---
abstract: 'We consider the torsion function for the Dirichlet Laplacian $-\Delta$, and for the Schrödinger operator $- \Delta + V$ on an open set $\Omega\subset {\mathbb{R}}^m$, with Lebesgue measure $0<|\Omega|<\infty$, with a real-valued, non-negative, measurable potential $V.$ We investigate the phenomena of vanishing efficiency and localisation, and large efficiency for the torsion function and the first Dirichlet eigenfunction.'
author:
- |
    [M. van den Berg]{}\
    School of Mathematics, University of Bristol\
    Fry Building, Woodland Road\
    Bristol BS8 1UG, United Kingdom\
    `mamvdb@bristol.ac.uk`\
    \
    [D. Bucur]{}\
    Laboratoire de Mathématiques, Université Savoie Mont Blanc\
    UMR CNRS 5127\
    Campus Scientifique, 73376 Le-Bourget-Du-Lac, France\
    `dorin.bucur@univ-savoie.fr`\
    \
    [T. Kappeler]{}\
    Institut für Mathematik, Universität Zürich\
    Winterthurerstrasse 190, CH-8057 Zürich, Switzerland\
    `thomas.kappeler@math.uzh.ch`
date: 13 May 2020
title: On efficiency and localisation for the torsion function
---

3truecm

     [ Mathematics Subject Classification (2000)]{}: 35J05, 35J25, 35K05, 35P99.

**Keywords**: Torsion function, first Dirichlet eigenfunction, Schrödinger operator, Dirichlet boundary condition, localisation, efficiency.

Introduction and main results \[sec1\]
======================================

Let $\Omega$ be an open set in ${\mathbb{R}}^m$, with finite Lebesgue measure, $0 < |\Omega| < \infty$, and boundary $\partial\Omega$, and let $V:\Omega\rightarrow {\mathbb{R}}^+,\, {\mathbb{R}}^+= [0,\infty) $ be measurable. Let $$L=-\Delta+V,$$ be the Schrödinger operator acting in $L^2(\Omega)$. Throughout this paper we denote by $\Vert\cdot\Vert_p$ the standard $L^p$ norm, $1\le p\le \infty$. It is well-known ([@vdBC],[@vdB]) that the first eigenvalue of the Dirichlet Laplacian defined by $$\lambda_1(\Omega)=\inf_{\varphi\in
H_0^1(\Omega)\setminus\{0\}}\frac{\Vert \nabla\varphi\Vert_2^2}{
\Vert \varphi\Vert_2^2},$$ is bounded away from $0$. Furthermore $$\label{e3}
-\Delta v=1,\, \qquad v\in H_0^{ 1 }(\Omega),$$ has a unique solution denoted by $v_{\Omega}$. The function $v_{\Omega}$ is non-negative, pointwise increasing in $\Omega$, and satisfies, $$\label{e4}
\lambda_1(\Omega)^{-1}< \|v_{\Omega}\|_{\infty}\le
(4+3m\log 2)\lambda_1(\Omega)^{-1}.$$ The $m$-dependent constant in the right-hand side of has subsequently been improved ([@GS],[@HV]). We denote the sharp constant by $$\label{e5}
\mathfrak{c}_m=\sup\{\lambda_1(\Omega)\|v_{\Omega}\|_{\infty}:\Omega\,\textup{ open in }\,{\mathbb{R}}^m,\,|\Omega|<\infty\}.$$

Similarly, $Lv = 1$ has a unique solution $v_{\Omega, V}\in H^1_0(\Omega)$, referred to as the torsion function for $L$.

