# Estimates for Robin $p$-Laplacian eigenvalues of convex sets with prescribed
perimeter

Vincenzo Amato, Andrea Gentile, Alba Lia Masiello

###### Abstract

In this paper, we prove an upper bound for the first Robin eigenvalue of the
$p$-Laplacian with a positive boundary parameter and a quantitative version of
the reverse Faber-Krahn type inequality for the first Robin eigenvalue of the
$p$-Laplacian with negative boundary parameter, among convex sets with
prescribed perimeter.

The proofs are based on a comparison argument obtained by means of inner sets,
introduced by Payne, Weimberger [PW61] and Polya [P6́0].

MSC 2020: 46E30, 35A23, 35J92.
Keywords: Rearrangements, $p$-Laplacian, Robin boundary conditions.

Dipartimento di Matematica e Applicazioni “R. Caccioppoli”, Università degli
Studi di Napoli “Federico II”, Complesso Universitario Monte S. Angelo, via
Cintia - 80126 Napoli, Italy.

e-mail<EMAIL_ADDRESS>
e-mail<EMAIL_ADDRESS>
Mathematical and Physical Sciences for Advanced Materials and Technologies,
Scuola Superiore Meridionale, Largo San Marcellino 10, 80126 Napoli, Italy.

e-mail<EMAIL_ADDRESS>
##  1 Introduction

Let $\Omega$ be a bounded, open and convex set in
$\operatorname{\mathbb{R}}^{n}$. We consider the following problem

$\begin{cases}-\Delta_{p}u=-\text{div}({\left|\nabla u\right|}^{p-2}\nabla
u)=\lambda_{p,\beta}(\Omega){\left|u\right|}^{p-2}u\,&\text{in}\,\,\Omega\\\
\displaystyle{{\left|{\nabla u}\right|}^{p-2}\frac{\partial
u}{\partial\nu}+\beta{\left|u\right|}^{p-2}u=0}\,&\text{on}\,\,\partial\Omega,\end{cases}$
(1.1)

where $\beta\in\operatorname{\mathbb{R}}$.

The fundamental eigenvalue of the Robin Laplacian on $\Omega$ is defined by

$\lambda_{p,\beta}(\Omega)=\min_{\begin{subarray}{c}v\in W^{1,p}(\Omega)\\\
v\neq 0\end{subarray}}\frac{\displaystyle{\int_{\Omega}{\left|\nabla
v\right|}^{p}\,dx+\beta\int_{\partial\Omega}{\left|v\right|}^{p}\,d\mathcal{H}^{n-1}}}{\displaystyle{\int_{\Omega}{\left|v\right|}^{p}\,dx}}$
(1.2)

and a minimizer $u$ in (1.2) satisfies the equation (1.1) in the weak form if

$\int_{\Omega}{\left|\nabla u\right|}^{p-2}\nabla
u\nabla\varphi\,dx+\beta\int_{\partial\Omega}{\left|u\right|}^{p-2}u\varphi\,d\mathcal{H}^{n-1}=\lambda_{p,\beta}(\Omega)\int_{\Omega}{\left|u\right|}^{p-2}u\varphi\,dx,\quad\forall\varphi\in
W^{1,p}(\Omega).$

It is well-known (see for instance [Lin92] for the Dirichlet case) that the
first eigenvalue (1.2) is simple and that in the case of a ball, the
corresponding eigenfunction is radially symmetric ([Bha88]).

In this paper we want to study the different behaviour of eigenvalues in the
case $\beta>0$ and $\beta<0$. For sake of completeness, we recall that, in the
case of $\beta=0$, we recover Neumann boundary condition, for which the first
eigenvalue is zero and the associated eigenfunctions are constants.

It is well known that in the case $\beta>0$, Bossel [Bos86] and Daners [Dan06]
proved a Faber-Krahn inequality for the first eigenvalue of the Robin-
Laplacian in two and higher dimensional case, respectively. In particular,
they proved that among sets of given volume, the one which minimizes the first
Robin-eigenvalue is the ball, i.e.

$\lambda_{2,\beta}(\Omega^{\sharp})\leq\lambda_{2,\beta}(\Omega),$ (1.3)

where $\Omega^{\sharp}$ is the ball, centered at the origin, having the same
volume as $\Omega$. This result was generalized by Bucur, Daners and Giacomini
in [BD10, BG10, BG15] to the eigenvalues of the $p$-Laplacian with Robin
boundary conditions.

This Faber-Krahn inequality for fixed volume and the following rescaling
property [BFK17]

$\lambda_{p,\beta}(t\Omega)\leq\frac{1}{t}\lambda_{p,\beta}(\Omega)\leq\lambda_{p,\beta}(\Omega),\quad\forall
t>1,$ (1.4)

give a Faber-Krahn inequality for fixed perimeters, i.e.

$\lambda_{p,\beta}(\Omega^{\star})\leq\lambda_{p,\beta}(\Omega),$

where $\Omega^{\star}$ is the ball having the same perimeter as $\Omega$.

Our aim is to give a continuity bound to the ratio

$\frac{\lambda_{p,\beta}(\Omega)-\lambda_{p,\beta}(\Omega^{\star})}{\lambda_{p,\beta}(\Omega)},$

indeed we prove

######  Theorem 1.1.

Let $\beta$ be a positive parameter. Let $\Omega$ be a bounded, open and
convex set in $\operatorname{\mathbb{R}}^{n}$ and let $\Omega^{\star}$ be the
ball, centered at the origin, such that $P(\Omega)=P(\Omega^{\star})=\rho$.
Let $\lambda_{p,\beta}(\Omega)$ and $\lambda_{p,\beta}(\Omega^{\star})$ be the
first eigenvalues of the $p$-Laplacian operator with Robin boundary conditions
on $\Omega$ and $\Omega^{\star}$, let $v$ be a positive eigenfunction
associated to $\lambda_{p,\beta}(\Omega^{\star})$. Then

$\frac{\lambda_{p,\beta}(\Omega)-\lambda_{p,\beta}(\Omega^{\star})}{\lambda_{p,\beta}(\Omega)}\leq
C(n,p,\beta,\rho)\left(1-\frac{n^{\frac{n}{n-1}}\omega_{n}^{\frac{1}{n-1}}{\left|\Omega\right|}}{P(\Omega)^{\frac{n}{n-1}}}\right),$
(1.5)

where $\omega_{n}$ is the measure of the unitary ball in
$\operatorname{\mathbb{R}}^{n}$, and
$\displaystyle{C(n,p,\beta,\rho)=\frac{{\left\|v\right\|}_{\infty}^{p}{\left|\Omega^{\star}\right|}}{{\left\|v\right\|}^{p}_{p}}}$.

It is possible to give a uniform bound to the constant in (1.5) from above
with a constant independent of the parameter $\beta$ and the perimeter, indeed
it holds

$C(n,p,\beta,\rho)\leq
C(n,p):=\frac{{\left\|v_{\infty}\right\|}_{\infty}^{p}{\left|\Omega^{\star}\right|}}{{\left\|v_{\infty}\right\|}^{p}_{p}},$
(1.6)

where $v_{\infty}$ is the first Dirichlet eigenfunction. Thanks to the
rescaling property of Dirichlet eigenvalues and eigenfunctions, $C(n,p)$ does
not depends on the perimeter. For more details see Remark 3.1.

We observe that this result can be seen as a generalization to the Robin case
of the result in [BNT10], which holds true in the case of Dirichlet
eigenvalues of the $p$-Laplacian.

When $\beta$ is a negative parameter, the authors in [BFNT19] proved a reverse
Faber-Krahn inequality for the first eigenvalue of the Dirichlet-Laplacian
among convex sets of given perimeter. In particular, they proved that among
convex sets of given perimeter the ball $\Omega^{\star}$ maximizes the first
Robin eigenvalue of the $p$-Laplacian, i.e.

$\lambda_{p,\beta}(\Omega)\leq\lambda_{p,\beta}(\Omega^{\star}).$ (1.7)

For completeness’ sake, we quote that in [AFK17], the authors already proved
that the disc maximizes the first eigenvalue under a perimeter constraint,
among $C^{2}$ domains in $\operatorname{\mathbb{R}}^{2}$, while the question
remained open in arbitrary dimension.

This question is related to the conjecture of Bareket (see [Bar77]) claiming
that the ball maximizes $\lambda_{p,\beta}(\Omega)$ among all Lipschitz sets
with given volume. Freitas and Krejčiřík in [FK15] proved that the conjecture
is false, giving a counter-example based on the asymptotic behaviour of the
eigenvalues on a disc and an annulus of the same area when $\beta\to-\infty$.
They also proved that among sets of area equal to $1$, the conjecture is true,
provided $\beta$ is close to $0$.

