mathbb{R}^{n})}\big{)}^{-1}+2\left((\operatorname{diam}\Omega)^{n-1}+\frac{2n|\Omega|}{r_{\Omega}}\right)\right][u]_{\partial
G}^{\frac{1}{s+2}},$

which gives (4.6). ∎

Note, setting $[u]_{\partial G}=0$ in 7TQmUHhl implies that $\Omega$ (and,
therefore, $G$) must be symmetric with respect to the critical hyperplane for
every direction. This implies that they are both balls, thereby recovering the
main result of [Dip+22].

###  4.2. Proof of Theorem LABEL:oAZAv7vy

We will now use the results of the previous subsection to prove oAZAv7vy. In
fact, oAZAv7vy follows almost immediately from the following result, k4VRS3Fx.
The idea of k4VRS3Fx is to choose a centre for $\Omega$ by looking at the
intersection of $n$ orthogonal critical hyperplanes, then to prove that every
other critical hyperplane is quantifiably close to this centre in terms of the
semi-norm $[u]_{\partial G}$. The precise statement is as follows.

######  Proposition 4.5.

k4VRS3Fx Let $\Omega$ be an open bounded set with $C^{1}$ boundary and
satisfying the uniform interior ball condition with radius $r_{\Omega}>0$ and
suppose that the critical planes $\pi_{e_{i}}$ with respect to the coordinate
directions $e_{i}$ coincide with $\\{x_{i}=0\\}$ for every $i=1,\dots,n$.
Also, given $e\in\mathbb{S}^{n-1}$, denote by $\lambda_{e}$ the critical value
associated with $e$ as in (4.1).

Assume that

(4.10) $\displaystyle[u]^{\frac{1}{s+2}}_{\partial
G}\leqslant\frac{|\Omega|}{nC_{\star}}$

where $C_{\star}$ is the constant in 7TQmUHhl.

Then,

(4.11) $\displaystyle|\lambda_{e}|\leqslant C[u]_{\partial G}^{\frac{1}{s+2}}$

for all $e\in\mathbb{S}^{n-1}$ with

$\displaystyle C$ $\displaystyle:=$ $\displaystyle
C(n,s)\Big{(}C_{\ast}^{-1}\big{(}f(0)+\|u\|_{L_{s}(\mathbb{R}^{n})}\big{)}^{-1}+(\operatorname{diam}\Omega)^{n-1}+r_{\Omega}^{-1}|\Omega|\Big{)}$
$\displaystyle\qquad\times
R^{-2s}|\Omega|^{-1}(\operatorname{diam}\Omega)^{n+2s+3}\Big{(}1+R^{2s}[f]_{C^{0,1}([0,\|u\|_{L^{\infty}(\Omega)}])}\Big{)}$

and $C_{\ast}$ is as in zAPw0npM.

###### Proof.

Define $\Omega^{\mathbf{0}}:=\\{-x\text{ s.t. }x\in\Omega\\}$. Moreover, let
$Q_{i}:=Q_{\pi_{e_{i}}}$ be the reflections across each critical plane
$\pi_{e_{i}}$ and define $\Omega^{\mathbf{0}}$ recursively via
$\Omega^{\mathbf{0}}_{i+1}:=Q_{i+1}(\Omega^{\mathbf{0}}_{i})$ for $i=1,\dots
n-1$ with $\Omega^{\mathbf{0}}_{1}:=\Omega$. Observe that
$\Omega^{\mathbf{0}}=\Omega^{\mathbf{0}}_{n}$. Via the triangle inequality for
symmetric difference, it follows that

$\displaystyle|\Omega\triangle\Omega^{\mathbf{0}}|$
$\displaystyle\leqslant|\Omega\triangle Q_{n}(\Omega)|+|Q_{n}(\Omega)\triangle
Q_{n}(\Omega^{\mathbf{0}}_{n-1})|$

Since $|Q_{n}(\Omega)\triangle
Q_{n}(\Omega^{\mathbf{0}}_{n-1})|=|Q_{n}(\Omega\triangle\Omega^{\mathbf{0}}_{n-1})|=|\Omega\triangle\Omega^{\mathbf{0}}_{n-1}|$,
we have that

$\displaystyle|\Omega\triangle\Omega^{\mathbf{0}}|$
$\displaystyle\leqslant|\Omega\triangle
Q_{n}(\Omega)|+|\Omega\triangle\Omega^{\mathbf{0}}_{n-1}|.$

Iterating, we obtain

(4.12) $\displaystyle|\Omega\triangle\Omega^{\mathbf{0}}|$
$\displaystyle\leqslant\sum_{i=1}^{n}|\Omega\triangle Q_{i}(\Omega)|\leqslant
nC_{\star}[u]_{\partial G}^{\frac{1}{s+2}}$

by 7TQmUHhl.

Next, let us assume that $\lambda_{e}>0$ (the case $\lambda_{e}<0$ is
analogous). If $\Lambda_{e}>\operatorname{diam}\Omega$ then $x\cdot
e\geqslant\Lambda_{e}-\operatorname{diam}\Omega\geqslant 0$ for all
$x\in\Omega$, so $|\Omega\triangle\Omega^{\mathbf{0}}|=2|\Omega|$. However,
this is in contradiction with (4.10) and (4.12), so we must have that
$\Lambda_{e}\leqslant\operatorname{diam}\Omega$. Arguing as in Lemma 4.1 in
[Cir+18] using 7TQmUHhl instead of [Cir+18, Proposition 3.1 (a)] , we find
that

$\displaystyle|\Omega_{\lambda_{e}}|\lambda_{e}\leqslant(n+3)C_{\star}(\operatorname{diam}\Omega)[u]_{\partial
G}^{\frac{1}{s+2}}.$

Now, recalling the notation of Section 4, we have that

$\displaystyle|\Omega\triangle\Omega^{\mathbf{0}}|$
$\displaystyle=2|(\Omega\cap
H_{\lambda_{e}}^{\prime})\setminus\Omega_{\lambda_{e}}^{\prime}|=2\big{(}|\Omega|-2|\Omega_{\lambda_{e}}|\big{)},$

and therefore

$\displaystyle|\Omega_{\lambda_{e}}|=\frac{|\Omega|}{2}-\frac{1}{4}|\Omega\triangle\Omega^{\mathbf{0}}|\geqslant\frac{|\Omega|}{2}-\frac{1}{4}C_{\star}[u]_{\partial
G}^{\frac{1}{s+2}}\geqslant\frac{|\Omega|}{4}$

by 7TQmUHhl and (4.10). Thus, we conclude that

$\displaystyle|\lambda_{e}|$
$\displaystyle\leqslant\frac{C(n)C_{\star}(\operatorname{diam}\Omega)}{|\Omega|}[u]_{\partial
G}^{\frac{1}{s+2}},$

as desired. ∎

Now the proof of oAZAv7vy follows almost immediately from k4VRS3Fx.

###### Proof of oAZAv7vy.

The result follows by reasoning as in the proof of [Cir+18, Theorem 1.2], but
using k4VRS3Fx instead of [Cir+18, Lemma 4.1]. Notice that the dependence on
$r_{\Omega}$ appearing in k4VRS3Fx can be removed (and, in fact, does not
appear in oAZAv7vy). In fact, by the definition of $\Omega:=G+B_{R}(0)$, we
have that $\Omega$ automatically satisfies the uniform interior sphere
condition and we can take, e.g., $r_{\Omega}:=R/2$. ∎

##  5\. The role of boundary estimates in the attainment of the optimal
exponent

In this section, we give a broad discussion on some of the challenges the
nonlocality of the fractional Laplacian presents in obtaining the optimal
exponent. By way of an example, via the Poisson representation formula, we
show that estimates for a singular integral involving the reciprocal of the
distance to the boundary function play a key role in obtaining the anticipated
optimal result. This suggests, surprisingly, that fine geometric estimates for
the distance function close to the boundary are required to obtain the optimal
exponent.

Recall the notation at the beginning of Section 4 and also

$\displaystyle\mathbb{R}^{n}_{+}:=\\{x=(x_{1},\dots,x_{n})\in\mathbb{R}^{n}\;{\mbox{
s.t. }}x_{1}>0\\},$
$\displaystyle\mathbb{R}^{n}_{-}:=\\{x=(x_{1},\dots,x_{n})\in\mathbb{R}^{n}\;{\mbox{
s.t. }}x_{1}<0\\},$ $\displaystyle
B_{1}^{+}:=B_{1}\cap\mathbb{R}^{n}_{+},\qquad
B_{1}^{-}:=B_{1}\cap\mathbb{R}^{n}_{-},$
$\displaystyle\Omega^{+}:=\Omega\cap\mathbb{R}^{n}_{+}\qquad{\mbox{and}}\qquad\Omega^{-}:=\Omega\cap\mathbb{R}^{n}_{-}.$

Let $\Omega$ and $G$ be bounded and smooth sets. For the purposes of this
discussion, let us assume that $\Omega=G+B_{1/2}$ where $G$ is such that
$\mathbb{R}^{n}_{-}\cap G=B_{1/2}^{-}$,
$B_{1/2}^{+}\subset\mathbb{R}^{n}_{+}\cap\Omega$, and the critical plane
$\pi_{\lambda}$ corresponding to running the method of moving planes with
$e=-e_{1}$ is equal to $\\{x_{1}=0\\}$, see Figure 1.

Figure 1. Geometry of $\Omega$ as described in Section 5.

Suppose that $u\in C^{2}(\Omega)\cap L^{\infty}(\mathbb{R}^{n})$ satisfies the
torsion problem

$\displaystyle\left\\{\begin{array}[]{r@{ \ }l @{\quad\: \;}
l}(-\Delta)^{s}u&=1\hfil\quad\>&\text{in }\Omega,\\\ u&=0\hfil\quad\>&\text{in
}\mathbb{R}^{n}\setminus\Omega.\end{array}\right.$

In this case, if $v(x):=u(x)-u(-x_{1},x_{2},\dots,x_{n})$, then $v$ is
antisymmetric in $\mathbb{R}^{n}$ with respect to $\\{x_{1}=0\\}$ and
$s$-harmonic in $B_{1}$. Hence, by the Poisson kernel representation, (up to
normalization constants)

$\displaystyle v(x)$ $\displaystyle=\int_{\mathbb{R}^{n}\setminus
B_{1}}\bigg{(}\frac{1-|x|^{2}}{|y|^{2}-1}\bigg{)}^{s}\frac{v(y)}{|x-y|^{n}}\mathop{}\\!dy.$

Using the antisymmetry of $v$, we may rewrite this as

$\displaystyle v(x)$ $\displaystyle=\int_{\mathbb{R}^{n}_{+}\setminus
B_{1}^{+}}\bigg{(}\frac{1-|x|^{2}}{|y|^{2}-1}\bigg{)}^{s}\bigg{(}\frac{1}{|x-y|^{n}}-\frac{1}{|x^{\prime}-y|^{n}}\bigg{)}v(y)\mathop{}\\!dy$
$\displaystyle=\int_{\Omega^{+}\setminus
B_{1}^{+}}\bigg{(}\frac{1-|x|^{2}}{|y|^{2}-1}\bigg{)}^{s}\bigg{(}\frac{1}{|x-y|^{n}}-\frac{1}{|x^{\prime}-y|^{n}}\bigg{)}u(y)\mathop{}\\!dy,$

where $x^{\prime}$ denotes the reflection of the point $x$ with respect to
$\\{x_{1}=0\\}$.