In this paper we consider the [*efficiency*]{} of the torsion function, and its localisation. The notion of efficiency, or mean to max ratio, goes back to [@ps] and [@sperb], where it was introduced for the first Dirichlet eigenfunction. Here we state the analogous definition for the torsion function of a Schrödinger operator on an open set $\Omega \subset {\mathbb{R}}^m $ with $0 < | \Omega| < \infty$.

\[def1\] Let $\Omega$ be an open set in ${\mathbb{R}}^m$ with $0 < |\Omega | < \infty,$ and let $V: \Omega \to {\mathbb{R}}^+$ be measurable. The [*efficiency*]{} of $v_{\Omega,V}$ is $$\Phi(\Omega,V)=\frac{\|v_{\Omega,V}\|_1}{|\Omega|
\|v_{\Omega,V}\|_\infty}.$$ Let $(\Omega_n)$ be a sequence of open sets in ${\mathbb{R}}^m$ with $0<|\Omega_n|<\infty$, and let $V_n:\Omega_n\rightarrow {\mathbb{R}}^+$ be a sequence of measurable functions. We say that $(v_{\Omega_n,V_n})$ has vanishing efficiency if $\lim_{n\rightarrow\infty}\Phi(\Omega_n,V_n)=0$.

If $V=0$, then $\Phi(\Omega,0)\equiv \Phi(\Omega)$ coincides with the definition in [@DPGGLB].

Our first two results concern the comparison of $\Phi(\Omega,V)$ and $\Phi(\Omega)$.

\[the1\] If $\Omega$ is an open set in ${\mathbb{R}}^m$ with $0<|\Omega|<\infty$, and if $V: \Omega \to {\mathbb{R}}^+$ is measurable with $0\le V \le c$, then $$\label{e7}
\Phi(\Omega,V)\le 2^{2(3m+4)c/\lambda_1(\Omega)}\frac{8c+\lambda_1(\Omega)}{\lambda_1(\Omega)}\bigg(\frac{8c+\lambda_1(\Omega)}{8c}\bigg)^{8c/\lambda_1(\Omega)}\Phi(\Omega),$$ and $$\label{e8}
\Phi(\Omega,V)\ge 2^{-2(3m+4)c/\lambda_1(\Omega)}\frac{\lambda_1(\Omega)}{8c+\lambda_1(\Omega)}\bigg(\frac{8c}{8c+\lambda_1(\Omega)}\bigg)^{8c/\lambda_1(\Omega)}\Phi(\Omega).$$ Furthermore for fixed $\Omega$ the right-hand sides of and converge to $\Phi(\Omega)$ as $c\downarrow 0$.

\[the2\] Let $(\Omega_n)$ be a sequence of open sets in ${\mathbb{R}}^m$ with $0 < |\Omega_n|<\infty$, and let $V_n:\Omega_n\to {\mathbb{R}}^+$ be a sequence of measurable functions. If there exists $\eta<\infty$ such that $\sup_{n\in{\mathbb{N}}}\frac{\| V_n \|_{\infty}}{\lambda_1(\Omega_n)} \le \eta,$ then $\lim_{n\rightarrow\infty}\Phi(\Omega_n,V_n)=0$ if and only if $\lim_{n\rightarrow\infty}\Phi(\Omega_n)=0$.

In [@HLP] it was shown that if $m\ge 2$, then $$\inf\{\Phi(\Omega):\Omega\subset {\mathbb{R}}^m,\,\Omega\,\,\,\textup{open},\, 0 < |\Omega|<\infty\}=0,$$ and $$\label{e17a}
\sup\{\Phi(\Omega):\Omega\subset {\mathbb{R}}^m,\,\Omega\,\,\,\textup{open},\, 0 < |\Omega|<\infty\}=1.$$

The analogous result for Schrödinger operators reads as follows.