In [CL21] the authors proved a quantitative version of the reverse Faber-Krahn
(1.7) in the case $p=2$ among convex sets of fixed perimeter following the
Fuglede’s approach introduced in [Fug89].

In this paper, we recover the result in [CL21] obtaining a quantitative
version of (1.7) for all $p$, but using a different approach. Indeed,
following the method introduced by Payne and Weinberger in [PW61], we
establish a comparison using the so-called parallel coordinates method. In
particular, we firstly prove a lower bound in terms of perimeter and measure
of $\Omega$, that is

######  Theorem 1.2.

Let $\beta$ be a negative parameter. Let $\Omega$ be a bounded, open and
convex set in $\operatorname{\mathbb{R}}^{n}$ and let $\Omega^{\star}$ be the
ball, centered at the origin, such that $P(\Omega)=P(\Omega^{\star})=\rho$.
Let $\lambda_{p,\beta}(\Omega)$ and $\lambda_{p,\beta}(\Omega^{\star})$ be the
first eigenvalues of the $p$-Laplacian operator with Robin boundary conditions
on $\Omega$ and $\Omega^{\star}$, let $v$ be a positive eigenfunction
associated to $\lambda_{p,\beta}(\Omega^{\star})$. Then

$\frac{\lambda_{p,\beta}(\Omega^{\star})-\lambda_{p,\beta}(\Omega)}{{\left|\lambda_{p,\beta}(\Omega)\right|}}\geq
C(n,p,\beta,\rho)\left(1-\frac{n^{\frac{n}{n-1}}\omega_{n}^{\frac{1}{n-1}}{\left|\Omega\right|}}{P(\Omega)^{\frac{n}{n-1}}}\right),$
(1.8)

where $\omega_{n}$ is the measure of the unitary ball in
$\operatorname{\mathbb{R}}^{n}$, and
$\displaystyle{C(n,p,\beta,\rho)=\frac{v_{m}^{p}{\left|\Omega^{\star}\right|}}{{\left\|v\right\|}^{p}_{p}}}$
with $\displaystyle{v_{m}=\min_{\Omega^{\star}}v}$.

In this case, the constant $C(n,p,\beta,\rho)$ cannot be replaced by a
constant independent of $\beta$ and of the perimeter, as it is shown in Remark
3.2.

Then we prove the quantitative result as in [CL21].

######  Theorem 1.3.

Let $n\geq 2$, $\rho>0$ and $\beta<0$. Then there exists two positive
constants $C(n,p,\beta,\rho)>0$ and $\delta_{0}(n,p,\beta,\rho)>0$, such that,
for all $\Omega\subset\operatorname{\mathbb{R}}^{n}$ bounded and convex with
$P(\Omega)=\rho$ and
$\lambda_{p,\beta}(\Omega^{\star})-\lambda_{p,\beta}(\Omega)\leq\delta_{0}$,
it holds

$\lambda_{p,\beta}(\Omega^{\star})-\lambda_{p,\beta}(\Omega)\geq
C(n,p,\beta,\rho)g(\mathcal{A}_{\mathcal{H}}^{\star}(\Omega))$ (1.9)

where $\Omega^{\star}$ is a ball with the same perimeter of $\Omega$,
$\mathcal{A}_{\mathcal{H}}^{\star}$ is the Hausdorff asymmetry defined in
(2.3) and $g$ is defined in (2.7).

The paper is organized as follows: in Section 2 we recall some preliminary
results and useful tools for our aim; in Section 3 we provide the proof of our
main results.

##  2 Notations and Preliminaries

Throughout this article, $|\cdot|$ will denote the Euclidean norm in
$\mathbb{R}^{n}$, while $\cdot$ is the standard Euclidean scalar product for
$n\geq 2$. By $\mathcal{H}^{k}(\cdot)$, for $k\in[0,n)$, we denote the
$k$-dimensional Hausdorff measure in $\mathbb{R}^{n}$.

The measure and the perimeter of $\Omega$ in $\mathbb{R}^{n}$ will be denoted
by ${\left|\Omega\right|}$ and $P(\Omega)$, respectively, and, if
$P(\Omega)<\infty$, we say that $\Omega$ is a set of finite perimeter. In our
case, $\Omega$ is a bounded, open and convex set; this ensures us that
$\Omega$ is a set of finite perimeter and that
$P(\Omega)=\mathcal{H}^{n-1}(\partial\Omega)$. Moreover, if $\Omega$ is an
open set with Lipschitz boundary, it holds

######  Theorem 2.1 (Coarea formula).

Let $f:\Omega\to\operatorname{\mathbb{R}}$ be a Lipschitz function and let
$u:\Omega\to\operatorname{\mathbb{R}}$ be a measurable function. Then,

$\int_{\Omega}u(x)\lvert\nabla
f(x)\rvert\,dx=\int_{\mathbb{R}}dt\int_{(\Omega\cap
f^{-1}(t))}u(y)\,d\mathcal{H}^{n-1}(y).$ (2.1)

Some references for results relative to the sets of finite perimeter and for
the coarea formula are, for instance [AFP00, Mag12].

###  2.1 Asymmetry index $\mathcal{A}^{\star}(E)$

In the case $\beta<0$, we want to prove a quantitative result so we need an
index that gives us information about the shape of $\Omega$. We will consider
the Hausdorff asymmetry index, as already done in [CL21], so we recall some
basic notions about the Hausdorff distance.

We recall that if $E,F$ are any two convex sets in $\mathbb{R}^{n}$ the
Hausdorff distance between $E,F$ is defined as

$d_{\mathcal{H}}(E,F):=\inf\left\\{\varepsilon>0\,:\,E\subset F+\varepsilon
B,\,F\subset E+\varepsilon B\right\\},$

where $B$ is the unitary ball centered at the origin and $F+\varepsilon B$ is
the well-known Minkowski sum. For such set we define two isoperimetric deficit

$\mathcal{D}(E):=P(E)-P(E^{\sharp}),\quad\mathcal{M}(E):={\left|E^{\star}\right|}-{\left|E\right|}$
(2.2)

where $E^{\sharp}$ and $E^{\star}$ are the ball with the same measure and the
same perimeter of $E$, respectively.

Moreover, we will consider the following Hausdorff asymmetry indices

$\mathcal{A}_{\mathcal{H}}^{\star}(E)=\min_{x\in\operatorname{\mathbb{R}}^{n}}\left\\{d_{\mathcal{H}}(E,B_{r}(x)),\,P(\Omega)=P(B_{r}(x))\right\\}$
(2.3)

and

$\mathcal{A}_{\mathcal{H}}^{\sharp}(E)=\min_{x\in\operatorname{\mathbb{R}}^{n}}\left\\{d_{\mathcal{H}}(E,B_{r}(x)),\,{\left|\Omega\right|}={\left|B_{r}(x)\right|}\right\\}.$
(2.4)

Lemma 2.9 in [GLPT20] tells us how these two indices are related one to the
other

######  Lemma 2.2.

Let $n\geq 2$ and let $E\subset\operatorname{\mathbb{R}}^{n}$ be a bounded,
convex, with $\mathcal{D}(E)\leq\delta$ then

$\mathcal{A}_{H}^{\star}(E)\leq C(n)\mathcal{A}_{H}^{\sharp}(E).$ (2.5)

With these definitions, we can recall the quantitative isoperimetric
inequality proved in [Fug89, Fus15].

######  Theorem 2.3 (Fuglede).

Let $n\geq 2$, and let $E$ be a bounded open and convex set with
${\left|E\right|}=\omega_{n}$. There exists $\delta,C$, depending only on $n$,
such that if $\mathcal{D}(E)\leq\delta$ then then

$\mathcal{D}(E)\geq Cg\bigl{(}\mathcal{A}_{\mathcal{H}}^{\sharp}(E)\bigr{)},$
(2.6)

where $g$ is defined by

$g(s)=\begin{cases}s^{2}&\text{ if }n=2\\\ f^{-1}(s^{2})&\text{ if }n=3\\\
s^{\frac{n+1}{2}}&\text{ if }n\geq 4\end{cases}$ (2.7)

and $f(t)=\sqrt{t\log(\frac{1}{t})}$ for $0<t<e^{-1}$.