Now, by oAZAv7vy, when $[u]_{\partial G}$ is sufficiently small, then $\Omega$
is uniformly close to $B_{1}$, so we can suppose that
$G_{\lambda}^{\prime}\subset B_{3/4}$. Therefore, for all $x\in\partial G$ and
$y\in\mathbb{R}^{n}_{+}\setminus\Omega$,

$\displaystyle(1-|x|^{2})^{s}\bigg{(}\frac{1}{|x-y|^{n}}-\frac{1}{|x^{\prime}-y|^{n}}\bigg{)}$
$\displaystyle\geqslant Cx_{1}y_{1}.$

Also, by Corollary LABEL:zAPw0npM, we have that $u\geqslant
C\delta^{s}_{\partial\Omega}$ (with $C>0$ not depending on $u$), and thus

$\displaystyle v(x)$ $\displaystyle\geqslant Cx_{1}\int_{\Omega^{+}\setminus
B_{1}^{+}}\frac{y_{1}u(y)}{(|y|^{2}-1)^{s}}\mathop{}\\!dy\geqslant
Cx_{1}\int_{\Omega^{+}\setminus
B_{1}^{+}}\frac{y_{1}\delta_{\partial\Omega}^{s}(y)}{\delta_{\partial
B_{1}}^{s}(y)}\mathop{}\\!dy.$

Recall the discussion on the moving plane method on page 4 and, for
simplicity, assume that we are in the first case (the second case can be
treated similarly), so there exists $p\in(\partial G\cap\partial
G_{\lambda}^{\prime})\setminus\\{x_{1}=0\\}$. Hence,

(5.1) $\displaystyle\int_{\Omega^{+}\setminus
B_{1}^{+}}y_{1}\bigg{(}\frac{\delta_{\partial\Omega}(y)}{\delta_{\partial
B_{1}}(y)}\bigg{)}^{s}\mathop{}\\!dy$ $\displaystyle\leqslant
C\frac{v(p)}{p_{1}}\leqslant C[u]_{\partial G}.$

Now the point of obtaining (5.1) is that the left hand side is geometric (does
not depend on $u$), and (5.1) is sharp in the sense that the only terms that
have been ‘thrown away’ are bounded away from zero as $[u]_{\partial G}\to 0$.
This suggests that if $[u]_{\partial G}$ is of order $\varepsilon$ then the
behaviour of

$\displaystyle\int_{\Omega^{+}\setminus
B_{1}^{+}}y_{1}\bigg{(}\frac{\delta_{\partial\Omega}(y)}{\delta_{\partial
B_{1}}(y)}\bigg{)}^{s}\mathop{}\\!dy$

as a function of $\varepsilon$ will entirely determine the optimal exponent
$\overline{\beta}(s)$ (recall that the definition of $\overline{\beta}(s)$ is
given after the statement of oAZAv7vy in Section 1). This is surprising, since
in the local case, the problem is entirely an interior one in that the proof
relies only on interior estimates such as the Harnack inequality while, it
appears that in the nonlocal case, the geometry of the boundary may have a
significant effect on the value of $\overline{\beta}(s)$. If one hopes to
obtain that $\overline{\beta}(s)=1$ from (5.1) then it would be necessary to
show that

$\displaystyle\int_{\Omega^{+}\setminus
B_{1}^{+}}y_{1}\bigg{(}\frac{\delta_{\partial\Omega}(y)}{\delta_{\partial
B_{1}}(y)}\bigg{)}^{s}\mathop{}\\!dy\geqslant C|\Omega\setminus B_{1}^{+}|.$

If it were true that $y_{1}\delta^{s}_{\partial\Omega}\geqslant
C\delta^{s}_{\partial B_{1}}$ then this would follow immediately; however,
this is not the case as seen by sending $y\to\partial\mathbb{R}^{n}_{+}\cap
B_{1/2}$.

Moreover, even though we made several major assumptions on the geometry of
$\Omega$ to obtain (5.1), we believe that the inequality is indicative of the
more general situation. Indeed, one would expect that, under reasonable
assumptions on $\partial\Omega$, similar estimates to the ones employed above
would hold for Poisson kernels in general domains, so one may suspect that an
inequality of the form

$\displaystyle\int_{\Omega\triangle\Omega^{\prime}}\delta_{\pi_{\lambda}}(y)\bigg{(}\frac{\delta_{\partial\Omega}(y)}{\delta_{\partial\Omega^{\prime}}(y)}\bigg{)}^{s}\mathop{}\\!dy$
$\displaystyle\leqslant C[u]_{\partial G}$

should hold. It would be interesting in future articles to further explore
this methodology to try obtain improvements on the exponent in oAZAv7vy.

##  6\. Sharp investigation of a technique to improve the stability exponent

In the papers [Cab+18, Cir+18], it was proven that the only open bounded
sets222For the precise statements and the regularity assumptions on the
boundary of the set, see [Cir+18, Cab+18]. whose boundary has constant
nonlocal mean curvature are balls. Moreover, in [Cir+18], the method of moving
planes was employed to obtain a stability estimate for sets whose boundary has
_almost_ constant nonlocal mean curvature. The stability estimate that the
authors proved also seems to achieve the optimal exponent of $1$ which is
however obtained using the following technical result, see [Cir+18,
Proposition 3.1(b)]:

######  Lemma 6.1.

Z08Q7917 Let $\Omega\subset\mathbb{R}^{n}$ be an open bounded set with
$C^{2,\alpha}$ boundary for $\alpha>2s$ and $s\in(0,1/2)$. Let $\pi_{\lambda}$
be the critical hyperplane. Suppose that
$\operatorname{dist}(0,\pi_{\lambda})\leqslant 1/8$ and
$B_{r}\subset\Omega\subset B_{R}$ with $1/2\leqslant r\leqslant R\leqslant 2$.

Then,

(6.1) $\displaystyle\Big{|}\big{\\{}x\in\Omega\triangle\Omega^{\prime}\text{
s.t.
}\operatorname{dist}(x,\pi_{\lambda})\leqslant\gamma\big{\\}}\Big{|}\leqslant
C(n)\gamma(R-r)$

for all $0<\gamma\leqslant 1/4$.

Without this result, the exponent in the stability estimate in [Cir+18] would
have been $1/2$, so Z08Q7917 seems to play a major role in [Cir+18] in the
attempt of “doubling the exponent” that they obtained in previous estimates.
This feature is extremely relevant to our main result, oAZAv7vy, since we
hoped that a similar argument to Z08Q7917 would lead to an exponent that was
twice the one that we obtained. Unfortunately, we believe that, in the very
broad generality in which the result is stated in [Cir+18], Z08Q7917 may not
be true, and we present a counter-example (which may also impact some of the
statements in [Cir+18]) as well as a corrected statement of Z08Q7917.

###  6.1. A geometric lemma and a counter-example to Z08Q7917

The main result of this section is a geometric lemma which may be viewed as a
corrected version of Z08Q7917. Moreover, we will give an explicit family of
domains which demonstrate that our geometric lemma is sharp. This family will
also serve as a counter-example to Z08Q7917.

An essential component to the geometric lemma that is not present in the
assumptions of Z08Q7917 is a uniform bound on the boundary regularity of
$\partial\Omega$. To our knowledge, there is no ‘standard’ definition of such
a bound, so we will begin by concretely specifying what is meant by this.

Let $\Omega$ be an open subset of $\mathbb{R}^{n}$ with $C^{1}$ boundary. For
each $x\in\partial\Omega$, let

$\Pi_{x}(y):=(y-x)-((y-x)\cdot\nu(x))\nu(x),$

which is the projection of $\mathbb{R}^{n}$ onto $T_{x}\partial\Omega$, that
is, the tangent plane of $\partial\Omega$ at $x$. We have made a translation
in the definition of $\Pi_{x}$, so that $\Pi_{x}(x)=0$. For simplicity, we
will often identify $T_{x}\partial\Omega$ with $\mathbb{R}^{n-1}$.

######  Definition 6.2.

deZqwlaB Let $\Omega$ be an open subset of $\mathbb{R}^{n}$ with $C^{\alpha}$
boundary, with $\alpha>1$, and let $\rho$, $M>0$. We say that
$\partial\Omega\in C^{\alpha}_{M,\rho}$ if, for all $x\in\partial\Omega$,
there exists $\psi^{(x)}:B_{\rho}^{n-1}\to\mathbb{R}$ such that $\psi^{(x)}\in
C^{\alpha}(B_{\rho}^{n-1})$,
$\|\psi^{(x)}\|_{C^{\alpha}(B_{\rho}^{n-1})}\leqslant M$, and

$\displaystyle B_{\rho}(x)\cap\Omega=\big{\\{}y\in B_{\rho}(x)\text{ s.t.
}y\cdot\nu(x)<\psi^{(x)}(\Pi_{x}(y))\big{\\}}.$

We remark that, if $\alpha\in\mathbb{Z}$ in Definition LABEL:deZqwlaB, the
notation $C^{\alpha}_{M,\rho}$ means $C^{\alpha-1,1}_{M,\rho}$.

We now give the geometric lemma, the main result of this section.

######  Theorem 6.3.

m9q9X2hE Let $\Omega\subset\mathbb{R}^{n}$ be an open bounded set with
$\partial\Omega\in C^{\alpha}_{M,\rho}$, with $\alpha>1$, for some $M>0$ and
$\rho\in(0,1/4)$. Moreover, suppose that $B_{r}\subset\Omega\subset B_{R}$
with $1/2\leqslant r\leqslant R\leqslant 2$. Let $e\in\mathbb{S}^{n-1}$,
denote by $\pi_{\lambda}$ the critical hyperplane with respect to $e$, and
suppose that $\operatorname{dist}(0,\pi_{\lambda})\leqslant 1/8$.

Then,

(6.2) $\displaystyle\Big{|}\big{\\{}x\in\Omega\triangle\Omega^{\prime}\text{
s.t.
}\operatorname{dist}(x,\pi_{\lambda})\leqslant\gamma\big{\\}}\Big{|}\leqslant
C\gamma(R-r)^{1-\frac{1}{\alpha}}$

for all $0<\gamma\leqslant 1/4$. The constant $C$ depends only on $n$, $M$,
$\rho$, and $\alpha$.

In order to prove Theorem LABEL:m9q9X2hE, we require two preliminary lemmata.
The first lemma gives an elementary estimate of the left hand side of (6.2) in
terms of $R-r$ and an error term involving
$\operatorname{dist}(0,\pi_{\lambda})$.

######  Lemma 6.4.

o1pUKCHX Let $\Omega\subset\mathbb{R}^{n}$ be an open bounded set with $C^{1}$
boundary, and $B_{r}\subset\Omega\subset B_{R}$ with $1/2\leqslant r\leqslant
R\leqslant 2$. Let $e\in\mathbb{S}^{n-1}$, denote by $\pi_{\lambda}$ the
critical hyperplane with respect to $e$, and suppose that
$\operatorname{dist}(0,\pi_{\lambda})\leqslant 1/8$.

Then,

$\displaystyle\Big{|}\big{\\{}x\in\Omega\triangle\Omega^{\prime}\text{ s.t.
}|x_{1}-\lambda|\leqslant\gamma\big{\\}}\Big{|}\leqslant
C\gamma\big{(}R-r+\gamma|\lambda|\big{)}$

for all $0<\gamma\leqslant 1/4$. The constant $C$ depends only on $n$.

###### Proof.

First, observe that

(6.3) $\displaystyle\Omega\triangle\Omega^{\prime}\subset(B_{R}\cup
B_{R}^{\prime})\setminus(B_{r}\cap B_{r}^{\prime})$

where $B_{\rho}^{\prime}$ is the reflection of $B_{\rho}$ across the critical
hyperplane. Indeed, if $x\in\Omega\cup\Omega^{\prime}$ then $x\in\Omega\subset
B_{R}$ or $x\in\Omega^{\prime}\subset B_{R}^{\prime}$, so $x\in B_{R}\cup
B_{R}^{\prime}$. Moreover, if $x\in B_{r}\cap B_{r}^{\prime}$ then $x\in
B_{r}\subset\Omega$ and $x\in B_{r}^{\prime}\subset\Omega^{\prime}$, so
$B_{r}\cap B_{r}^{\prime}\subset\Omega\cap\Omega^{\prime}$. Then (6.3) follows
from the the fact that
$\Omega\triangle\Omega^{\prime}=(\Omega\cup\Omega^{\prime})\setminus(\Omega\cap\Omega^{\prime})$.