\[the3\] If $\Omega$ is a fixed open set in ${\mathbb{R}}^m$ with $0 < | \Omega | < \infty$, then $$\label{e51}
\sup\{\Phi(\Omega,V): ( V:\Omega\rightarrow {\mathbb{R}}^+, \textup{measurable})\}=1,$$ and $$\label{e52}
\inf\{\Phi(\Omega,V): ( V:\Omega\rightarrow {\mathbb{R}}^+, \textup{measurable})\}=\Phi(\Omega).$$ In fact, $$\label{e50}
1>\Phi(\Omega,c) \ge 1-\frac{2^{(m+4)/2}}{|\Omega|}\int_{\Omega}dx\,e^{-c^{1/2}d_{\Omega}(x)/2}, \qquad \forall c\ge 0,$$ where $d_\Omega : \Omega \to {\mathbb{R}}^+$ is the distance function to the boundary, $$d_{\Omega}(x)=\min\{|x-y|:y\in {\mathbb{R}}^m\setminus\Omega\}.$$

We denote the spectrum of $L$ with Dirichlet boundary conditions by $$\{\lambda_1(\Omega,V)\le \lambda_2(\Omega,V)\le \cdots\},$$ accumulating at infinity only. We denote a corresponding $L^2$-orthonormal basis of eigenfunctions by $\{\varphi_{1,\Omega,V}, \varphi_{2,\Omega,V},\cdots\}.$ If the first Dirichlet eigenvalue $\lambda_1(\Omega,V)$ of $L$ has multiplicity $1$, then its corresponding eigenspace is one-dimensional, and $\varphi_{1,\Omega,V}$ is uniquely defined up to a sign. Since $\varphi_{1,\Omega,V}$ does not change sign we may choose $\varphi_{1,\Omega,V}>0.$ If $\lambda_1(\Omega,V)$ is simple, then we define the efficiency of $\varphi_{1, \Omega, V}$ by $$E(\Omega,V)=\frac{\|\varphi_{1, \Omega, V}\|_1}{| \Omega | \, \| \varphi_{1, \Omega, V} \|_{\infty}} .$$ If $V=0$, then we write $$E(\Omega)=E(\Omega,0) \ , \qquad \lambda_1(\Omega) = \lambda_1(\Omega,0).$$ We note that if $\Omega$ is connected, then $\lambda_1(\Omega)$ is simple.

By [@vdBFNT Theorem 1.2] and [@vdB1 Theorem 1] it is possible for $m\ge 2$ to construct, for any $\varepsilon\in(0,1)$, an open connected set $\Omega_{\varepsilon}\subset {\mathbb{R}}^m$ such that both $$\frac{\lambda_1(\Omega_{\varepsilon})\|v_{\Omega_{\varepsilon}}\|_1}{|\Omega_{\varepsilon}|}>1-\varepsilon,$$ and $$\lambda_1(\Omega_{\varepsilon})\|v_{\Omega_{\varepsilon}}\|_{\infty}<1+\varepsilon.$$ This implies that $$\Phi(\Omega_{\varepsilon}) \ge \frac{1-\varepsilon}{1+\varepsilon},\qquad\forall\varepsilon\in(0,1),$$ which in turn implies . We were unable to prove that $E(\Omega_{\varepsilon})>1-\varepsilon,\,\forall\varepsilon\in(0,1)$. Nevertheless have the following:

\[the7\] If $m\ge 2$, then $$\sup\{E(\Omega):\Omega\subset {\mathbb{R}}^m,\,\Omega\,\,\,\textup{open and connected},\,0 < |\Omega|<\infty\}=1.$$

By examining the proof of Theorem \[the7\] in Section \[sec5\] we see that for any $m\ge 2$, and $\varepsilon \in (0,1)$ there exists an open, bounded and connected set $\Omega_{\varepsilon}\subset {\mathbb{R}}^m$ such that (i) $E(\Omega_{\varepsilon})\ge 1-\varepsilon$, and (ii) $\lambda_1(\Omega_{\varepsilon})|\Omega_{\varepsilon}|^{2/m} $ is large for $\varepsilon$ small. In Theorem \[the9\] below we show that is a general phenomenon. That is if $\Omega$ is any open and connected set in $ {\mathbb{R}}^m, \,m\ge 2,$ such the mean to max ratio of the first Dirichlet eigenfunction is close to $1$, then the eigenfunction is close to its maximum on most of $\Omega$, and $\lambda_1(\Omega)|\Omega|^{2/m}$ is large. We have a similar phenomenon for the torsion function. Throughout we denote for $\Omega$ open with $0<|\Omega|<\infty$, and $u\in L^1(\Omega),$ $$\fint_{\Omega}u := \frac{1}{|\Omega|} \int_{\Omega} u.$$