We are interested in a modified version of this theorem, in terms of
$\mathcal{M}(\Omega)$, so we have

######  Lemma 2.4.

Let $\Omega\subset\operatorname{\mathbb{R}}^{n}$ be a bounded, open and convex
set and let $\Omega^{\star}$ be the ball satisfying
$P(\Omega)=P(\Omega^{\star})=\rho$. Then, there exist $\delta,C$, depending
only on $n$ and $\rho$, such that, if

$\mathcal{M}(\Omega)={\left|\Omega^{\star}\right|}-{\left|\Omega\right|}\leq\delta$
(2.8)

then

$\mathcal{M}(\Omega)\geq
Cg\bigl{(}\mathcal{A}_{\mathcal{H}}^{\star}(\Omega)\bigr{)}$

where $g$ is the function defined in (2.7).

###### Proof.

Let us start by setting $\delta<\frac{{\left|\Omega^{\star}\right|}}{2}$, so

${\left|\Omega\right|}>\frac{{\left|\Omega^{\star}\right|}}{2}$ (2.9)

and let us assume ${\left|\Omega\right|}=\omega_{n}$.

By (2.9) and the differentiability of the function $h(t)=t^{\frac{n-1}{n}}$,
we have

${\left|\Omega^{\star}\right|}^{\frac{n-1}{n}}-{\left|\Omega\right|}^{\frac{n-1}{n}}=h^{\prime}(\xi)\left({\left|\Omega^{\star}\right|}-{\left|\Omega\right|}\right)\leq\frac{n-1}{\omega_{n}^{\frac{1}{n}}n}\left({\left|\Omega^{\star}\right|}-{\left|\Omega\right|}\right).$

Hence, if $\mathcal{M}(\Omega)\leq\delta$, then

${\left|\Omega^{\star}\right|}^{\frac{n-1}{n}}-{\left|\Omega\right|}^{\frac{n-1}{n}}=\frac{P(\Omega)}{n\omega^{\frac{1}{n}}}-{\left|\Omega\right|}^{\frac{n-1}{n}}=\frac{\mathcal{D}(\Omega)}{n\omega^{\frac{1}{n}}}\leq\tilde{\delta}(n,\rho).$

So we can apply (2.6)

$P(\Omega)\geq
n\omega_{n}\Bigl{(}1+\gamma(n)g\bigl{(}\mathcal{A}_{\mathcal{H}}^{\sharp}(\Omega)\bigr{)}\Bigr{)}$

to obtain

$\displaystyle{\left|\Omega^{\star}\right|}-{\left|\Omega\right|}$
$\displaystyle=\frac{P(\Omega)^{\frac{n}{n-1}}}{n^{\frac{n}{n-1}}\omega_{n}^{\frac{1}{n-1}}}-{\left|\Omega\right|}$
$\displaystyle\geq\omega_{n}\left(1+\gamma(n)g\bigl{(}\mathcal{A}_{\mathcal{H}}^{\sharp}(\Omega)\bigr{)}\right)^{\frac{n}{n-1}}-\omega_{n}$
$\displaystyle\geq\frac{n\omega_{n}\gamma(n)}{n-1}g\bigl{(}\mathcal{A}_{\mathcal{H}}^{\sharp}(\Omega)\bigr{)},$

where in the last step we used Bernoulli’s inequality

$(1+x)^{r}\geq 1+rx\qquad x\geq-1,\,r\geq 1.$

Applying Lemma 2.2 we finally have

${\left|\Omega^{\star}\right|}-{\left|\Omega\right|}\geq
C(n)g\bigl{(}\mathcal{A}_{\mathcal{H}}^{\star}(\Omega)\bigr{)}.$

The general case will be recovered by rescaling and by the following
inequality for $Y=\omega_{n}{\left|\Omega\right|}^{-\frac{1}{n}}\Omega$

${\left|Y^{\star}\right|}-{\left|Y\right|}\leq
2\omega_{n}^{n}{\left|\Omega^{\star}\right|}^{-1}\left({\left|\Omega^{\star}\right|}-{\left|\Omega\right|}\right).\qed$
(2.10)

###  2.2 Quermassintegrals

Let us recall some basic facts about convex sets. Let
$K\subset\operatorname{\mathbb{R}}^{n}$ be a non-empty, bounded, convex set,
let $B$ be the unitary ball centered at the origin and $\rho>0$. We can write
the Steiner formula for the Minkowski sum $K+\rho B$ as

${\left|K+\rho B\right|}=\sum_{i=0}^{n}\binom{n}{i}W_{i}(K)\rho^{i}.$ (2.11)

The coefficients $W_{i}(K)$ are known in literature as quermassintegrals of
$K$. In particular, $W_{0}(K)={\left|K\right|},nW_{1}(K)=P(K)$ and
$W_{n}(K)=\omega_{n}$ where $\omega_{n}$ is the measure of $B$.

If $K$ has $C^{2}$ boundary, the quermassintegrals can be written in terms of
principal curvatures of $K$. More precisely, denoting with $H_{j}$ the $j$-th
normalized elementary symmetric function of the principal curvature
$\kappa_{1},\ldots,\kappa_{n-1}$ $\partial K$, i.e.

$H_{0}=1\qquad\qquad H_{j}=\binom{n-1}{j}^{-1}\sum_{1\leq
i_{1}<\ldots<i_{j}\leq n-1}\kappa_{i_{1}}\ldots\kappa_{i_{j}}\qquad
j=1,\ldots,n-1,$

then the quermassintegrals can be written as

$W_{i}(K)=\frac{1}{n}\int_{\partial K}H_{i-1}\,d\mathcal{H}^{n-1}\qquad
i=1,\ldots,n.$ (2.12)

Moreover, the Steiner formula holds true also for quermassintegrals, that is

$W_{j}(K+\rho B)=\sum_{i=0}^{n-j}\binom{n-j}{i}W_{j+i}(K)\rho^{i}\qquad
j=0,\ldots,n-1.$ (2.13)

For $j=1$ we have

$\displaystyle P(K+\rho B_{1})$
$\displaystyle=n\sum_{i=0}^{n-1}\binom{n-1}{i}W_{i+1}(K)\rho^{i}$
$\displaystyle=P(K)+n(n-1)W_{2}(K)\rho+\ldots+nW_{n}(K)\rho^{n},$

from which follows

$\lim_{\rho\to 0}\frac{P(K+\rho B)-P(K)}{\rho}=n(n-1)W_{2}(K),$ (2.14)

and if $\partial K$ is of class $C^{2}$ formula (2.12) gives

$\lim_{\rho\to 0}\frac{P(K+\rho B)-P(K)}{\rho}=(n-1)\int_{\partial
K}H_{1}\,d\mathcal{H}^{n-1}.$

Furthermore, Aleksandrov-Fenchel inequalities hold true

$\biggl{(}\frac{W_{j}(K)}{\omega_{n}}\biggr{)}^{\frac{1}{n-j}}\geq\biggl{(}\frac{W_{i}(K)}{\omega_{n}}\biggr{)}^{\frac{1}{n-i}}\qquad
0\leq i<j\leq n-1,$ (2.15)

where equality hold if and only if $K$ is a ball. When $i=0$ and $j=1$,
formula (2.15) reduce to the classical isoperimetric inequality, i.e.

$P(K)\geq n\omega_{n}^{\frac{1}{n}}{\left|K\right|}^{\frac{n-1}{n}}.$

In the following we will use (2.15) for $i=1$ and $j=2$, that is

$W_{2}(K)\geq
n^{-\frac{n-2}{n-1}}\omega_{n}^{\frac{1}{n-1}}P(K)^{\frac{n-2}{n-1}}.$ (2.16)

###  2.3 Some useful lemmas

Let $\Omega\subset\operatorname{\mathbb{R}}^{n}$ be a convex set, for
$t\in[0,r_{\Omega}]$ we denote by

$\Omega_{t}=\Set{x\in\Omega:d(x)>t}$

where $d(x)$ is the distance of $x\in\Omega$ from the boundary of $\Omega$ and
$r_{\Omega}$ is the inradius of $\Omega$. By the Brunn-Minkowski Theorem
([Sch13, Theorem 7.4.5]) and the concavity of the distance function, the map

$t\mapsto P(\Omega_{t})^{\frac{1}{n-1}}$

is concave in $[0,r_{\Omega}]$, hence absolutely continuous in
$(0,r_{\Omega})$. Moreover, there exists its right derivative at $0$ and it is
negative, since $P(\Omega_{t})^{\frac{1}{n-1}}$ is strictly monotone
decreasing.