From (6.3), it immediately follows that

$\displaystyle\Big{|}\big{\\{}x\in\Omega\triangle\Omega^{\prime}\text{ s.t.
}|x_{1}-\lambda|\leqslant\gamma\big{\\}}\Big{|}$
$\displaystyle\leqslant\Big{|}\big{\\{}x\in(B_{R}\cup
B_{R}^{\prime})\setminus(B_{r}\cap B_{r}^{\prime})\text{ s.t.
}|x_{1}-\lambda|\leqslant\gamma\big{\\}}\Big{|}.$

Without loss of generality, we may assume that $\lambda>0$. Then

$\displaystyle\Big{|}\big{\\{}x\in(B_{R}\cup B_{R}^{\prime})$
$\displaystyle\setminus(B_{r}\cap B_{r}^{\prime})\text{ s.t.
}|x_{1}-\lambda|\leqslant\gamma\big{\\}}\Big{|}$
$\displaystyle=2\Big{|}\big{\\{}x\in B_{R}\setminus B_{r}^{\prime}\text{ s.t.
}\lambda-\gamma<x_{1}<\lambda\big{\\}}\Big{|}$
$\displaystyle=2\Big{(}\Big{|}B_{R}\cap\\{\lambda-\gamma<x_{1}<\lambda\big{\\}}\Big{|}-\Big{|}B_{r}^{\prime}\cap\big{\\{}\lambda-\gamma<x_{1}<\lambda\big{\\}}\Big{|}\Big{)}.$

Hence, via the co-area formula, we obtain that

$\displaystyle\Big{|}\big{\\{}x\in(B_{R}\cup B_{R}^{\prime})$
$\displaystyle\setminus(B_{r}\cap B_{r}^{\prime})\text{ s.t.
}|x_{1}-\lambda|\leqslant\gamma\big{\\}}\Big{|}$
$\displaystyle=2\int_{\lambda-\gamma}^{\lambda}\left[\mathcal{H}^{n-1}\left(B_{\sqrt{R^{2}-t^{2}}}^{n-1}\right)-\mathcal{H}^{n-1}\left(B_{\sqrt{r^{2}-(2\lambda-t)^{2}}}^{n-1}\right)\right]\mathop{}\\!dt$
$\displaystyle=2\omega_{n-1}\int_{\lambda-\gamma}^{\lambda}\left[\big{(}R^{2}-t^{2}\big{)}^{\frac{n-1}{2}}-\big{(}r^{2}-(2\lambda-t)^{2}\big{)}^{\frac{n-1}{2}}\right]\mathop{}\\!dt.$

Next, if $0<a<\tau<b$ then

$\displaystyle\bigg{|}\frac{\mathop{}\\!d}{\mathop{}\\!d\tau}\tau^{\frac{n-1}{2}}\bigg{|}$
$\displaystyle=\frac{n-1}{2}\tau^{\frac{n-3}{2}}\leqslant\frac{n-1}{2}\begin{cases}a^{-1/2},&\text{if
}n=2\\\ b^{\frac{n-3}{2}},&\text{if }n\geqslant 3.\end{cases}$

Since $\lambda>0$ and $2\lambda-t\in(\lambda,\lambda+\gamma)\subset(0,3/8)$,
we have that

$\displaystyle C^{-1}\leqslant r^{2}-(2\lambda-t)^{2}\leqslant
R^{2}-t^{2}\leqslant C\qquad\text{ for all }t\in(\lambda-\gamma,\lambda)$

with $C>0$ a universal constant. Hence,

$\displaystyle\Big{|}\big{(}R^{2}-t^{2}\big{)}^{\frac{n-1}{2}}-\big{(}r^{2}-(2\lambda-t)^{2}\big{)}^{\frac{n-1}{2}}\Big{|}$
$\displaystyle\leqslant
C\big{|}\big{(}R^{2}-t^{2}\big{)}-\big{(}r^{2}-(2\lambda-t)^{2}\big{)}\big{|}$
$\displaystyle\leqslant C\big{(}R-r+\lambda(\lambda-t)\big{)}.$

Gathering these pieces of information, we conclude that

$\displaystyle\Big{|}\big{\\{}x\in\Omega\triangle\Omega^{\prime}\text{ s.t.
}|x_{1}-\lambda|\leqslant\gamma\big{\\}}\Big{|}$ $\displaystyle\leqslant
C\int_{\lambda-\gamma}^{\lambda}\big{(}R-r+\lambda(\lambda-t)\big{)}\mathop{}\\!dt\leqslant
C\gamma\big{(}R-r+\gamma\lambda\big{)},$

as required. ∎

The purpose of the second lemma will be to reduce the proof of m9q9X2hE to the
case of graphs of functions.

######  Lemma 6.5.

GNZAnsDn Let $\Omega\subset\mathbb{R}^{n}$ be an open bounded set with
$\partial\Omega\in C^{\alpha}_{M,\rho}$, with $\alpha>1$, for some $M>0$ and
$\rho\in(0,1/4)$.

Then, there exists $\varepsilon_{0}\in(0,1)$ such that, for all
$\varepsilon\in(0,\varepsilon_{0}]$, it holds that if
$B_{1-\varepsilon/2}\subset\Omega\subset B_{1+\varepsilon/2}$ then there
exists $\psi_{\varepsilon}:=\psi:B_{3/4}^{n-1}\to(0,+\infty)$ such that
$\psi\in C^{\alpha}(B_{3/4}^{n-1})$,

(6.4)
$\displaystyle\big{\|}\psi-\sqrt{1-|x^{\prime}|^{2}}\big{\|}_{C^{1}(B_{3/4}^{n-1})}\leqslant
C\varepsilon^{1-\frac{1}{\alpha}},$

and

(6.5)
$\displaystyle\Omega\cap\big{(}B_{3/4}^{n-1}\times(0,+\infty)\big{)}=\\{x\in\mathbb{R}^{n}\text{
s.t. }0<x_{n}<\psi(x^{\prime})\\}.$

The constant $C$ depends only on $n$, $\alpha$, $\rho$, and $M$.

###### Proof.

Let
$x_{\star}\in\partial\Omega\cap\big{(}B_{3/4}^{n-1}\times(0,+\infty)\big{)}$
and $\psi^{(x_{\star})}$ be the function given in deZqwlaB corresponding to
the point $x_{\star}$. Next, let $A$ be a rigid motion such that
$Ax_{\star}=0$, $T_{x_{\star}}\partial\Omega$ is mapped t
$\mathbb{R}^{n-1}\times\\{0\\}$, and $\nu(x_{\star})$ is mapped to $e_{n}$. To
avoid any confusion, we will always use the variable $x$ to denote points in
the original (unrotated) coordinates and $y$ to denote points in the new
rotated coordinates, i.e. $y=Ax$. By deZqwlaB, in the $y$ coordinates,

$\displaystyle y=(y^{\prime},\psi^{(x_{\star})}(y^{\prime}))\qquad\text{ for
all }y\in B_{\rho}\cap\partial\Omega$

where $y^{\prime}=(y_{1},\dots,y_{n-1})$.

Next, if $\partial\Omega$ is contained in the closure of
$B_{1+\varepsilon/2}\setminus B_{1-\varepsilon/2}$, we have that

$\displaystyle\bigg{(}1-\frac{\varepsilon}{2}\bigg{)}^{2}-|y^{\prime}|^{2}\leqslant\big{(}\psi^{(x_{\star})}(y^{\prime})\big{)}^{2}\leqslant\bigg{(}1+\frac{\varepsilon}{2}\bigg{)}^{2}-|y^{\prime}|^{2}$

for all $y^{\prime}\in B^{n-1}_{\rho}$.

Additionally, by Bernoulli’s inequality, we have that

$\displaystyle
0\leqslant\sqrt{\bigg{(}1+\frac{\varepsilon}{2}\bigg{)}^{2}-|y^{\prime}|^{2}}-\sqrt{1-|y^{\prime}|^{2}}$
$\displaystyle\leqslant\frac{1}{2\sqrt{1-|y^{\prime}|^{2}}}\bigg{(}\varepsilon+\frac{\varepsilon^{2}}{4}\bigg{)}\leqslant
C\varepsilon$

with $C=C(\rho)>0$. Similarly, we also have that

$\displaystyle-C\varepsilon\leqslant\sqrt{\bigg{(}1-\frac{\varepsilon}{2}\bigg{)}^{2}-|y^{\prime}|^{2}}-\sqrt{1-|y^{\prime}|^{2}}\leqslant
0.$

Hence, it follows that

$\displaystyle\big{|}|\psi^{(x_{\star})}(y^{\prime})|-\sqrt{1-|y^{\prime}|^{2}}\big{|}$
$\displaystyle\leqslant C\varepsilon\qquad\text{for all }y^{\prime}\in
B^{n-1}_{\rho}.$

Thus, by interpolation, we have that

$\displaystyle\big{\|}$
$\displaystyle|\psi^{(x_{\star})}(y^{\prime})|-\sqrt{1-|y^{\prime}|^{2}}\big{\|}_{C^{1}(B^{n-1}_{\rho})}$
$\displaystyle\leqslant
C\big{\|}|\psi^{(x_{\star})}(y^{\prime})|-\sqrt{1-|y^{\prime}|^{2}}\big{\|}_{C^{\alpha}(B^{n-1}_{\rho})}\big{\|}|\psi^{(x_{\star})}(y^{\prime})|-\sqrt{1-|y^{\prime}|^{2}}\big{\|}_{L^{\infty}(B^{n-1}_{\rho})}^{1-\frac{1}{\alpha}}$
$\displaystyle\leqslant C\varepsilon^{1-\frac{1}{\alpha}}$

using also that $\partial\Omega\in C^{\alpha}_{M,\rho}$.

Note that we have left the $y$ variable in the equations above to emphasise
that we are still using the rotated $y$ coordinates. Now, returning to the
original $x$ coordinates, we observe that, by the above computation, we have
that $\partial\Omega$ is uniformly close to $\partial B_{1}$ in the $C^{1}$
sense, so

$\displaystyle\nu(x)\cdot e_{n}\geqslant C>0\qquad\text{ for all
}x\in\partial\Omega\cap\big{(}B_{3/4}^{n-1}\times(0,+\infty)\big{)}$

provided that $\varepsilon$ is sufficiently small. Thus, it follows that
$\partial\Omega$ is given by a graph with respect to the $e_{n}$ direction,
that is, the claim in (6.5) holds for some $\psi:B_{3/4}^{n-1}\to(0,+\infty)$.
We can see that $\psi\in C^{\alpha}(B_{3/4}^{n-1})$ since $\partial\Omega$ is
$C^{\alpha}$ and we obtain the claim in (6.4) by an identical interpolation
argument to the one above. ∎

We may now give the proof of m9q9X2hE.

###### Proof of m9q9X2hE.

Without loss of generality, we may assume that $e=e_{1}$ and $\lambda>0$. By
o1pUKCHX, it is enough to prove that

(6.6) $\displaystyle\lambda\leqslant
C(n,M,\rho,\alpha)(R-r)^{1-\frac{1}{\alpha}}.$

Since $\lambda\leqslant 2$, if there exists $C=C(n,M,\rho,\alpha)>0$ such that
$R-r\geqslant C$ then we are done, so we may assume that $R-r=\varepsilon$
with $\varepsilon$ arbitrarily small. Moreover, by rescaling, we may further
assume that $r=1-\varepsilon/2$ and $R=1+\varepsilon/2$. Furthermore, let
$\psi:B_{3/4}^{n-1}$ be the function given by GNZAnsDn.

First, let us consider Case 1 of the method of moving planes. In this case, we
obtain a point
$p=(p_{1},\dots,p_{n})\in(\partial\Omega\cap\partial\Omega^{\prime}_{\lambda})\setminus\pi_{\lambda}$
(recall that, in this notation, $\Omega^{\prime}_{\lambda}=\Omega^{\prime}\cap
H_{\lambda}$, so $p_{1}<\lambda$). Hence, we have that
$r^{2}\leqslant|p|^{2}\leqslant(r+\varepsilon)^{2}$ and
$r^{2}\leqslant|Q_{\pi_{\lambda}}(p)|^{2}\leqslant(r+\varepsilon)^{2}$ (recall
that $Q_{\pi_{\lambda}}$ reflects a point across $\pi_{\lambda}$), from which
it follows that

$\displaystyle\lambda(\lambda-
p_{1})\leqslant\varepsilon(2r+\varepsilon)\leqslant C\varepsilon.$

If $\lambda-p_{1}\geqslant 1/4$ then we are done, so we may assume that
$\lambda-p_{1}<1/4$.

Moreover, by rotating with respect to $(x_{2},\dots,x_{n})$, we may assume
without loss of generality that $p_{2}=\dots=p_{n-1}=0$ and $p_{n}>0$. In
particular, this implies that $p^{\prime}=(p_{1},\dots,p_{n-1})\in
B_{3/4}^{n-1}$.