\[the9\] Let $m\ge 2$ and let $\Omega$ be a non-empty open set in ${\mathbb{R}}^m$ with finite Lebesgue measure, $|\Omega|<+\infty$.

-   If $u\in H_0^1(\Omega)\cap L^{\infty}(\Omega)\cap\{u\ne 0\}$, and if $$\fint_{\Omega}u\ge \frac{2\|u\|_{\infty}}{m+2}\,$$ then $$\label{p22}
    \bigg(\frac{\omega_m}{|\Omega |}\bigg)^{(2-m)/2}
    \bigg(\|u\|_{\infty}-\fint_{\Omega}u\bigg)\int_{\Omega}|\nabla u|^2\ge \frac{4m^2}{(m+2)^2}\omega_m\|u\|^3_{\infty}.$$ Equality occurs if and only if $\Omega$ is a ball, and $u$ is a multiple of the torsion function.

-   $$\label{p22a}
    \|v_{\Omega}\|_{\infty}\le
    \frac{(m+2)^2}{4m^2} \bigg(\frac{|\Omega|}{\omega_m}\bigg)^{2/m}(1-\Phi(\Omega)),$$

    $$\label{p22b}
    \lambda_1(\Omega)\ge
    \frac{4m^2}{(m+2)^2}  \bigg(\frac{\omega_m}{|\Omega |}\bigg)^{2/m}(1-\Phi(\Omega))^{-1}.$$

-   If $\Omega$ is connected, then $$\label{p22c}
    \lambda_1(\Omega)\ge
    \frac{4m^2}{(m+2)^2}  \bigg(\frac{\omega_m}{|\Omega |}\bigg)^{2/m}(1-E(\Omega))^{-1}.$$

We now introduce the notion of $(\kappa, \delta)$-localisation of the torsion function of a Schrödinger operator on a non-empty open set $\Omega$ with finite Lebesgue measure. We refer to [@grebenkov] for a related notion of localisation for Dirichlet eigenfunctions.

\[def2\] Let $\Omega$ be an open set in ${\mathbb{R}}^m$ with $0 <|\Omega|<\infty$, and let $V:\Omega \to {\mathbb{R}}^+$ be measurable. Let $0 < \kappa<1, 0<\delta < 1.$ We say that $v_{\Omega, V}$ $(\kappa, \delta)$-localises on a measurable subset $A \subset \Omega$ if $$\frac{|A|}{|\Omega|} \le \delta,\qquad
\frac{\int_A v_{\Omega, V}}{\|v_{\Omega, V}\|_1}
\ge \kappa.$$ We say that $v_{\Omega, V}$ $(\kappa, \delta)$-localises if there exists a measurable subset $A \subset \Omega$ so that $v_{\Omega, V}$ $(\kappa, \delta)$-localises on $A$.

In Example \[exa1\] in Section \[sec3\] we analyse a one parameter family of Schrödinger operators in one dimension, where $(\kappa,\delta)$-localisation occurs for certain parameter choices.