######  Lemma 2.5.

Let $\Omega$ be a bounded, convex, open set in
$\operatorname{\mathbb{R}}^{n}$. Then for almost every $t\in(0,r_{\Omega})$

$-\frac{d}{dt}P(\Omega_{t})\geq n(n-1)W_{2}(\Omega_{t})$ (2.17)

and the equality holds if $\Omega$ is a ball.

###### Proof.

For every $\rho\in(0,t)$ it holds

$\Omega_{t}+\rho B_{1}\subset\Omega_{t-\rho}$

and if $\Omega$ is a ball, the two sets coincide. The monotonicity of the
perimeter with respect to the inclusion of convex sets and formula (2.14)
give, for almost every $t\in(0,r_{\Omega})$,

$\displaystyle-\frac{d}{dt}P(\Omega_{t})$ $\displaystyle=\lim_{\rho\to
0^{+}}\frac{P(\Omega_{t-\rho})-P(\Omega_{t})}{\rho}$
$\displaystyle\geq\lim_{\rho\to 0^{+}}\frac{P(\Omega_{t}+\rho
B_{1})-P(\Omega_{t})}{\rho}=n(n-1)W_{2}(\Omega_{t})$

∎

Combining the chain rule, the previous lemma and the fact that ${\left|\nabla
d(x)\right|}=1$ almost everywhere, we obtain

######  Lemma 2.6.

Let $f\colon[0,+\infty)\to[0,+\infty)$ be a strictly increasing $C^{1}$
function with $f(0)=0$. Set $u(x)=f(d(x))$ and

$E_{t}=\Set{x\in\Omega\,:\,u(x)>t}=\Omega_{f^{-1}(t)}$

then

$-\frac{d}{dt}P(E_{t})\geq(n-1)\frac{W_{2}(E_{t})}{{\left|\nabla
u\right|}_{u=t}}$ (2.18)

###  2.4 Monotonicity of eigenfunctions

It is well-known (by definition) that for fixed $1<p<+\infty$ and
$\Omega\subset\operatorname{\mathbb{R}}^{n}$, the map

$\beta\in\operatorname{\mathbb{R}}\mapsto\lambda_{p,\beta}(\Omega)$

is increasing and it holds (see for instance [Lê06])

$\lambda_{p,N}<\lambda_{p,\beta}<\lambda_{p,D}\qquad\forall\beta\in(0,+\infty)$

and

$\lim_{\beta\to
0^{+}}\lambda_{p,\beta}=\lambda_{p,N}=0\qquad\lim_{\beta\to+\infty}\lambda_{p,\beta}=\lambda_{p,D}$

where $\lambda_{p,N}$ and $\lambda_{p,D}$ are the first eigenvalue of the
Neumann $p$-Laplacian and Dirichlet $p$-Laplacian respectively.

Now we want to observe that, if we consider a ball $B_{R}$ and we fix the
value of the eigenfunction in the origin, the map

$\beta\to v_{\beta},\quad v_{\beta}(0)=C$

is decreasing, in the sense that

$\text{if }\beta_{1}<\beta_{2}\Rightarrow v_{p,\beta_{1}}(x)\geq
v_{p,\beta_{2}}(x)\qquad\forall x\in B_{R}$

First of all, let us recall that the first eigenvalue of $p$-laplacian is
simple and that the corresponding eigenfunction is radially symmetric, i.e.
there exists $h\colon[0,R]\to\operatorname{\mathbb{R}}$ such that
$u(x)=h({\left|x\right|})$. In the following, we will denote with
$v_{p,\beta}$ both the eigenfunction and the function $h$.

######  Lemma 2.7.

Let $0<r<R$, $1<p<+\infty$, $\beta\in\operatorname{\mathbb{R}}$. Let us denote
by $\lambda_{p,\beta}$ the first eigenvalue of the $p$-Laplacian on the ball
$B_{R}$ defined in (1.2) with boundary parameter $\beta$ and let $v_{p,\beta}$
be the corresponding eigenfunction, then $v_{p,\beta}\rvert_{B_{r}}$ is the
first eigenfunction of the $p$-Laplacian on the ball $B_{r}$ with boundary
parameter

$\displaystyle{\gamma=-\frac{\lvert
v^{\prime}_{p,\beta}\rvert^{p-2}(r)v^{\prime}_{p,\beta}(r)}{v_{p,\beta}^{p-1}(r)}}.$

###### Proof.

Let us suppose $p=2$, the general case is analogous. For sake of simplicity,
we denote by $\lambda_{\beta}:=\lambda_{2,\beta}$ and
$v_{\beta}:=v_{2,\beta}$.

By the radiality of $v_{\beta}$, we can infer that it is a solution to

$\begin{cases}-\Delta v_{\beta}=\lambda_{\beta}v_{\beta}&\text{in }B_{r}\\\
\displaystyle{\frac{\partial v_{\beta}{}}{\partial\nu}+\gamma
v_{\beta}=0}&\text{on }\partial B_{r}.\end{cases}$

Let us suppose by contradiction that $\lambda_{\beta}$ is not the first
eigenvalue of the Robin Laplacian with boundary parameter $\gamma$. So we can
choose

$\lambda_{\gamma}<\lambda_{\beta}$ (2.19)

and $w_{\gamma}$ respectively first eigenvalue and the first eigenfunction of
the Robin Laplacian with boundary parameter $\gamma$, that is

$\lambda_{\gamma}=\frac{\displaystyle{\int_{B_{r}}{\left|\nabla
w_{\gamma}\right|}^{2}\,dx+\gamma\int_{\partial
B_{r}}w_{\gamma}^{2}\,d\mathcal{H}^{n-1}}}{\displaystyle{\int_{B_{r}}w_{\gamma}^{2}\,dx}}=\min_{\psi\in
W^{1,2}(B_{r})}\frac{\displaystyle{\int_{B_{r}}{\left|\nabla\psi\right|}^{2}\,dx+\gamma\int_{\partial
B_{r}}\psi^{2}\,d\mathcal{H}^{n-1}}}{\displaystyle{\int_{B_{r}}\psi^{2}\,dx}}$
(2.20)

We know that $w_{\gamma}$ is unique up to a multiplicative constant, so we can
choose the constant such that $v_{\beta}=w_{\gamma}$ on $\partial B_{r}$.

Let us consider the function

$f=\begin{cases}w_{\gamma}(x)&\text{if }x\in B_{r}\\\ v_{\beta}(x)&\text{if
}x\in B_{R}\setminus B_{r},\end{cases}$

we can use it as test function in the definition of $\lambda_{\beta}$. In
particular

$\displaystyle\lambda_{\beta}$
$\displaystyle\leq\frac{\displaystyle{\int_{B_{R}}{\left|\nabla
f\right|}^{2}+\beta\int_{\partial
B_{r}}f^{2}\,d\mathcal{H}^{n-1}}}{\displaystyle{\int_{B_{R}}f^{2}\,dx}}$
$\displaystyle=\frac{\displaystyle{\int_{B_{r}}{\left|\nabla
w_{\gamma}\right|}^{2}+\int_{B_{R}\setminus B_{r}}{\left|\nabla
v_{\beta}\right|}^{2}+\beta\int_{\partial
B_{R}}v_{\beta}^{2}\,d\mathcal{H}^{n-1}}}{\displaystyle{\int_{B_{r}}w_{\gamma}^{2}\,dx+\int_{B_{R}\setminus
B_{r}}v_{\beta}^{2}\,dx}}.$

If we add and subtract $\displaystyle{\gamma\int_{\partial
B_{r}}w_{\gamma}\,d\mathcal{H}^{n-1}}$ and
$\displaystyle{\int_{B_{r}}{\left|\nabla u\right|}^{2}}$, we recall that
$v_{\beta}$ and $w_{\gamma}$ are eigenfunction, so (1.2) hold, and we recall
that $v_{\beta}$ and $w_{\gamma}$ coincides on $\partial B_{r}$, we have