Now, on one hand, if
$\psi_{\lambda}(x^{\prime}):=\psi(x^{\prime})-\psi(2\lambda-
x_{1},x_{2},\dots,x_{n-1})$ then $\psi_{\lambda}\geqslant 0$ for
$x^{\prime}\in B_{3/4}^{n-1}\cap\\{x_{1}<\lambda\\}$ and
$\psi_{\lambda}(p^{\prime})=0$, so $\partial_{1}\psi_{\lambda}(p^{\prime})=0$.
On the other hand, by GNZAnsDn, if $x^{\prime\prime}=(x_{2},\dots,x_{n-1})$,

$\displaystyle\partial_{1}\psi_{\lambda}(p^{\prime})$
$\displaystyle=\partial_{1}\big{(}\psi_{\lambda}-\sqrt{1-|x^{\prime}|^{2}}+\sqrt{1-(2\lambda-
x_{1})^{2}-|x^{\prime\prime}|^{2}}\big{)}(p^{\prime})$
$\displaystyle\qquad+\partial_{1}\big{(}\sqrt{1-|x^{\prime}|^{2}}-\sqrt{1-(2\lambda-
x_{1})^{2}-|x^{\prime\prime}|^{2}}\big{)}(p^{\prime})$ $\displaystyle\leqslant
C\varepsilon^{1-\frac{1}{\alpha}}-\bigg{(}\frac{p_{1}}{\sqrt{1-p_{1}^{2}}}+\frac{2\lambda-
p_{1}}{\sqrt{1-(2\lambda-p_{1})^{2}}}\bigg{)}.$

Hence, we have that

(6.7) $\frac{p_{1}}{\sqrt{1-p_{1}^{2}}}+\frac{2\lambda-
p_{1}}{\sqrt{1-(2\lambda-p_{1})^{2}}}\leqslant
C\varepsilon^{1-\frac{1}{\alpha}}.$

Moreover, we claim that

(6.8) $\frac{p_{1}}{\sqrt{1-p_{1}^{2}}}+\frac{2\lambda-
p_{1}}{\sqrt{1-(2\lambda-
p_{1})^{2}}}\geqslant\frac{2\lambda}{\sqrt{1-\lambda^{2}}}.$

To prove this, we consider the function

$f(t):=\frac{\lambda-t}{\sqrt{1-(\lambda-t)^{2}}}+\frac{\lambda+t}{\sqrt{1-(\lambda+t)^{2}}}$

for $t\in\left(-\frac{1}{2},\frac{1}{2}\right)$. Notice that $f$ is even, and
therefore we restrict our analysis to the interval
$\left[0,\frac{1}{2}\right)$.

By a direct calculation,

$\displaystyle f^{\prime}(t)$ $\displaystyle=$
$\displaystyle-\frac{(\lambda-t)^{2}}{(1-(\lambda-t)^{2})^{3/2}}-\frac{1}{\sqrt{1-(\lambda-t)^{2}}}+\frac{1}{\sqrt{1-(\lambda+t)^{2}}}+\frac{(\lambda+t)^{2}}{(1-(\lambda+t)^{2})^{3/2}}$
$\displaystyle=$
$\displaystyle-\frac{1}{(1-(\lambda-t)^{2})^{3/2}}+\frac{1}{(1-(\lambda+t)^{2})^{3/2}}.$

Now we define $\phi(\tau):=(1-\tau)^{-3/2}$. Since
$(\lambda-t)^{2}\leqslant(\lambda+t)^{2}\leqslant 17/32<1$, we can write

$\displaystyle
f^{\prime}(t)=\int_{(\lambda-t)^{2}}^{(\lambda+t)^{2}}\phi^{\prime}(\tau)\,d\tau=\frac{3}{2}\int_{(\lambda-t)^{2}}^{(\lambda+t)^{2}}(1-\tau)^{-5/2}\,d\tau\geqslant
0.$

Therefore, for all $t\in\left(0,\frac{1}{2}\right)$ (and thus for all
$t\in\left(-\frac{1}{2},\frac{1}{2}\right)$), we have that

$f(t)\geqslant f(0)=\frac{2\lambda}{\sqrt{1-\lambda^{2}}}.$

Hence, setting $t=\lambda-p_{1}\in\left(-\frac{1}{2},\frac{1}{2}\right)$, we
obtain the desired inequality in (6.8).

In particular, from (6.8) we have that

$\frac{p_{1}}{\sqrt{1-p_{1}^{2}}}+\frac{2\lambda-p_{1}}{\sqrt{1-(2\lambda-
p_{1})^{2}}}\geqslant C\lambda,$

for some $C>0$. Putting together this and (6.7), we deduce that the claim in
(6.6) holds.

For Case 2 of the method of moving planes, we obtain a point
$p\in\partial\Omega\cap\pi_{\lambda}$ at which $\partial\Omega$ is orthogonal
to $\pi_{\lambda}$. In this case, we have that
$\partial_{1}\psi_{\lambda}(p^{\prime})\geqslant 0$. From here the proof is
identical to Case 1. ∎

We will now give an explicit family of domains which demonstrate that the
result obtained in m9q9X2hE is sharp. This also serves as a counter-example to
Z08Q7917.

######  Theorem 6.6.

There exists a smooth $1$-parameter family $\\{\Omega_{\varepsilon}\\}$ of
open bounded subsets of $\mathbb{R}^{2}$ such that

  * •

$\partial\Omega_{\varepsilon}\in C^{\alpha}_{M,1/8}$, with $\alpha>1$, for
some $M>0$ independent of $\varepsilon$;

  * •

$B_{1-C_{1}\varepsilon}\subset\Omega_{\varepsilon}\subset
B_{1+C_{1}\varepsilon}$ for a universal constant $C_{1}>0$;

  * •

if $\pi_{\lambda}$ is the critical hyperplane with respect to $e_{1}$ and
$\gamma\in(0,1/4)$, then

(6.9)
$\displaystyle\Big{|}\big{\\{}x\in\Omega_{\varepsilon}\triangle\Omega_{\varepsilon}^{\prime}\text{
s.t.
}\operatorname{dist}(x,\pi_{\lambda})\leqslant\gamma\big{\\}}\Big{|}\geqslant
C_{2}\varepsilon^{1-\frac{1}{\alpha}}$

as $\varepsilon\to 0^{+}$. The constant $C_{2}>0$ depends on $n$ and $\gamma$.

Figure 2. Diagram of $\Omega_{\varepsilon}$ in Theorem 6.6.

###### Proof.

We will define $\Omega_{\varepsilon}$ by specifying its boundary, see Figure
2. More precisely, in the region
$\mathbb{R}^{2}\setminus\big{(}(0,1/2)\times(-3/2,-1/2)\big{)}$, let
$\partial\Omega_{\varepsilon}=\partial B_{1}$. For the definition of
$\partial\Omega_{\varepsilon}$ in $(0,1/2)\times(-3/2,-1/2)$, let $\eta~{}\in
C^{\infty}_{0}(\mathbb{R})$ be such that

$\displaystyle\begin{cases}\eta(\tau)=2\tau&\text{in }(-1/4,1/4),\\\
\eta=0&\text{in }\mathbb{R}\setminus(-1,1),\\\
\eta(-\tau)=-\eta(\tau)&\text{in }\mathbb{R},\\\ |\eta|\leqslant 1&\text{in
}\mathbb{R}.\end{cases}$

Let also

$\displaystyle\eta_{\varepsilon}(\tau):=\varepsilon\eta\left(\frac{\tau-\varepsilon^{1-\frac{1}{\alpha}}}{\varepsilon^{1/\alpha}}\right)\qquad{\mbox{and}}\qquad\psi_{\varepsilon}(\tau):=-\sqrt{1-\tau^{2}}-\eta_{\varepsilon}(\tau)$

and define the remaining portion of $\partial\Omega_{\varepsilon}$ by

$\displaystyle\partial\Omega_{\varepsilon}\cap\Big{(}(0,1/2)\times(-3/2,-1/2)\Big{)}=\big{\\{}(x,\psi_{\varepsilon}(x))\text{
s.t. }x\in(0,1/2)\big{\\}}.$

We observe that $\partial\Omega_{\varepsilon}$ is smooth and that

$\displaystyle\tau^{2}+(\psi_{\varepsilon}(\tau))^{2}$
$\displaystyle=1+2\eta_{\varepsilon}(\tau)\sqrt{1-\tau^{2}}+(\eta_{\varepsilon}(\tau))^{2}\leqslant
1+C\varepsilon.$

Hence, by Bernoulli’s inequality,

$\displaystyle\sqrt{\tau^{2}+\psi_{\varepsilon}(\tau)^{2}}-1$
$\displaystyle\leqslant C\varepsilon,$

and similarly, we can show that
$\sqrt{\tau^{2}+\psi_{\varepsilon}(\tau)^{2}}-1\geqslant-C\varepsilon$, so
$B_{1-C\varepsilon}\subset\Omega_{\varepsilon}\subset B_{1+C\varepsilon}$.
Moreover, we have that $\|\psi_{\varepsilon}\|_{C^{\alpha}((0,1/2))}\leqslant
M$ for some $M>0$ sufficiently large and independent of $\varepsilon$, so
$\partial\Omega_{\varepsilon}\in C^{\alpha}_{M,1/8}$.

Now all that is left to be shown is (6.9). First, we claim that the critical
parameter $\lambda_{\varepsilon}$ satisfies

(6.10)
$\displaystyle\lambda_{\varepsilon}\geqslant\varepsilon^{1-\frac{1}{\alpha}}.$

To prove this, for all $\mu\in\mathbb{R}$, we define
$\psi_{\varepsilon,\mu}(\tau):=\psi_{\varepsilon}(\tau)-\psi_{\varepsilon}(2\mu-\tau)$.
Furthermore, we consider

$\tau_{\varepsilon}:=\varepsilon^{1-\frac{1}{\alpha}}-\frac{1}{4}\varepsilon^{\frac{1}{\alpha}}\qquad{\mbox{and}}\qquad\mu_{\varepsilon}:=\varepsilon^{1-\frac{1}{\alpha}}.$

We have that

$\displaystyle\psi_{\varepsilon,\mu_{\varepsilon}}(\tau_{\varepsilon})$
$\displaystyle=$
$\displaystyle-\Big{(}\sqrt{1-\tau_{\varepsilon}^{2}}-\sqrt{1-(2\mu_{\varepsilon}-\tau_{\varepsilon})^{2}}\Big{)}-\varepsilon\left(\eta\left(\frac{\tau_{\varepsilon}-\varepsilon^{1-\frac{1}{\alpha}}}{\varepsilon^{1/\alpha}}\right)-\eta\left(\frac{2\mu_{\varepsilon}-\tau_{\varepsilon}-\varepsilon^{1-\frac{1}{\alpha}}}{\varepsilon^{1/\alpha}}\right)\right)$
$\displaystyle=$
$\displaystyle-\Big{(}\sqrt{1-\tau_{\varepsilon}^{2}}-\sqrt{1-(2\mu_{\varepsilon}-\tau_{\varepsilon})^{2}}\Big{)}-2\varepsilon\bigg{(}\frac{\tau_{\varepsilon}-\varepsilon^{1-\frac{1}{\alpha}}}{\varepsilon^{1/\alpha}}-\frac{2\mu_{\varepsilon}-\tau_{\varepsilon}-\varepsilon^{1-\frac{1}{\alpha}}}{\varepsilon^{1/\alpha}}\bigg{)}$
$\displaystyle=$
$\displaystyle-\frac{4\mu_{\varepsilon}(\mu_{\varepsilon}-\tau_{\varepsilon})}{\sqrt{1-\tau_{\varepsilon}^{2}}+\sqrt{1-(2\mu_{\varepsilon}-\tau_{\varepsilon})^{2}}}+4\varepsilon^{1-\frac{1}{\alpha}}\big{(}\mu_{\varepsilon}-\tau_{\varepsilon}\big{)}$
$\displaystyle=$
$\displaystyle-4\mu_{\varepsilon}(\mu_{\varepsilon}-\tau_{\varepsilon})\left(\frac{1}{2}+o(1)\right)+4\varepsilon^{1-\frac{1}{\alpha}}\big{(}\mu_{\varepsilon}-\tau_{\varepsilon}\big{)}$
$\displaystyle=$ $\displaystyle
2\varepsilon^{1-\frac{1}{\alpha}}\big{(}\mu_{\varepsilon}-\tau_{\varepsilon}\big{)}+o\Big{(}\varepsilon^{1-\frac{1}{\alpha}}\big{(}\mu_{\varepsilon}-\tau_{\varepsilon}\big{)}\Big{)}$
$\displaystyle=$ $\displaystyle\frac{1}{2}\varepsilon+o(\varepsilon)$
$\displaystyle\geqslant$ $\displaystyle\frac{1}{4}\varepsilon,$

as soon as $\varepsilon$ is sufficiently small.

Hence, we have that
$\psi_{\varepsilon,\mu_{\varepsilon}}(\tau_{\varepsilon})>0$, which implies
that the reflected region of $\Omega_{\varepsilon,\mu}$ must have left the
region $\Omega_{\varepsilon}$ by the time we have reached
$\mu=\mu_{\varepsilon}$, so we conclude that the critical time
$\lambda_{\varepsilon}$ satisfies $\lambda_{\varepsilon}>\mu_{\varepsilon}$,
which gives (6.10).

In fact, since $\eta$ is zero outside of $(-1,1)$, it follows that

$\displaystyle\lambda_{\varepsilon}\leqslant
C\varepsilon^{1-\frac{1}{\alpha}},$

so, in particular, $\lambda_{\varepsilon}\to 0$ as $\varepsilon\to 0^{+}$.

Finally, to complete the proof of (6.9), we will show that

(6.11)
$\Big{|}\big{\\{}x\in\Omega_{\varepsilon}\triangle\Omega_{\varepsilon}^{\prime}\text{
s.t.
}\operatorname{dist}(x,\pi_{\lambda})\leqslant\gamma\big{\\}}\Big{|}\geqslant
C\lambda_{\varepsilon}.$

Indeed, let
$\mathscr{U}_{\varepsilon}:=\big{(}\big{(}\Omega_{\varepsilon}\cap\\{\lambda-\gamma<x_{1}<\lambda\\}\big{)}\setminus\Omega_{\varepsilon,\lambda}^{\prime}\big{)}\cap\\{x_{2}>0\\}$,
see Figure 2. Observe that

$\displaystyle\Big{|}\big{(}\Omega_{\varepsilon}\cap\\{\lambda-\gamma<x_{1}<\lambda\\}\big{)}\setminus\Omega_{\varepsilon,\lambda}^{\prime}\Big{|}\geqslant|\mathscr{U}_{\varepsilon}|.$

Moreover, using that $\lambda_{\varepsilon}\to 0$, we have that

$\displaystyle|\mathscr{U}_{\varepsilon}|$ $\displaystyle=$
$\displaystyle\int_{\lambda_{\varepsilon}-\gamma}^{\lambda_{\varepsilon}}\Big{(}\sqrt{1-\tau^{2}}-\sqrt{1-(2\lambda_{\varepsilon}-\tau)^{2}}\Big{)}\mathop{}\\!d\tau$
$\displaystyle=$
$\displaystyle\int_{\lambda_{\varepsilon}-\gamma}^{\lambda_{\varepsilon}}\frac{4\lambda_{\varepsilon}(\lambda_{\varepsilon}-\gamma)}{\sqrt{1-\tau^{2}}+\sqrt{1-(2\lambda_{\varepsilon}-\tau)^{2}}}\mathop{}\\!d\tau$
$\displaystyle\geqslant$ $\displaystyle
4\lambda_{\varepsilon}\int_{\lambda_{\varepsilon}-\gamma}^{\lambda_{\varepsilon}-\frac{\gamma}{2}}\frac{\lambda_{\varepsilon}-\gamma}{2}\mathop{}\\!d\tau$
$\displaystyle\geqslant$
$\displaystyle\frac{\gamma^{2}}{2}\lambda_{\varepsilon},$

which establishes (6.11), and thus completes the proof of Theorem 6.6. ∎

###  6.2. An application of m9q9X2hE: improvement of the exponent in oAZAv7vy

We will now show how to use m9q9X2hE to improve the stability exponent in
m9q9X2hE. The key step is to use m9q9X2hE to achieve the following modified
version of 7TQmUHhl.

######  Proposition 6.7.

Prop:new per improvement Let $\Omega$ be an open bounded set with $C^{1}$
boundary and satisfying the uniform interior ball condition with radius
$r_{\Omega}>0$ and $G$ be an open bounded set with $C^{1}$ boundary such that
$\Omega=G+B_{R}$ for some $R>0$. Let $f\in C^{0,1}_{\mathrm{loc}}(\mathbb{R})$
be such that $f(0)\geqslant 0$ and $u$ satisfies (1.8) in the weak sense.

For $e\in\mathbb{S}^{n-1}$, let $\Omega^{\prime}$ denote the reflection of
$\Omega$ with respect to the critical hyperplane $\pi_{\lambda}$.

In addition, suppose that $\partial\Omega\in C^{\alpha}_{M,\rho}$, with
$\alpha>1$, for some $M>0$ and $\rho\in(0,1/4)$, and that

(6.12) $B_{\rho_{i}}\subset\Omega\subset B_{\rho_{e}}\quad\text{ with
}\quad\frac{1}{2}\leqslant\rho_{i}\leqslant\rho_{e}\leqslant 2$

and

(6.13) $\rho_{e}-\rho_{i}\geqslant
4^{\frac{\alpha(s+2)}{(\alpha-1)(s+1)}}[u]_{\partial
G}^{\frac{\alpha}{(\alpha-1)(s+1)}}.$

Then,

$\displaystyle|\Omega\triangle\Omega^{\prime}|\leqslant\widetilde{C}_{\star}(\rho_{e}-\rho_{i})^{\frac{\alpha-1}{\alpha(s+2)}}\,[u]_{\partial
G}^{\frac{1}{s+2}},$

where $\widetilde{C}_{\star}$ is some explicit constant depending on $n$, $s$,
$\alpha$, $M$, $\rho$, $\operatorname{diam}\Omega$, $R$,
$[f]_{C^{0,1}([0,\|u\|_{L^{\infty}(\Omega)}])}$, and
$\|u\|_{L_{s}(\mathbb{R}^{n})}$.

###### Proof.

The proof of Proposition LABEL:Prop:new_per_improvement is a suitable
modification of the one of 7TQmUHhl. More precisely, we recall (4.8) and we
use that $f(0)\geqslant 0$ to obtain that

(6.14) $\begin{split}\Big{|}\big{\\{}x\in(\Omega\cap
H_{\lambda}^{\prime})\setminus\Omega_{\lambda}^{\prime}\text{ s.t.
}x_{1}\delta_{\partial\Omega}^{s}(x)>\gamma\big{\\}}\Big{|}\leqslant\;&\frac{CC_{\ast}^{-1}\big{(}f(0)+\|u\|_{L_{s}(\mathbb{R}^{n})}\big{)}^{-1}}{\gamma}[u]_{\partial
G}\\\
\leqslant\;&\frac{CC_{\ast}^{-1}\|u\|_{L_{s}(\mathbb{R}^{n})}^{-1}}{\gamma}[u]_{\partial
G},\end{split}$

where $C_{\ast}$ is as in zAPw0npM.

Furthermore, given $\gamma$, $\beta\in\left(0,\frac{1}{4}\right]$, we see that

$\displaystyle\Big{|}\big{\\{}x\in(\Omega\cap
H_{\lambda}^{\prime})\setminus\Omega_{\lambda}^{\prime}\text{ s.t.
}x_{1}\delta_{\partial\Omega}^{s}(x)\leqslant\gamma\big{\\}}\Big{|}$
$\displaystyle=$ $\displaystyle\Big{|}\big{\\{}x\in(\Omega\cap
H_{\lambda}^{\prime})\setminus\Omega_{\lambda}^{\prime}\text{ s.t.
}x_{1}\delta_{\partial\Omega}^{s}(x)\leqslant\gamma,x_{1}<\beta\big{\\}}\Big{|}$
$\displaystyle\qquad+\Big{|}\big{\\{}x\in(\Omega\cap
H_{\lambda}^{\prime})\setminus\Omega_{\lambda}^{\prime}\text{ s.t.
}x_{1}\delta_{\partial\Omega}^{s}(x)\leqslant\gamma,x_{1}\geqslant\beta\big{\\}}\Big{|}$
$\displaystyle\leqslant$ $\displaystyle\Big{|}\big{\\{}x\in\Omega^{+}\text{
s.t. }x_{1}<\beta\big{\\}}\Big{|}+\left|\left\\{x\in\Omega\text{ s.t.
}\delta_{\partial\Omega}(x)\leqslant\left(\frac{\gamma}{\beta}\right)^{\frac{1}{s}}\right\\}\right|.$

Now, using m9q9X2hE we have that

$\displaystyle\Big{|}\big{\\{}x\in\Omega^{+}\text{ s.t.
}x_{1}<\beta\big{\\}}\Big{|}$ $\displaystyle\leqslant
C\,(\rho_{e}-\rho_{i})^{1-\frac{1}{\alpha}}\beta,$

for some $C>0$ depending only on $n$, $M$, $\rho$ and $\alpha$.

Also, using [Cir+23, Lemma 5.2] and [Tam76], we obtain that

$\displaystyle\left|\left\\{x\in\Omega\text{ s.t.
}\delta_{\partial\Omega}(x)\leqslant\left(\frac{\gamma}{\beta}\right)^{\frac{1}{s}}\right\\}\right|$
$\displaystyle\leqslant\frac{2n|\Omega|}{r_{\Omega}}\left(\frac{\gamma}{\beta}\right)^{\frac{1}{s}}.$

Gathering these pieces of information, we obtain that

$\displaystyle\Big{|}\big{\\{}x\in(\Omega\cap
H_{\lambda}^{\prime})\setminus\Omega_{\lambda}^{\prime}\text{ s.t.
}x_{1}\delta_{\partial\Omega}^{s}(x)\leqslant\gamma\big{\\}}\Big{|}\leqslant
C(\rho_{e}-\rho_{i})^{1-\frac{1}{\alpha}}\beta+\frac{2n|\Omega|}{r_{\Omega}}\left(\frac{\gamma}{\beta}\right)^{\frac{1}{s}}.$

From this and (6.14), we now deduce that

$\displaystyle|\Omega\triangle\Omega^{\prime}|$ $\displaystyle=$
$\displaystyle 2|(\Omega\cap
H_{\lambda}^{\prime})\setminus\Omega^{\prime}_{\lambda}|$
$\displaystyle\leqslant$ $\displaystyle
CC_{\ast}^{-1}\|u\|_{L_{s}(\mathbb{R}^{n})}^{-1}\frac{[u]_{\partial
G}}{\gamma}+C(\rho_{e}-\rho_{i})^{1-\frac{1}{\alpha}}\beta+\frac{2n|\Omega|}{r_{\Omega}}\left(\frac{\gamma}{\beta}\right)^{\frac{1}{s}}$
$\displaystyle\leqslant$ $\displaystyle C\left[\frac{[u]_{\partial
G}}{\gamma}+(\rho_{e}-\rho_{i})^{1-\frac{1}{\alpha}}\beta+\left(\frac{\gamma}{\beta}\right)^{\frac{1}{s}}\right].$

Minimizing the expression in the last line in the variables $(\beta,\gamma)$
gives

$\gamma\beta=\frac{[u]_{\partial
G}}{(\rho_{e}-\rho_{i})^{1-\frac{1}{\alpha}}}\qquad{\mbox{ and
}}\qquad\beta=\frac{[u]_{\partial
G}^{\frac{1}{s+2}}}{(\rho_{e}-\rho_{i})^{{\frac{(\alpha-1)(s+1)}{\alpha(s+2)}}}},$

that is,

$\gamma=\frac{[u]_{\partial
G}^{\frac{s+1}{s+2}}}{(\rho_{e}-\rho_{i})^{\frac{\alpha-1}{\alpha(s+2)}}}\qquad{\mbox{
and }}\qquad\beta=\frac{[u]_{\partial
G}^{\frac{1}{s+2}}}{(\rho_{e}-\rho_{i})^{{\frac{(\alpha-1)(s+1)}{\alpha(s+2)}}}}.$

Notice that the assumption in (6.13) guarantees that, with these choices,
$\gamma$, $\beta\in\left(0,\frac{1}{4}\right]$.