$\displaystyle\lambda_{\beta}$
$\displaystyle\leq\frac{\displaystyle{\lambda_{\gamma}\int_{B_{r}}w_{\gamma}^{2}\,dx+\lambda_{\beta}\int_{B_{R}}v_{\beta}^{2}\,dx-\lambda_{\beta}\int_{B_{r}}v_{\beta}^{2}\,dx}}{\displaystyle{\int_{B_{r}}w_{\gamma}^{2}\,dx+\int_{B_{R}\setminus
B_{r}}v_{\beta}^{2}\,dx}}$
$\displaystyle=\frac{\displaystyle{\lambda_{\gamma}\int_{B_{r}}w_{\gamma}^{2}\,dx+\lambda_{\beta}\int_{B_{R}\setminus
B_{r}}v_{\beta}^{2}\,dx}}{\displaystyle{\int_{B_{r}}w_{\gamma}^{2}\,dx+\int_{B_{R}\setminus
B_{r}}v_{\beta}^{2}\,dx}}$
$\displaystyle<\lambda_{\beta}\frac{{\displaystyle{\int_{B_{r}}w^{2}\,dx+\int_{B_{R}\setminus
B_{r}}w^{2}\,dx}}}{{\displaystyle{\int_{B_{r}}w^{2}\,dx+\int_{B_{R}\setminus
B_{r}}w^{2}\,dx}}}=\lambda_{\beta},$

where in the last formula we use (2.19), and this is an absurd. ∎

######  Proposition 2.8.

Let $R>0$, $1<p<+\infty$ and $\beta_{1}<\beta_{2}$. Let us denote by
$\lambda_{p,\beta_{1}}$ and $\lambda_{p,\beta_{2}}$ the eigenvalues defined in
(1.2) and let $v_{p,\beta_{1}}$ and $v_{p,\beta_{2}}$ be the corresponding
eigenfunctions normalized such that $v_{p,\beta_{1}}(0)=v_{p,\beta_{2}}(0)>0$,
then

$v_{p,\beta_{1}}(x)\geq v_{p,\beta_{2}}(x)\qquad\forall x\in B_{R}.$ (2.21)

###### Proof.

Let us suppose $p=2$, the general case is analogous. For sake of simplicity,
we denote by $\lambda_{\beta_{i}}:=\lambda_{2,\beta_{i}}$ and
$v_{\beta_{i}}:=v_{2,\beta_{i}}$.

Since both $v_{\beta_{1}}$ and $v_{\beta_{2}}$ are radial, we can write the
laplacian in polar coordinates, that is

$r^{n-1}\Delta u(r)=\bigl{(}r^{n-1}u^{\prime}(r)\bigr{)}\qquad r\in[0,R].$
(2.22)

Therefore function $u_{\beta_{i}}$, for $i=1,2$, satisfies

$v_{\beta_{i}}^{\prime}(r)=-\frac{1}{r^{n-1}}\biggl{[}\int_{0}^{r}s^{n-1}\lambda_{\beta_{i}}v_{\beta_{i}}\,ds\biggr{]}.$
(2.23)

Since $\lambda_{\beta_{1}}<\lambda_{\beta_{2}}$ and
$v_{\beta_{1}}(0)=v_{\beta_{2}}(0)$, by continuity there exists $\delta>0$
such that

$\lambda_{\beta_{1}}v_{\beta_{1}}(s)<\lambda_{\beta_{2}}v_{\beta_{2}}(s)\qquad\forall
s\in(0,\delta),$

and by (2.23)

$v_{\beta_{1}}^{\prime}(s)>v_{\beta_{2}}^{\prime}(s)\qquad s\in(0,\delta).$

By classical ODE comparison result, we obtain

$v_{\beta_{1}}(s)>v_{\beta_{2}}(s)\qquad s\in(0,\delta).$ (2.24)

Let us define

$A=\\{r>\delta\,:\,v_{\beta_{1}}(s)=v_{\beta_{2}}(s)\\},$

we want to prove that $A$ is empty, so by (2.24) we get the claim. Let us
suppose by contradiction that $A\neq\varnothing$, hence there exists

$t=\inf A.$

By continuity $v_{\beta_{1}}(t)=v_{\beta_{2}}(t)$, and this, combined with
(2.24), leads to

$v_{\beta_{1}}^{\prime}(t)<v_{\beta_{2}}^{\prime}(t).$ (2.25)

Let us set

$\gamma_{i}=-\frac{v_{\beta_{i}}^{\prime}(t)}{v_{\beta_{i}}(t)},$

by (2.25) we have $\gamma_{1}>\gamma_{2}$.

By Lemma 2.7, $v_{\beta_{1}}$ and $v_{\beta_{2}}$ are the first eigenfunction
of the Robin Laplacian in $B_{t}$ respectively with eigenvalue
$\lambda_{\gamma_{1}}$ and $\lambda_{\gamma_{2}}$ with parameter $\gamma_{1}$
and $\gamma_{2}$. Therefore by monotonicity of eigenvalues with respect to the
boundary parameter, we get

$\lambda_{\beta_{1}}=\lambda_{\gamma_{1}}>\lambda_{\gamma_{2}}=\lambda_{\beta_{2}}$

that is an absurd. ∎

######  Remark 2.1.

We highlight that, if we fix the value $v_{p,\beta}(0)$, (2.21) implies that

$\beta\mapsto{\left\|v_{p,\beta}\right\|}_{L^{p}}$

is non increasing, while the map

$\beta\mapsto
C(n,p,\beta,\rho)=\frac{v_{p,\beta}(0){\left|\Omega^{\star}\right|}}{{\left\|v_{p,\beta}\right\|}_{p}^{p}}$

is non decreasing, for all $\beta\in\operatorname{\mathbb{R}}$.

##  3 Proofs of main results

###### Proof of Theorem 1.1.

The quantity

$\frac{\lambda_{p,\beta}(\Omega)-\lambda_{p,\beta}(\Omega^{\star})}{\lambda_{p,\beta}(\Omega)},$

is bounded from above by 1, so inequality (1.5) is trivial when

$\left(1-\frac{n^{\frac{n}{n-1}}\omega_{n}^{\frac{1}{n-1}}{\left|\Omega\right|}}{P(\Omega)^{\frac{n}{n-1}}}\right)\geq\frac{1}{C(n,p,\beta,\rho)}=\frac{{\left\|v\right\|}_{\infty}^{p}{\left|\Omega^{\star}\right|}}{{\left\|v\right\|}_{p}^{p}}.$

We can assume

$\left(1-\frac{n^{\frac{n}{n-1}}\omega_{n}^{\frac{1}{n-1}}{\left|\Omega\right|}}{P(\Omega)^{\frac{n}{n-1}}}\right)<\frac{1}{C(n,p,\beta,\rho)}.$
(3.1)

Let $v$ be the solution to

$\begin{cases}-\Delta_{p}v=\lambda_{p,\beta}(\Omega^{\star}){\left|v\right|}^{p-2}v\,&\text{in}\,\,\Omega^{\star}\\\
\displaystyle{{\left|{\nabla v}\right|}^{p-2}\frac{\partial
v}{\partial\nu}+\beta{\left|v\right|}^{p-2}v=0}\,&\text{on}\,\,\partial\Omega^{\star},\end{cases}$
(3.2)

it is well known that $v$ is positive and radially symmetric. So the function

$g(t)={\left|\nabla v\right|}_{v=t}$

is well defined for all $t\in(v_{m},v_{M})$, where
$v_{m}=\min_{\Omega^{\star}}v$ and $v_{M}=\max_{\Omega^{\star}}v$.

Let us define $u(x)=G(d(x))$, $x\in\Omega$, where

$G^{-1}(t)=\int_{v_{m}}^{t}\frac{1}{g(s)}\,ds.\qquad v_{m}<t<v_{M}.$

By construction, $u\in W^{1,2}(\Omega)$ and

$\displaystyle\min_{\Omega}u=G(0)=v_{m},$
$\displaystyle{\left\|u\right\|}_{\infty}\leq v_{M},$
$\displaystyle{\left|\nabla
u\right|}_{u=t}={\left|G^{\prime}(d(x))\right|}_{G=t}=g(t)={\left|\nabla
v\right|}_{v=t}.$

Let

$E_{t}=\\{x\in\Omega:\,u(x)>t\\},\,\,\,B_{t}=\\{x\in\Omega^{\star}:\,v(x)>t\\}.$

By Lemma 2.6 and formula (2.16) we have

$-\frac{d}{dt}P(E_{t})\geq(n-1)\frac{W_{2}(E_{t})}{g(t)}\geq(n-1)n^{-\frac{n-2}{n-1}}\omega_{n}^{\frac{1}{n-1}}\frac{(P(E_{t}))^{\frac{n-2}{n-1}}}{g(t)}$

while for $v$ it holds

$-\frac{d}{dt}P(B_{t})=(n-1)\frac{W_{2}(B_{t})}{g(t)}=(n-1)n^{-\frac{n-2}{n-1}}\omega_{n}^{\frac{1}{n-1}}\frac{(P(B_{t}))^{\frac{n-2}{n-1}}}{g(t)}$

and $P(E_{0})=P(B_{0})$. Then, by classical comparison theorems for
differential inequalities,

$P(E_{t})\leq P(B_{t}),\quad 0\leq t\leq{\left\|u\right\|}_{\infty}.$ (3.3)

Denoting by $\mu(t)={\left|E_{t}\right|}$ and by
$\nu(t)={\left|B_{t}\right|}$, the coarea formula (2.1) ensures us that

$\displaystyle-\mu^{\prime}(t)=$
$\displaystyle\int_{u=t}\frac{1}{{\left|\nabla
u\right|}}\,d\mathcal{H}^{n-1}=\frac{P(E_{t})}{g(t)}\leq\frac{P(B_{t})}{g(t)}$
$\displaystyle=$ $\displaystyle\int_{v=t}\frac{1}{{\left|\nabla
v\right|}}\,d\mathcal{H}^{n-1}=-\nu^{\prime}(t),$

for $t\in(0,{\left\|u\right\|}_{\infty})$. The first equality holds true since
${\left|\nabla u\right|}\neq 0$ in $\Set{v_{m}<u<{\left\|u\right\|}_{\infty}}$
(see [BZ88]). So the function $\nu-\mu$ is decreasing in
$[0,{\left\|u\right\|}_{\infty}]$, and

$\displaystyle\int_{\Omega}u^{p}\,dx=$
$\displaystyle\int_{0}^{{\left\|u\right\|}_{\infty}}pt^{p-1}\mu(t)\,dt=\int_{0}^{v_{M}}pt^{p-1}\nu(t)\,dt-\int_{0}^{v_{M}}pt^{p-1}(\nu(t)-\mu(t))\,dt$
$\displaystyle\geq$ $\displaystyle\int_{\Omega^{\star}}v^{p}\,dx-
v_{M}^{p}({\left|\Omega^{\star}\right|}-{\left|\Omega\right|}).$

Moreover, by (3.3), we get

$\int_{u=t}{\left|\nabla
u\right|}^{p-1}\,d\mathcal{H}^{n-1}=g(t)^{p-1}P(E_{t})\leq
g(t)^{p-1}P(B_{t})=\int_{v=t}{\left|\nabla
v\right|}^{p-1}\,d\mathcal{H}^{n-1},$

so, if we integrate from $0$ to ${\left\|u\right\|}_{\infty}$,

$\int_{\Omega}{\left|\nabla
u\right|}^{p}\,dx\leq\int_{0}^{{\left\|u\right\|}_{\infty}}\int_{v=t}{\left|\nabla
v\right|}^{p-1}\,d\mathcal{H}^{n-1}\,dt\leq\int_{\Omega^{\star}}{\left|\nabla
v\right|}^{p}.$

We also observe that by construction both $u$ and $v$ are constant on
$\partial\Omega$, so

$\beta\int_{\partial\Omega}u^{p}\,d\mathcal{H}^{n-1}=\beta
v_{m}^{p}P(\Omega)=\beta\int_{\partial\Omega^{\star}}v^{p}\,d\mathcal{H}^{n-1}.$

We finally get

$\displaystyle\lambda_{p,\beta}(\Omega)$
$\displaystyle\leq\frac{\displaystyle{\int_{\Omega}{\left|\nabla
u\right|}^{p}\,dx+\beta\int_{\partial\Omega}u^{p}\,d\mathcal{H}^{n-1}}}{\displaystyle{\int_{\Omega}u^{p}\,dx}}\leq\frac{\displaystyle{\int_{\Omega^{\star}}{\left|\nabla
v\right|}^{p}\,dx+\beta\int_{\partial\Omega^{\star}}v^{p}\,d\mathcal{H}^{n-1}}}{\displaystyle{\int_{\Omega^{\star}}v^{p}\,dx-
v_{M}^{p}({\left|\Omega^{\star}\right|}-{\left|\Omega\right|})}}$ (3.4)
$\displaystyle=\lambda_{p,\beta}(\Omega^{\star})\,\frac{1}{\displaystyle{1-C(n,p,\beta,\rho)\left(1-\frac{{\left|\Omega\right|}}{{\left|\Omega^{\star}\right|}}\right)}}.$

The claim follows from (3.1) as the quantity

$1-C(n,p,\beta,\rho)\left(1-\frac{{\left|\Omega\right|}}{{\left|\Omega^{\star}\right|}}\right)$

is non-negative. ∎

######  Remark 3.1.

The constant
$\displaystyle{C(n,p,\beta,\rho)=\frac{{\left\|v\right\|}_{\infty}^{p}{\left|\Omega^{\star}\right|}}{{\left\|v\right\|}_{p}^{p}}}$
depends on the perimeter of the set $\Omega$ and on $\beta$. Thanks to
Proposition 2.8 and Remark 2.1 it is possible to bound the constant
$C(n,p,\beta,\rho)$ from above with a constant independent of the perimeter
and of $\beta$. Indeed $\forall\beta>0$, if we denote by $v_{p,\infty}$ the
first Dirichlet eigenfunction normalized in such a way
$v_{p,\beta}(0)=v_{p,\infty}(0)$, we have

$C(n,p,\beta,\rho)\leq\frac{v_{p,\infty}(0){\left|\Omega^{\star}\right|}}{{\left\|v_{p,\infty}\right\|}_{L^{p}}^{p}}=:C(n,p)$

that is independent of the perimeter thanks to the rescaling properties of the
Dirichlet $p$-Laplacian eigenfunction.

###### Proof of Theorem 1.2.

Let $v$ be a positive eigenfunction associated to
$\lambda_{p,\beta}(\Omega^{\star})$, then $v$ is a $p$ sub-harmonic function.
We denote by $v_{m}=v(0)=\min_{\Omega^{\star}}v$ and by
$v_{M}=\max_{\Omega^{\star}}v$.

Let us consider the function $\tilde{v}=v_{M}-v$, that is a positive function
with zero trace, and $g(t)={\left|\nabla v\right|}_{v=t}$, $v_{m}<t<v_{M}$. We
set $\tilde{u}(x)=G(d(x)),\,x\in\Omega$, where
$\displaystyle{G^{-1}(t)=\int_{0}^{t}\frac{1}{g(s)}\,ds}$ with
$0<t<v_{M}-v_{m}$. By construction, $\tilde{u}\in W^{1,p}_{0}(\Omega)$. Now we
can set $u=v_{M}-\tilde{u}$, and we have:

$\begin{gathered}u_{M}=\max_{\Omega}u=v_{M}\\\
u_{m}=\min_{\Omega}u=v_{M}-\max_{\Omega}\tilde{u}\geq
v_{M}-\max_{\Omega^{\star}}\tilde{v}=\min_{\Omega^{\star}}v=v_{m}\\\
{\left|\nabla u\right|}_{u=t}={\left|\nabla v\right|}_{v=t}=g(t)\quad
u_{m}<t<u_{M}.\end{gathered}$ (3.5)

Let

$\displaystyle\widetilde{E}_{t}$
$\displaystyle=\\{x\in\Omega\,:\,\tilde{u}(x)>t\\},$
$\displaystyle\widetilde{B}_{t}$
$\displaystyle=\\{x\in\Omega\,:\,\tilde{v}(x)>t\\},$ (3.6) $\displaystyle
E_{t}$
$\displaystyle=\\{x\in\Omega\,:\,u(x)>t\\}=\Omega\setminus\overline{\widetilde{E}}_{v_{M}-t},$
$\displaystyle B_{t}$
$\displaystyle=\\{x\in\Omega\,:\,v(x)>t\\}=\Omega\setminus\overline{\widetilde{B}}_{v_{M}-t}.$

By Lemma 2.6 and formula (2.16) we have

$-\frac{d}{dt}P(\widetilde{E}_{t})\geq(n-1)\frac{W_{2}(\widetilde{E}_{t})}{g(t)}\geq(n-1)n^{-\frac{n-2}{n-1}}\omega_{n}^{\frac{1}{n-1}}\frac{(P(\widetilde{E}_{t}))^{\frac{n-2}{n-1}}}{g(t)}$

while for $v$ it holds

$-\frac{d}{dt}P(\widetilde{B}_{t})=(n-1)\frac{W_{2}(\widetilde{B}_{t})}{g(t)}=(n-1)n^{-\frac{n-2}{n-1}}\omega_{n}^{\frac{1}{n-1}}\frac{(P(\widetilde{B}_{t}))^{\frac{n-2}{n-1}}}{g(t)}$

and $P(E_{0})=P(B_{0})$. Then, by classical comparison theorems for
differential inequalities,