Thus, we conclude that

$|\Omega\triangle\Omega^{\prime}|\leqslant
C\,(\rho_{e}-\rho_{i})^{\frac{\alpha-1}{\alpha(s+2)}}\,[u]_{\partial
G}^{\frac{1}{s+2}}.$

This completes the proof of Proposition LABEL:Prop:new_per_improvement. ∎

Proposition LABEL:Prop:new_per_improvement leads to the following statement,
which is the counterpart of Proposition LABEL:k4VRS3Fx.

######  Proposition 6.8.

k4VRS3FxBIS Let $\Omega$ be an open bounded set with $C^{1}$ boundary and
satisfying the uniform interior ball condition with radius $r_{\Omega}>0$ and
suppose that the critical planes $\pi_{e_{i}}$ with respect to the coordinate
directions $e_{i}$ coincide with $\\{x_{i}=0\\}$ for every $i=1,\dots,n$.
Also, given $e\in\mathbb{S}^{n-1}$, denote by $\lambda_{e}$ the critical value
associated with $e$ as in (4.1).

In addition, suppose that $\partial\Omega\in C^{\alpha}_{M,\rho}$, with
$\alpha>1$, for some $M>0$ and $\rho\in(0,1/4)$, and that the assumptions in
(6.12) and (6.13) are satisfied.

Assume that

$\displaystyle[u]^{\frac{1}{s+2}}_{\partial
G}\leqslant\frac{|\Omega|}{n\widetilde{C}_{\star}}$

where $\widetilde{C}_{\star}$ is the constant in Proposition
LABEL:Prop:new_per_improvement.

Then,

$\displaystyle|\lambda_{e}|\leqslant
C\,(\rho_{e}-\rho_{i})^{\frac{\alpha-1}{\alpha(s+2)}}[u]_{\partial
G}^{\frac{1}{s+2}}$

for all $e\in\mathbb{S}^{n-1}$, where $C$ is some explicit constant depending
on $n$, $s$, $\alpha$, $M$, $\rho$, $\operatorname{diam}\Omega$, $R$,
$[f]_{C^{0,1}([0,\|u\|_{L^{\infty}(\Omega)}])}$, and
$\|u\|_{L_{s}(\mathbb{R}^{n})}$.

We omit the proof of Proposition LABEL:k4VRS3FxBIS, since it follows the same
line as that of Proposition LABEL:k4VRS3Fx.

Now, up to a translation, we can suppose that the critical planes
$\pi_{e_{i}}$ with respect to the coordinate directions $e_{i}$ coincide with
$\\{x_{i}=0\\}$ for every $i=1,\dots,n$. Hence, following [Cir+18, Proof of
Theorem 1.2], we can define

(6.15) $\rho_{e}:=\max_{\partial\Omega}|x|\qquad{\mbox{ and
}}\qquad\rho_{i}:=\min_{\partial\Omega}|x|,$

and work with $\rho_{e}-\rho_{i}$ (which clearly gives an upper bound for
$\rho(\Omega)$) to achieve the desired stability estimates.

In the following result, we also assume that

(6.16) $\rho_{e}=1.$

Notice that such an assumption is always satisfied, up to a dilation.

######  Theorem 6.9.

thm:improvement Let $G$ be an open bounded subset of $\mathbb{R}^{n}$ and
$\Omega=G+B_{R}$ for some $R>0$. Furthermore, let $\Omega$ and $G$ have
$C^{1}$ boundary, and let $f\in C^{0,1}_{\mathrm{loc}}(\mathbb{R})$ be such
that $f(0)\geqslant 0$. Suppose that $u$ satisfies (1.8) in the weak sense.

In addition, let $\partial\Omega\in C^{\alpha}_{M,\rho}$, with $\alpha>1$, for
some $M>0$ and $\rho\in(0,1/4)$.

Let $\rho_{e}$ and $\rho_{i}$ be as in (6.15) and (6.16).

Then,

$\displaystyle\rho(\Omega)\leqslant\rho_{e}-\rho_{i}$
$\displaystyle\leqslant\widetilde{C}[u]_{\partial
G}^{\frac{\alpha}{1+\alpha(s+1)}},$

where $\widetilde{C}$ is some explicit constant depending on $n$, $s$,
$\alpha$, $M$, $\rho$, $\operatorname{diam}\Omega$, $R$,
$[f]_{C^{0,1}([0,\|u\|_{L^{\infty}(\Omega)}])}$, and
$\|u\|_{L_{s}(\mathbb{R}^{n})}$.

###### Proof.

Without loss of generality, we can assume that (6.13) is satisfied, as
otherwise thm:improvement trivially holds.

An inspection of the proof of oAZAv7vy shows that $\rho_{e}$ and $\rho_{i}$
defined in (6.15) satisfy

$\displaystyle\rho_{e}-\rho_{i}$ $\displaystyle\leqslant C[u]_{\partial
G}^{\frac{1}{s+2}}.$

Thus, if $[u]_{\partial G}\leqslant(2C)^{-(s+2)}$, we obtain that
$\rho_{e}-\rho_{i}\leqslant 1/2$, and therefore, in light of the assumption in
(6.16), we have that $1/2\leqslant\rho_{i}\leqslant\rho_{e}\leqslant 2$,
namely (6.12) is satisfied.

Hence, we can employ Proposition LABEL:k4VRS3FxBIS and obtain that

(6.17) $|\lambda_{e}|\leqslant
C\,(\rho_{e}-\rho_{i})^{\frac{\alpha-1}{\alpha(s+2)}}[u]_{\partial
G}^{\frac{1}{s+2}}.$

The desired result follows by reasoning as in the proof of oAZAv7vy, but using
(6.17) instead of (4.11). ∎

##  7\. The exponent for a family of ellipsoids

Throughout this section, we will denote a point
$x=(x_{1},\dots,x_{n})\in\mathbb{R}^{n}$ by $x=(x_{1},x^{\prime})$ where333Not
to be confused with the notation of the method of moving planes where an
apostrophe referred to a reflection across a hyperplane.
$x^{\prime}=(x_{2},\dots,x_{n})\in\mathbb{R}^{n-1}$.

######  Proposition 7.1.

0LlyfQlx Suppose that $n\geqslant 2$ is an integer, $s\in(0,1)$, and
$\varepsilon>0$. Define

(7.1)
$\displaystyle\Omega_{\varepsilon}:=\bigg{\\{}(x_{1},x^{\prime})\in\mathbb{R}^{n}\text{
s.t. }\frac{x_{1}^{2}}{(1+\varepsilon)^{2}}+|x^{\prime}|^{2}<1\bigg{\\}}.$

Then, for $0<\varepsilon<1/4$, there exists a 1-parameter family
$G_{\varepsilon}\subset\Omega_{\varepsilon}$ with smooth boundary such that

$\displaystyle G_{\varepsilon}+B_{1/2}=\Omega_{\varepsilon}.$

Moreover, let $u_{\varepsilon}\in C^{2}(\Omega_{\varepsilon})\cap
C^{s}(\mathbb{R}^{n})$ be the unique solution to

(7.4) $\displaystyle\left\\{\begin{array}[]{r@{ \ }l @{\quad\: \;}
l}(-\Delta)^{s}u_{\varepsilon}&=1\hfil\quad\>&\text{in
}\Omega_{\varepsilon},\\\ u_{\varepsilon}&=0\hfil\quad\>&\text{in
}\mathbb{R}^{n}\setminus\Omega_{\varepsilon}.\end{array}\right.$

Then,

(7.5) $\displaystyle\lim_{\varepsilon\to
0^{+}}\frac{[u_{\varepsilon}]_{\partial
G_{\varepsilon}}}{\rho(\Omega_{\varepsilon})}$
$\displaystyle=s\gamma_{n,s}\bigg{(}\frac{3}{4}\bigg{)}^{s-1},$

where

(7.6)
$\displaystyle\gamma_{n,s}:=\frac{2^{-2s}\Gamma\big{(}\frac{n}{2}\big{)}}{\Gamma\left(\frac{n+2s}{2}\right)\Gamma(1+s)}.$

The existence of the 1-parameter family $G_{\varepsilon}$ as in 0LlyfQlx is an
easy corollary of the following geometric lemma.

######  Lemma 7.2.

jUJs9Upe Let $n\geqslant 2$ be an integer and suppose that $\Omega$ is an open
bounded subset of $\mathbb{R}^{n}$ that satisfies the uniform interior ball
condition with radius $r_{\Omega}>0$. If $\delta:\Omega\to\mathbb{R}$ is the
distance to the boundary of $\Omega$ and $\Omega^{\rho}:=\\{x\in\Omega\text{
s.t. }\delta(x)>\rho\\}$ then for all $\rho\in(0,r_{\Omega})$,

$\displaystyle\Omega^{\rho}+B_{\rho}=\Omega.$

Moreover, if the boundary of $\Omega$ is $C^{k}$ for some integer $k\geqslant
2$ then, for all $\rho\in(0,r_{\Omega})$, the boundary of $\Omega^{\rho}$ is
also $C^{k}$.

###### Proof.

We will begin by showing that $\Omega^{\rho}+B_{\rho}\subset\Omega$. If
$z\in\Omega^{\rho}+B_{\rho}$ then $z=x+y$ for some $x\in\Omega^{\rho}$ and
$y\in B_{\rho}$. Let $R>0$ be the largest value possible such that
$B_{R}(x)\subset\Omega$. Then $R=\delta(x)$, so $R>\rho$. Hence,

$\displaystyle|z-x|=|y|<\rho<R,$

so $z\in B_{R}(x)\subset\Omega$.

Now, let us show that $\Omega\subset\Omega^{\rho}+B_{\rho}$. Let $z\in\Omega$.
If $z\in\Omega^{\rho}$ then $z=z+0\in\Omega^{\rho}+B_{\rho}$ and we are done,
so we may assume that $z\in\Omega\setminus\Omega^{\rho}$. Now, let $R>0$ be
the largest value possible such that $B_{R}(z)\subset\Omega$ and let
$p\in\partial\Omega$ be a point at which $B_{R}(z)$ touches $\partial\Omega$.
Define the unit vector

$\displaystyle\nu(p):=\frac{p-z}{|p-z|}.$

Next, since $B_{R}(z)$ is an interior touching ball, there exists some
$\bar{z}\in\Omega$ such that $B_{r_{\Omega}}(\bar{z})$ also touches
$\partial\Omega$ at $p$ and $p$, $z$, and $\bar{z}$ are collinear. Now, let
$\mu\in(0,\min\\{r_{\Omega}-\rho,R\\})$ and define

$\displaystyle x:=z-(\rho-R+\mu)\nu(p).$

Since $z=\bar{z}+(r_{\Omega}-R)\nu(p)$, it follows that
$|x-\bar{z}|=|r_{\Omega}-\rho-\mu|<r_{\Omega}$, so $x\in
B_{r_{\Omega}}(\bar{z})$. Hence, $x\in\Omega$. Moreover,
$|x-p|=\rho+\mu>\rho$, so $x\in\Omega^{\rho}$ and $|z-x|=\rho-R+\mu<\rho$, so
$z-x\in B_{\rho}$. Thus, we have that $z=x+(z-x)\in\Omega^{\rho}+B_{\rho}$ as
required. This completes the first part of the proof.

The fact that the boundary of $\Omega^{\rho}$ is $C^{k}$ for any
$\rho\in(0,r_{\Omega})$ is a simple consequence of [GT01, Lemma 14.16].
Indeed, following the proof of [GT01, Lemma 14.16], we see that $\delta\in
C^{k}(\Gamma^{r_{\Omega}})$ where

$\Gamma^{r_{\Omega}}:=\\{x\in\Omega\text{ s.t. }\delta(x)<r_{\Omega}\\}.$

Then the boundary of $\Omega^{\rho}$ is simply the level set
$\Omega\cap\\{\delta=\rho\\}$ which is contained in $\Gamma^{r_{\Omega}}$, so
$\partial\Omega^{\rho}$ is $C^{k}$. ∎

From jUJs9Upe, we immediately obtain the proof of the first part of 0LlyfQlx:

######  Corollary 7.3.

FOHX1qV4 Suppose that $n\geqslant 2$ is an integer, $s\in(0,1)$,
$\varepsilon>0$, and $\Omega_{\varepsilon}$ be as in (7.1).