$P(\widetilde{E}_{t})\leq P(\widetilde{B}_{t}),\quad 0\leq t\leq v_{M}-v_{m}.$
(3.7)

Hence

$P(E_{t})=P(\widetilde{E}_{v_{M}-t})\leq
P(\widetilde{B}_{v_{M}-t})=P(B_{t}),\quad v_{m}\leq t\leq v_{M}.$ (3.8)

Moreover, denoted by $\tilde{\mu}(t)=\lvert\widetilde{E}_{t}\rvert$ and
$\tilde{\nu}(t)=\lvert\widetilde{B}_{t}\rvert$, the coarea formula (2.1)
ensures us that

$\displaystyle-\tilde{\mu}^{\prime}(t)=$
$\displaystyle\int_{\tilde{u}=t}\frac{1}{{\left|\nabla\tilde{u}\right|}}\,d\mathcal{H}^{n-1}=\frac{P(\widetilde{E}_{t})}{g(t)}\leq\frac{P(\widetilde{B}_{t})}{g(t)}$
$\displaystyle=$
$\displaystyle\int_{\tilde{v}=t}\frac{1}{{\left|\nabla\tilde{v}\right|}}\,d\mathcal{H}^{n-1}=-\tilde{\nu}^{\prime}(t),\qquad
0\leq t<v_{M}-v_{m}.$

Moreover, setting
$\mu(t)={\left|E_{t}\right|}={\left|\Omega\right|}-\tilde{\mu}(v_{M}-t)$ and
by
$\nu(t)={\left|B_{t}\right|}={\left|\Omega^{\star}\right|}-\tilde{\nu}(v_{M}-t)$,
we have $-\mu^{\prime}(t)\leq-\nu^{\prime}(t)$ in $[v_{m},v_{M}]$. So the
function $\nu-\mu$ is decreasing in $[v_{m},v_{M}]$, and

$\displaystyle\int_{\Omega}u^{p}\,dx=$
$\displaystyle\int_{0}^{u_{M}}pt^{p-1}\mu(t)\,dt=\int_{0}^{v_{M}}pt^{p-1}\nu(t)\,dt-\int_{0}^{v_{M}}pt^{p-1}(\nu(t)-\mu(t))\,dt$
$\displaystyle=$
$\displaystyle\int_{0}^{v_{M}}pt^{p-1}\nu(t)\,dt-\int_{0}^{v_{m}}pt^{p-1}(\nu(t)-\mu(t))\,dt-\int_{v_{m}}^{v_{M}}pt^{p-1}(\nu(t)-\mu(t))\,dt$
$\displaystyle\leq$ $\displaystyle\int_{\Omega^{\star}}v^{p}\,dx-
v_{m}^{p}({\left|\Omega^{\star}\right|}-{\left|\Omega\right|})=\int_{\Omega^{\star}}v^{p}\,dx\left[1-\frac{v_{m}^{p}}{{\left\|v\right\|}_{p}^{p}}({\left|\Omega^{\star}\right|}-{\left|\Omega\right|})\right].$

Moreover, by (3.8), we get

$\int_{\tilde{u}=t}{\left|\nabla\tilde{u}\right|}^{p-1}\,d\mathcal{H}^{n-1}=g(t)^{p-1}P(\widetilde{E}_{t})\leq
g(t)^{p-1}P(\widetilde{B}_{t})=\int_{\tilde{v}=t}{\left|\nabla\tilde{v}\right|}^{p-1}\,d\mathcal{H}^{n-1},$

so, if we integrate from $0$ to ${\left\|u\right\|}_{\infty}$,

$\displaystyle\int_{\Omega}{\left|\nabla
u\right|}^{p}\,dx=\int_{\Omega}{\left|\nabla\tilde{u}\right|}^{p}\,dx=$
$\displaystyle\int_{0}^{{\left\|\tilde{u}\right\|}_{\infty}}\int_{\tilde{u}=t}{\left|\nabla\tilde{u}\right|}^{p-1}\,d\mathcal{H}^{n-1}\,dt$
$\displaystyle\leq$
$\displaystyle\int_{0}^{{\left\|\tilde{u}\right\|}_{\infty}}\int_{\tilde{v}=t}{\left|\nabla\tilde{v}\right|}^{p-1}\,d\mathcal{H}^{n-1}\,dt=\int_{\Omega^{\star}}{\left|\nabla\tilde{v}\right|}^{p}=\int_{\Omega^{\star}}{\left|\nabla
v\right|}^{p}.$

We also observe that by construction both $u$ and $v$ are constant on
$\partial\Omega$, so

$\beta\int_{\partial\Omega}u^{p}\,d\mathcal{H}^{n-1}=\beta
u_{M}^{p}P(\Omega)=\beta
v_{M}^{p}P(\Omega)=\beta\int_{\partial\Omega^{\star}}v^{p}\,d\mathcal{H}^{n-1}.$

We finally get

$\displaystyle\lambda_{p,\beta}(\Omega)$
$\displaystyle\leq\frac{\displaystyle{\int_{\Omega}{\left|\nabla
u\right|}^{p}\,dx+\beta\int_{\partial\Omega}u^{p}\,d\mathcal{H}^{n-1}}}{\displaystyle{\int_{\Omega}u^{p}\,dx}}\leq\frac{\displaystyle{\int_{\Omega^{\star}}{\left|\nabla
v\right|}^{p}\,dx+\beta\int_{\partial\Omega^{\star}}v^{p}\,d\mathcal{H}^{n-1}}}{\displaystyle{\int_{\Omega^{\star}}v^{p}\,dx\Bigl{[}1-\frac{v_{m}^{p}}{{\left\|v\right\|}_{p}^{p}}({\left|\Omega^{\star}\right|}-{\left|\Omega\right|})\Bigr{]}}}$
(3.9)
$\displaystyle=\lambda_{p,\beta}(\Omega^{\star})\,\frac{1}{\displaystyle{1-\frac{v_{m}^{p}}{{\left\|v\right\|}_{p}^{p}}\,({\left|\Omega^{\star}\right|}-{\left|\Omega\right|})}}.$

Hence, by direct calculation

$\frac{\lambda_{p,\beta}(\Omega^{\star})-\lambda_{p,\beta}(\Omega)}{{\left|\lambda_{p,\beta}(\Omega)\right|}}\geq\frac{v_{m}^{p}}{{\left\|v\right\|}_{p}^{p}}({\left|\Omega^{\star}\right|}-{\left|\Omega\right|})\qed$
(3.10)

######  Remark 3.2.

Unlike the constant in Theorem 1.1, the constant
$\displaystyle{C(n,p,\beta,\rho)=\frac{v_{m}^{p}{\left|\Omega^{\star}\right|}}{{\left\|v\right\|}_{p}^{p}}}$
cannot be bounded from below with a constant independent of the perimeter and
of $\beta$. Indeed, for example if $n=p=2$ and $P(\Omega)=2\pi$, we have that

$v_{\beta}(x)=I_{0}\left(\sqrt{-\lambda_{\beta}(B_{1})}{\left|x\right|}\right),$

where $I_{0}$ is the modified Bessel function.