Then, for $0<\varepsilon<1/4$, there exists a 1-parameter family
$G_{\varepsilon}\subset\Omega_{\varepsilon}$ with smooth boundary such that

$\displaystyle G_{\varepsilon}+B_{1/2}=\Omega_{\varepsilon}.$

###### Proof.

It is easy to check that $\Omega_{\varepsilon}$ satisfies the uniform interior
ball condition with uniform radius given by
$r_{\Omega_{\varepsilon}}=(1+\varepsilon)^{-1}$. Hence, since
$\varepsilon<1/4$, we have that $r_{\Omega_{\varepsilon}}>1/2$, so if
$G_{\varepsilon}:=\Omega_{\varepsilon}^{\rho}$ with $\rho=1/2$ (i.e.
$G_{\varepsilon}:=\\{x\in\Omega_{\varepsilon}\text{ s.t.
}\operatorname{dist}(x,\partial\Omega_{\varepsilon})>1/2\\}$) then jUJs9Upe
implies that $G_{\varepsilon}$ satisfies all the required properties. ∎

The remainder of this section will be spent proving the second part of
0LlyfQlx. In theory, this is relatively simple: since $\Omega_{\varepsilon}$
is an ellipsoid, the solution to the torsion problem (7.4) is known
explicitly, see [AJS21], and is given by

$\displaystyle
u_{\varepsilon}(x):=\gamma_{n,s,\varepsilon}\bigg{(}1-\frac{x_{1}^{2}}{(1+\varepsilon)^{2}}-|x^{\prime}|^{2}\bigg{)}^{s}_{+}$

where

$\displaystyle\gamma_{n,s,\varepsilon}:=\frac{\gamma_{n,s}}{(1+\varepsilon){}_{2}F_{1}\big{(}\frac{n+2s}{2},\frac{1}{2};\frac{n}{2};1-(1+\varepsilon)^{2}\big{)}}$

and $\gamma_{n,s}$ is given by (7.6).

Here above ${}_{2}F_{1}$ is the Hypergeometric function, see [AS92] for more
details. Note that ${}_{2}F_{1}(a,b;c;0)=1$, so $\gamma_{n,s,0}=\gamma_{n,s}$,
and $\gamma_{n,s}$ is precisely the constant for which
$u_{0}(x):=\gamma_{n,s}(1-|x|^{2})^{s}_{+}$ satisfies $(-\Delta)^{s}u_{0}=1$
in $B_{1}$.

Now, it is easy to check that $\rho(\Omega_{\varepsilon})=\varepsilon$, so to
prove the second part of 0LlyfQlx, one simply has to find the first order
expansion of $[u_{\varepsilon}]_{\partial G_{\varepsilon}}$ in $\varepsilon$.
However, in practice, this is not trivial for a few reasons: one,
parametrizations of $G_{\varepsilon}$ are algebraically complicated; two,
$[u_{\varepsilon}]_{\partial G_{\varepsilon}}$ is defined in terms of a
supremum, which, in general, does not commute well with limits; and three, the
supremum is over a quotient whose numerator and denominator both depend on
$\varepsilon$ and whose denominator is not bounded away from zero.

We will now state a technical result that will allow us to postpone addressing
these difficulties and to proceed directly to the proof of 0LlyfQlx.

######  Lemma 7.4.

M2fW7tQX Let $\varepsilon\in[0,1)$ and define
$a_{\varepsilon},b_{\varepsilon}:[0,+\infty)\to\mathbb{R}$ and
$\phi_{\varepsilon}:B_{1}^{n-1}\to\mathbb{R}^{n}$ as follows

(7.7)
$\begin{split}a_{\varepsilon}(\tau):=\;&1+\varepsilon-\frac{1}{2\sqrt{1+\big{(}(1+\varepsilon)^{2}-1\big{)}\tau^{2}}},\\\
b_{\varepsilon}(\tau):=\;&1-\frac{1+\varepsilon}{2\sqrt{1+\big{(}(1+\varepsilon)^{2}-1\big{)}\tau^{2}}},\\\
{\mbox{and}}\qquad\phi_{\varepsilon}(r):=\;&\Big{(}a_{\varepsilon}(|r|)\sqrt{1-|r|^{2}},b_{\varepsilon}(|r|)r\Big{)}.\end{split}$

Then,

$\displaystyle\bigg{|}$
$\displaystyle\frac{|(u_{\varepsilon}\circ\phi_{\varepsilon})(r)-(u_{\varepsilon}\circ\phi_{\varepsilon})(\tilde{r})|}{\varepsilon|\phi_{\varepsilon}(r)-\phi_{\varepsilon}(\tilde{r})|}-\frac{1}{2}s\gamma_{n,s}\left(\frac{3}{4}\right)^{s-1}\frac{\big{|}|r|^{2}-|\tilde{r}|^{2}\big{|}}{|\phi_{0}(r)-\phi_{0}(\tilde{r})|}\bigg{|}\leqslant
C\varepsilon.$

The constant $C>0$ depends only on $n$ and $s$.

In the statement of M2fW7tQX and in what follows, we use the notation
$B^{n-1}_{1}$ for the unit ball in $\mathbb{R}^{n-1}$.

We will withhold the proof of M2fW7tQX until the end of the section. Given
M2fW7tQX, we can now complete the proof of 0LlyfQlx.

###### Proof 0LlyfQlx.

The proof of the first statement regarding the existence of $G_{\varepsilon}$
is the subject of FOHX1qV4, so we will focus on proving (7.5). Since the
ellipsoid is symmetric with respect to reflections across the plane
$\\{x_{1}=0\\}$, it follows that $G_{\varepsilon}$ is too. Hence, we have that

$\displaystyle[u_{\varepsilon}]_{\partial G_{\varepsilon}}$
$\displaystyle=\sup_{\begin{subarray}{c}x,y\in\partial G_{\varepsilon}^{+}\\\
x\neq
y\end{subarray}}\bigg{\\{}\frac{|u_{\varepsilon}(x)-u_{\varepsilon}(y)|}{|x-y|}\bigg{\\}}.$

Here, we are using the notation that, given $A\subset\mathbb{R}^{n}$,
$A^{+}:=A\cap\\{x_{1}>0\\}$.

Next, the upper half ellipsoid $\Omega_{\varepsilon}^{+}$ can be parametrized
by $\bar{\phi}_{\varepsilon}:B_{1}^{n-1}\to\Omega_{\varepsilon}^{+}$ defined
by $\bar{\phi}_{\varepsilon}(r):=((1+\varepsilon)\sqrt{1-|r|^{2}},r)$.
Moreover, the outward pointing normal to $B_{1}$ is given by

$\displaystyle\nu_{\varepsilon}(x)$
$\displaystyle:=\frac{1}{\sqrt{\frac{x_{1}^{2}}{(1+\varepsilon)^{4}}+|x^{\prime}|^{2}}}\bigg{(}\frac{x_{1}}{(1+\varepsilon)^{2}},x^{\prime}\bigg{)},$

so a parametrization of $\partial G_{\varepsilon}^{+}$ is
$\phi_{\varepsilon}:B_{1}^{n-1}\to\mathbb{R}^{n}$ defined by
$\phi_{\varepsilon}:=\bar{\phi}_{\varepsilon}-\frac{1}{2}(\nu_{\varepsilon}\circ\bar{\phi}_{\varepsilon})$,
that is,

$\displaystyle\phi_{\varepsilon}(r)$
$\displaystyle=\big{(}(1+\varepsilon)\sqrt{1-|r|^{2}},r\big{)}-\frac{1+\varepsilon}{2\sqrt{1+((1+\varepsilon)^{2}-1)|r|^{2}}}\bigg{(}\frac{\sqrt{1-|r|^{2}}}{1+\varepsilon},r\bigg{)}$
$\displaystyle=\big{(}a_{\varepsilon}(|r|)\sqrt{1-|r|^{2}},b_{\varepsilon}(|r|)r\big{)}$

in the notation of M2fW7tQX.

Hence,

$\displaystyle[u_{\varepsilon}]_{\partial G_{\varepsilon}}$
$\displaystyle=\sup_{\begin{subarray}{c}r,\tilde{r}\in B_{1}^{n-1}\\\
r\neq\tilde{r}\end{subarray}}\left\\{\frac{|(u_{\varepsilon}\circ\phi_{\varepsilon})(r)-(u_{\varepsilon}\circ\phi_{\varepsilon})(\tilde{r})|}{|\phi_{\varepsilon}(r)-\phi_{\varepsilon}(\tilde{r})|}\right\\}.$

Next, using M2fW7tQX and the second triangle inequality in the $\sup$ norm
(over $r,\tilde{r}\in B_{1}^{n-1}$), we obtain that

$\displaystyle\bigg{|}\frac{[u_{\varepsilon}]_{\partial
G_{\varepsilon}}}{\varepsilon}-$
$\displaystyle\frac{1}{2}s\gamma_{n,s}\left(\frac{3}{4}\right)^{s-1}\sup_{\begin{subarray}{c}r,\tilde{r}\in
B_{1}^{n-1}\\\
r\neq\tilde{r}\end{subarray}}\bigg{\\{}\frac{\big{|}|r|^{2}-|\tilde{r}|^{2}\big{|}}{|\phi_{0}(r)-\phi_{0}(\tilde{r})|}\bigg{\\}}\bigg{|}$
$\displaystyle\leqslant\sup_{\begin{subarray}{c}r,\tilde{r}\in B_{1}^{n-1}\\\
r\neq\tilde{r}\end{subarray}}\bigg{|}\frac{|(u_{\varepsilon}\circ\phi_{\varepsilon})(r)-(u_{\varepsilon}\circ\phi_{\varepsilon})(\tilde{r})|}{\varepsilon|\phi_{\varepsilon}(r)-\phi_{\varepsilon}(\tilde{r})|}-\frac{1}{2}s\gamma_{n,s}\left(\frac{3}{4}\right)^{s-1}\frac{\big{|}|r|^{2}-|\tilde{r}|^{2}\big{|}}{|\phi_{0}(r)-\phi_{0}(\tilde{r})|}\bigg{|}$
$\displaystyle\leqslant C\varepsilon,$

so, by sending $\varepsilon\to 0^{+}$, we find that

(7.8) $\displaystyle\lim_{\varepsilon\to
0^{+}}\frac{[u_{\varepsilon}]_{\partial G_{\varepsilon}}}{\varepsilon}$
$\displaystyle=\frac{1}{2}s\gamma_{n,s}\left(\frac{3}{4}\right)^{s-1}\sup_{\begin{subarray}{c}r,\tilde{r}\in
B_{1}^{n-1}\\\
r\neq\tilde{r}\end{subarray}}\bigg{\\{}\frac{\big{|}|r|^{2}-|\tilde{r}|^{2}\big{|}}{|\phi_{0}(r)-\phi_{0}(\tilde{r})|}\bigg{\\}}.$

Finally, we claim that

(7.9) $\displaystyle\sup_{\begin{subarray}{c}r,\tilde{r}\in B_{1}^{n-1}\\\
r\neq\tilde{r}\end{subarray}}\bigg{\\{}\frac{\big{|}|r|^{2}-|\tilde{r}|^{2}\big{|}}{|\phi_{0}(r)-\phi_{0}(\tilde{r})|}\bigg{\\}}=2.$

Indeed, if

$\displaystyle
Q:=\frac{\big{(}|r|^{2}-|\tilde{r}|^{2}\big{)}^{2}}{|\phi_{0}(r)-\phi_{0}(\tilde{r})|^{2}}$

then

$\displaystyle
Q=\frac{4\big{(}|r|^{2}-|\tilde{r}|^{2}\big{)}^{2}}{\big{(}\sqrt{1-|r|^{2}}-\sqrt{1-|\tilde{r}|^{2}}\big{)}^{2}+|r-\tilde{r}|^{2}}\leqslant\frac{4\big{(}|r|^{2}-|\tilde{r}|^{2}\big{)}^{2}}{(\sqrt{1-|r|^{2}}-\sqrt{1-|\tilde{r}|^{2}})^{2}+\big{(}|r|-|\tilde{r}|\big{)}^{2}}.$

Next, let $t,\tilde{t}\in(0,\pi/2)$ be such that $|r|=\cos t$ and
$|\tilde{r}|=\cos\tilde{t}$. Then

$\displaystyle Q$
$\displaystyle\leqslant\frac{4(\cos^{2}t-\cos^{2}\tilde{t})^{2}}{(\sin
t-\sin\tilde{t})^{2}+(\cos t-\cos\tilde{t})^{2}}.$

Then, via elementary trigonometric identities,

$\displaystyle(\cos^{2}t-\cos^{2}\tilde{t})^{2}=4\sin^{2}(t+\tilde{t})\sin^{2}\bigg{(}\frac{t-\tilde{t}}{2}\bigg{)}\cos^{2}\bigg{(}\frac{t-\tilde{t}}{2}\bigg{)}$
and $\displaystyle(\sin t-\sin\tilde{t})^{2}+(\cos
t-\cos\tilde{t})^{2}=4\sin^{2}\bigg{(}\frac{t-\tilde{t}}{2}\bigg{)},$

so

$\displaystyle Q\leqslant
4\sin^{2}(t+\tilde{t})\cos^{2}\bigg{(}\frac{t-\tilde{t}}{2}\bigg{)}\leqslant
4,$

which shows that the left-hand side of (7.9) is less than or equal to the
right-hand side of (7.9). To show that the reverse inequality is also true,
consider $r:=e_{1}/\sqrt{2}$ and $\tilde{r}:=(1/\sqrt{2}+h)e_{1}$. Then

$\displaystyle\frac{\big{|}|r|^{2}-|\tilde{r}|^{2}\big{|}}{|\phi_{0}(r)-\phi_{0}(\tilde{r})|}$
$\displaystyle=\frac{2|h|\left(h+\sqrt{2}\right)}{\sqrt{1-h\sqrt{2}-\sqrt{1-2\sqrt{2}h-2h^{2}}}}\to
2$

as $h\to 0$ which completes the proof of (7.9). Thus, recalling that
$\rho(\Omega_{\varepsilon})=\varepsilon$, (7.5) follows from (7.8) and (7.9).
∎

We will now give the proof of M2fW7tQX. We will do this over two lemmata.