We recall that for $z$ sufficiently large (see [AS64, Section 9.7]), we have

$I_{0}(z)\sim\frac{e^{z}}{\sqrt{2\pi
z}}\biggl{[}1+\frac{1}{8z}+\frac{9}{2!(8z)^{2}}+\ldots\biggr{]}$

therefore

$\lVert
v_{\beta}(x)\rVert_{L^{2}}\sim\frac{e^{\beta}}{\beta^{2}}\xrightarrow{\beta\to-\infty}+\infty$

and

$C(2,2,\beta,2\pi)=\frac{(v_{\beta})_{m}^{2}{\left|B_{1}\right|}}{{\left\|v_{\beta}\right\|}^{2}_{2}}\sim\frac{\beta^{2}}{e^{-\beta}}\xrightarrow{\beta\to-\infty}0.$

######  Remark 3.3.

We want to highlight that the constant $C$ depends actually just on $n,p$ and
$\rho^{\frac{p-1}{n-1}}\beta$. Indeed, for all
$\Omega\subset\operatorname{\mathbb{R}}^{n}$ bounded and convex set with
$P(\Omega)=\rho$, we can consider

$\Omega_{1}=\biggl{(}\frac{n\omega_{n}}{\rho}\biggr{)}^{\frac{1}{n-1}}\Omega\qquad\Omega_{1}^{*}=\biggl{(}\frac{n\omega_{n}}{\rho}\biggr{)}^{\frac{1}{n-1}}\Omega^{\star}\qquad
t:=\biggl{(}\frac{\rho}{n\omega_{n}}\biggr{)}^{\frac{1}{n-1}}$

so $P(\Omega_{1})=P(\Omega_{1}^{\star})=n\omega_{n}$ and we have

$\displaystyle\frac{\lambda_{p,\beta}(\Omega^{\star})-\lambda_{p,\beta}(\Omega)}{-\lambda_{p,\beta}(\Omega)}$
$\displaystyle=\frac{\lambda_{p,\beta}(t\Omega_{1}^{\star})-\lambda_{p,\beta}(t\Omega_{1})}{-\lambda_{p,\beta}(t\Omega_{1})}$
$\displaystyle=\frac{\lambda_{p,t^{p-1}\beta}(\Omega_{1}^{*})-\lambda_{p,t^{p-1}\beta}(\Omega_{1})}{\lambda_{p,t^{p-1}\beta}(\Omega_{1})}$
$\displaystyle\geq
C\bigl{(}n,p,n\omega_{n},t^{p-1}\beta\bigr{)}\biggl{(}1-\frac{n^{\frac{n}{n-1}}\omega_{n}^{\frac{1}{n-1}}{\left|\Omega\right|}}{P(\Omega)^{\frac{n}{n-1}}}\biggr{)}$
$\displaystyle=C\Bigl{(}n,p,n\omega_{n},\Bigl{(}\frac{\rho}{n\omega_{n}}\Bigr{)}^{\frac{p-1}{n-1}}\beta\Bigr{)}\biggl{(}1-\frac{n^{\frac{n}{n-1}}\omega_{n}^{\frac{1}{n-1}}{\left|\Omega\right|}}{P(\Omega)^{\frac{n}{n-1}}}\biggr{)}$
$\displaystyle=C\bigl{(}n,p,\rho^{\frac{p-1}{n-1}}\beta\bigr{)}\biggl{(}1-\frac{n^{\frac{n}{n-1}}\omega_{n}^{\frac{1}{n-1}}{\left|\Omega\right|}}{P(\Omega)^{\frac{n}{n-1}}}\biggr{)}$

###### Proof of Theorem 1.3.

Using (3.10), the isoperimetric inequality and

${\left|\lambda_{p,\beta}(\Omega)\right|}\geq{\left|\beta\right|}\frac{P(\Omega)}{{\left|\Omega\right|}},$

we have

$\displaystyle\lambda_{p,\beta}(\Omega^{\star})-\lambda_{p,\beta}(\Omega)$
$\displaystyle\geq{\left|\lambda_{p,\beta}(\Omega)\right|}\frac{v_{m}^{p}}{{\left\|v\right\|}_{p}^{p}}({\left|\Omega^{\star}\right|}-{\left|\Omega\right|})$
(3.11)
$\displaystyle\geq{\left|\beta\right|}\frac{P(\Omega)}{{\left|\Omega\right|}}\frac{v_{m}^{p}}{{\left\|v\right\|}_{p}^{p}}({\left|\Omega^{\star}\right|}-{\left|\Omega\right|})$
$\displaystyle\geq{\left|\beta\right|}\frac{n^{\frac{n}{n-1}}\omega_{n}^{\frac{1}{n-1}}}{\rho^{\frac{1}{n-1}}}\frac{v_{m}^{p}}{{\left\|v\right\|}_{p}^{p}}({\left|\Omega^{\star}\right|}-{\left|\Omega\right|}).$

Now, if we suppose that
$\lambda_{p,\beta}(\Omega^{\star})-\lambda_{p,\beta}(\Omega)\leq\delta_{0}$,
by (3.11) we have

${\left|\Omega^{\star}\right|}-{\left|\Omega\right|}\leq
K(n,\rho,p,\beta)\delta_{0}.$

So by Lemma 2.4 we conclude

$\lambda_{p,\beta}(\Omega^{\star})-\lambda_{p,\beta}(\Omega)\geq
C(n,\rho,p,\beta)\,g(\mathcal{A}_{\mathcal{H}}^{\star}(\Omega)).\qed$

######  Remark 3.4.

Our result, Theorem 1.3, applies only when

$\lambda_{p,\beta}(\Omega^{\star})-\lambda_{p,\beta}(\Omega)\leq\delta_{0}.$

It is possible to get rid of this constraint, obtaining a weaker result.

In order to obtain it, we need the quantitative version of the isoperimetric
inequality proved in [FMP08]

$P(\Omega)\geq
P(\Omega^{\sharp})\bigl{(}1+\gamma(n)\alpha(\Omega)^{2}\bigr{)}\qquad\text{where
}\alpha(\Omega)=\min\Set{\frac{{\left|\Omega\triangle
B_{r}\right|}}{{\left|\Omega\right|}}\,}{\,{\left|B_{r}\right|}={\left|\Omega\right|}},$
(3.12)

and the following result [EFT05, Lemma 4.2]: there exists a constant $C(n)$
such that if $C,W$ are open and convex sets such that
${\left|C\right|}={\left|W\right|}$ and ${\left|C\triangle
W\right|}<\frac{{\left|C\right|}}{2}$, it holds

$d_{\mathcal{H}}(C,W)\leq
C(n)\bigl{[}\operatorname{\text{diam}}(C)+\operatorname{\text{diam}}(W)\bigr{]}\biggl{(}\frac{{\left|C\triangle
W\right|}}{{\left|C\right|}}\biggr{)}^{\frac{1}{n}}.$ (3.13)

Moreover we have to recall that

${\left|\Omega^{\star}\right|}=\frac{P(\Omega)^{\frac{n}{n-1}}}{n^{\frac{n}{n-1}}\omega_{n}^{\frac{1}{n-1}}}\qquad\text{
and }\qquad
P(\Omega^{\sharp})=n\omega_{n}^{\frac{1}{n}}{\left|\Omega\right|}^{\frac{n-1}{n}}$

and we have

$\displaystyle 1-\frac{{\left|\Omega\right|}}{\lvert\Omega^{\star}\rvert}$
$\displaystyle=1-\frac{n^{\frac{n}{n-1}}\omega_{n}^{\frac{1}{n-1}}{\left|\Omega\right|}}{P(\Omega)}$
$\displaystyle\geq
1-\frac{1}{\displaystyle{\bigl{(}1+\gamma(n)\alpha^{2}(\Omega)\bigr{)}^{\frac{n}{n-1}}}}$
$\displaystyle\geq
1-\frac{1}{\displaystyle{\bigl{(}1+\gamma(n)\alpha^{2}(\Omega)\bigr{)}}}$
$\displaystyle=\frac{\gamma(n)\alpha^{2}(\Omega)}{1+\gamma(n)\alpha^{2}(\Omega)},$

where we used Bernoulli’s inequality

$(1+x)^{r}\geq 1+rx\qquad\forall x\geq-1,\,\forall r\geq 0.$

Since $0<\alpha^{2}(\Omega)<4$ we have

$1-\frac{{\left|\Omega\right|}}{\lvert\Omega^{\star}\rvert}\geq\frac{\gamma(n)}{1+4\gamma(n)}\,\alpha^{2}(\Omega)=C(n)\alpha^{2}(\Omega).$

If $\Omega^{\sharp}$ is a ball that realizes the minimum in (3.12), then using
(3.13) we obtain

$C(n)\alpha^{2}(\Omega)\geq
C(n)\,\frac{\displaystyle{d_{\mathcal{H}}(\Omega,\Omega^{\sharp})^{2n}}}{\displaystyle{\bigl{[}\operatorname{\text{diam}}(\Omega)+\operatorname{\text{diam}}(\Omega^{\sharp})\bigr{]}^{2n}}}$

and so

$\frac{\lambda_{p,\beta}(\Omega^{\star})-\lambda_{p,\beta}(\Omega)}{{\left|\lambda_{p,\beta}(\Omega)\right|}}\geq
C(n,p,\beta\rho)C(n)\frac{\displaystyle{d_{\mathcal{H}}(\Omega,\Omega^{\sharp})^{2n}}}{\displaystyle{\bigl{[}\operatorname{\text{diam}}(\Omega)+\operatorname{\text{diam}}(\Omega^{\sharp})\bigr{]}^{2n}}}.$

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