######  Lemma 7.5.

c3Ge7XSV Let $\varepsilon\in[0,1)$ and $a_{\varepsilon}$, $b_{\varepsilon}$,
and $\phi_{\varepsilon}$ be as in M2fW7tQX.

Then, for all $r,\tilde{r}\in B_{1}^{n-1}$ with $r\neq\tilde{r}$,

(7.10)
$\displaystyle\bigg{|}\frac{|\phi_{0}(r)-\phi_{0}(\tilde{r})|}{|\phi_{\varepsilon}(r)-\phi_{\varepsilon}(\tilde{r})|}-1\bigg{|}\leqslant
C\varepsilon.$

The constant $C>0$ depends only on $n$.

###### Proof.

Observe that

(7.11)
$\begin{split}\bigg{|}\frac{|\phi_{0}(r)-\phi_{0}(\tilde{r})|}{|\phi_{\varepsilon}(r)-\phi_{\varepsilon}(\tilde{r})|}-1\bigg{|}&=\bigg{|}\frac{|\phi_{0}(r)-\phi_{0}(\tilde{r})|-|\phi_{\varepsilon}(r)-\phi_{\varepsilon}(\tilde{r})|}{|\phi_{\varepsilon}(r)-\phi_{\varepsilon}(\tilde{r})|}\bigg{|}\\\
&\leqslant\frac{|(\phi_{0}-\phi_{\varepsilon})(r)-(\phi_{0}-\phi_{\varepsilon})(\tilde{r})|}{|\phi_{\varepsilon}(r)-\phi_{\varepsilon}(\tilde{r})|}.\end{split}$

Next, observe that

$\displaystyle\phi_{0}(r)$
$\displaystyle=A_{\varepsilon}(|r|)\phi_{\varepsilon}(r)$

where, for each $\tau\in\mathbb{R}$, $A_{\varepsilon}(\tau)$ is the $n\times
n$ matrix given by

$\displaystyle A_{\varepsilon}(\tau)$
$\displaystyle:=\begin{pmatrix}\frac{1}{2a_{\varepsilon}(\tau)}&0\\\
0&\frac{1}{2b_{\varepsilon}(\tau)}I_{n-1}\end{pmatrix}.$

Here $I_{k}$ denotes the $k\times k$ identity matrix and we are thinking of
$\phi_{\varepsilon}$ and $\phi_{0}$ as column vectors.

It will be useful in the future to note that $a_{\varepsilon}$ and
$b_{\varepsilon}$ are strictly increasing for $\tau\geqslant 0$ (see (7.17)
below for $a_{\varepsilon}$, the computation for $b_{\varepsilon}$ is
analogous), so, for all $\tau\in[0,1)$,

(7.12) $\begin{split}&\frac{1}{2}+\varepsilon\leqslant
a_{\varepsilon}(\tau)\leqslant
1+\varepsilon-\frac{1}{2(1+\varepsilon)}\leqslant\frac{1}{2}+C\varepsilon\\\
{\mbox{and }}\quad&\frac{1}{2}(1-\varepsilon)\leqslant
b_{\varepsilon}(\tau)\leqslant\frac{1}{2}.\end{split}$

In particular, when $\varepsilon$ is small, $A_{\varepsilon}(\tau)$ is well-
defined since $a_{\varepsilon},b_{\varepsilon}>0$. Hence, it follow from
(7.11) that

(7.13)
$\begin{split}&\bigg{|}\frac{|\phi_{0}(r)-\phi_{0}(\tilde{r})|}{|\phi_{\varepsilon}(r)-\phi_{\varepsilon}(\tilde{r})|}-1\bigg{|}\\\
\leqslant\;&\frac{|(I_{n}-A_{\varepsilon}(|r|))\phi_{\varepsilon}(r)-(I_{n}-A_{\varepsilon}(|\tilde{r}|))\phi_{\varepsilon}(\tilde{r})|}{|\phi_{\varepsilon}(r)-\phi_{\varepsilon}(\tilde{r})|}\\\
\leqslant\;&\frac{|(I_{n}-A_{\varepsilon}(|r|))\phi_{\varepsilon}(r)-(I_{n}-A_{\varepsilon}(|r|))\phi_{\varepsilon}(\tilde{r})|}{|\phi_{\varepsilon}(r)-\phi_{\varepsilon}(\tilde{r})|}+\frac{|(A_{\varepsilon}(|\tilde{r}|)-A_{\varepsilon}(|r|))\phi_{\varepsilon}(\tilde{r})|}{|\phi_{\varepsilon}(r)-\phi_{\varepsilon}(\tilde{r})|}\\\
\leqslant\;&\|I_{n}-A_{\varepsilon}(|r|)\|_{\textrm{OP}}+\frac{\|A_{\varepsilon}(|\tilde{r}|)-A_{\varepsilon}(|r|)\|_{\mathrm{OP}}}{|\phi_{\varepsilon}(r)-\phi_{\varepsilon}(\tilde{r})|}|\phi_{\varepsilon}(\tilde{r})|,\end{split}$

where $\|\cdot\|_{\mathrm{OP}}$ denotes the matrix operator norm.

It follows from (7.12) that

(7.14) $\displaystyle\|I_{n}-A_{\varepsilon}(\tau)\|_{\textrm{OP}}$
$\displaystyle=\max\bigg{\\{}\bigg{|}1-\frac{1}{2a_{\varepsilon}(\tau)}\bigg{|},\bigg{|}1-\frac{1}{2b_{\varepsilon}(\tau)}\bigg{|}\bigg{\\}}\leqslant
C\varepsilon.$

Moreover, if $\|\cdot\|_{\mathrm{F}}$ denotes the Frobenius norm, then

$\displaystyle\|A_{\varepsilon}(|\tilde{r}|)-A_{\varepsilon}(|r|)\|_{\mathrm{OP}}^{2}$
$\displaystyle\leqslant
C\|A_{\varepsilon}(|\tilde{r}|)-A_{\varepsilon}(|r|)\|_{\mathrm{F}}^{2}$
$\displaystyle=C\left[\left(\frac{1}{a_{\varepsilon}(|r|)}-\frac{1}{a_{\varepsilon}(|\tilde{r}|)}\right)^{2}+\left(\frac{1}{b_{\varepsilon}(|r|)}-\frac{1}{b_{\varepsilon}(|\tilde{r}|)}\right)^{2}\right]$
$\displaystyle=C\left(\frac{\big{(}a_{\varepsilon}(|r|)-a_{\varepsilon}(|\tilde{r}|)\big{)}^{2}}{a_{\varepsilon}(|r|)^{2}a_{\varepsilon}(|\tilde{r}|)^{2}}+\frac{\big{(}b_{\varepsilon}(|r|)-b_{\varepsilon}(|\tilde{r}|)\big{)}^{2}}{b_{\varepsilon}(|r|)^{2}b_{\varepsilon}(|\tilde{r}|)^{2}}\right)$
$\displaystyle\leqslant
C\Big{(}\big{(}a_{\varepsilon}(|r|)-a_{\varepsilon}(|\tilde{r}|)\big{)}^{2}+\big{(}b_{\varepsilon}(|r|)-b_{\varepsilon}(|\tilde{r}|)\big{)}^{2}\Big{)},$

where we used (7.12) to obtain the final inequality.

We also have that

$\displaystyle|\phi_{\varepsilon}(r)-\phi_{\varepsilon}(\tilde{r})|^{2}$
$\displaystyle=\Big{(}a_{\varepsilon}(|r|)\sqrt{1-|r|^{2}}-a_{\varepsilon}(|\tilde{r}|)\sqrt{1-|\tilde{r}|^{2}}\Big{)}^{2}+|b_{\varepsilon}(|r|)r-b_{\varepsilon}(|\tilde{r}|)\tilde{r}|^{2}$
$\displaystyle\geqslant\Big{(}a_{\varepsilon}(|r|)\sqrt{1-|r|^{2}}-a_{\varepsilon}(|\tilde{r}|)\sqrt{1-|\tilde{r}|^{2}}\Big{)}^{2}+\big{(}b_{\varepsilon}(|r|)|r|-b_{\varepsilon}(|\tilde{r}|)|\tilde{r}|\big{)}^{2}.$

We claim that

(7.15)
$\begin{split}&\big{|}a_{\varepsilon}(|r|)-a_{\varepsilon}(|\tilde{r}|)\big{|}\leqslant
C\varepsilon\big{|}a_{\varepsilon}(|r|)\sqrt{1-|r|^{2}}-a_{\varepsilon}(|\tilde{r}|)\sqrt{1-|\tilde{r}|^{2}}\big{|}\\\
\text{and}\qquad&\big{|}b_{\varepsilon}(|r|)-b_{\varepsilon}(|\tilde{r}|)\big{|}\leqslant
C\varepsilon\big{|}b_{\varepsilon}(|r|)|r|-b_{\varepsilon}(|\tilde{r}|)|\tilde{r}|\big{|}.\end{split}$

We will prove the inequality in (7.15) involving $a_{\varepsilon}$, the proof
of the inequality involving $b_{\varepsilon}$ is entirely analogous. Let
$\bar{a}_{\varepsilon}(\tau):=a_{\varepsilon}(\tau)\sqrt{1-\tau^{2}}$. Then,
by Cauchy’s Mean Value Theorem,

(7.16)
$\displaystyle\bigg{|}\frac{a_{\varepsilon}(|r|)-a_{\varepsilon}(|\tilde{r}|)}{\bar{a}_{\varepsilon}(|r|)-\bar{a}_{\varepsilon}(|\tilde{r}|)}\bigg{|}$
$\displaystyle\leqslant\bigg{\|}\frac{a_{\varepsilon}^{\prime}}{\bar{a}_{\varepsilon}^{\prime}}\bigg{\|}_{L^{\infty}((0,1))}.$

On one hand,

$\displaystyle a^{\prime}_{\varepsilon}(\tau)$
$\displaystyle=\frac{1}{2}\varepsilon(\varepsilon+2)\tau\big{(}1+\big{(}(1+\varepsilon)^{2}-1\big{)}\tau^{2}\big{)}^{-3/2},$

so, for all $\tau\in[0,1)$,

(7.17) $\displaystyle C^{-1}\varepsilon\tau\leqslant
a^{\prime}_{\varepsilon}(\tau)\leqslant C\varepsilon\tau.$

On the other hand, using (7.12) and (7.17), we have that

$\displaystyle\bar{a}_{\varepsilon}^{\prime}(\tau)$
$\displaystyle=a_{\varepsilon}^{\prime}(\tau)\sqrt{1-\tau^{2}}-\frac{\tau
a_{\varepsilon}(\tau)}{\sqrt{1-\tau^{2}}}\leqslant C(\varepsilon-1)\tau<0.$

Hence,