The following definition defines localisation of the torsion function for sequences $(\Omega_n,V_n)$, and complements Definition \[def2\] above.

\[def3\] Let $(\Omega_n)$ be a sequence of open sets in ${\mathbb{R}}^m$ with $0 < |\Omega_n|<\infty$, and let $V_n:\Omega_n\to {\mathbb{R}}^+,$ be a sequence of measurable functions. Furthermore, let $$\label{e11}
\mathfrak{A}((\Omega_n))=\bigg\{(A_n): (\forall n\in {\mathbb{N}})(A_n\subset\Omega_n, A_n\, \textup{measurable}), \lim_{n\rightarrow\infty}\frac{|A_n|}{|\Omega_n|}=0\bigg\}$$ and let $$\label{e12}
\kappa=\sup\bigg\{\limsup_{n\rightarrow\infty}\frac{\int_{A_{n}}v_{\Omega_n,V_n}}{\|v_{\Omega_n,V_n}\|_1}:(A_n)\in\mathfrak{A}((\Omega_n))\bigg\}.$$ We say that $(v_{\Omega_n,V_n})$ $\kappa$-[*localises*]{} if $\kappa>0$, $(v_{\Omega_n,V_n})$ [*localises*]{} if holds with $\kappa=1$.

In Example \[exa2\] in Section \[sec4\] we analyse a sequence of open sets in ${\mathbb{R}}^2$ where localisation occurs for certain parameter choices. For one specific parameter choice the sequence of torsion functions is $\kappa$-localising with $0<\kappa<1$.

The theorem below asserts that localisation implies vanishing efficiency.

\[the5\] Let $(\Omega_n)$ be a sequence of open sets in ${\mathbb{R}}^m$ with $0 < |\Omega_n|<\infty$ and let $V_n:\Omega_n\to {\mathbb{R}}^+$ be a sequence of measurable functions. If $(v_{\Omega_n,V_n})$ localises, then $(v_{\Omega_n,V_n})$ has vanishing efficiency, $\lim_{n\rightarrow\infty}\Phi(\Omega_n,V_n)=0.$

It follows from Theorem 1.1 in [@DPGGLB] that, when restricting $\Omega$ to be open, bounded and convex in ${\mathbb{R}}^m,\, m\ge 1$, one has $$\label{e70}
\frac{2}{m(m+2)}\le \inf\{\Phi(\Omega): \Omega\subset {\mathbb{R}}^m,\,\Omega\,\,\,\textup{open,\,convex},\,|\Omega|<\infty\}.$$ We see by that any sequence $(\Omega_n)$ of elongating open, bounded and convex sets in ${\mathbb{R}}^m$ has non-vanishing efficiency, and so by Theorem \[the5\], $(v_{\Omega_n,0})$ is not localising. This is in contrast with the results of [@vdBBDPG], where localisation of the first Dirichlet eigenfunction in $L^2(\Omega_n)$ was obtained for a wide class of elongating, open, bounded, convex sets in ${\mathbb{R}}^m$. Further examples demonstrating different behaviour of the torsion function and the first eigenfunction of the Dirichlet Laplacian around their respective maxima for elongated convex planar domains have been constructed in [@Beck].

Theorem \[the8\] below, asserts that (under suitable hypotheses), vanishing efficiency for the torsion function implies localisation for the first Dirichlet eigenfunction. We make two further definitions.

\[def4\] Let $\Omega \subset {\mathbb{R}}^m$ be an open set with $0 <|\Omega|<\infty$, and let $V:\Omega\rightarrow {\mathbb{R}}^+$ be measurable, such that $\lambda_1(\Omega,V)$ has multiplicity $1$. Let $0 < \kappa<1,\, 0< \delta < 1.$ We say that $\varphi_{1, \Omega,V}$ $(\kappa, \delta)$-localises if there exists a measurable subset $A \subset \Omega$ with $$\frac{|A|}{|\Omega|} < \delta,\qquad
\frac{\int_A \varphi_{1, \Omega,V}}
{\|\varphi_{1, \Omega, V}\|_1}
\ge \kappa.$$

Following Definition \[def3\] we also introduce the notion of localisation for a sequence of first Dirichlet eigenfunctions.

\[def5\] Let $(\Omega_n)$ be a sequence of open sets in ${\mathbb{R}}^m$ with $0 < |\Omega_n|<\infty,\,n\in{\mathbb{N}}$, and let $V_n:\Omega_n\rightarrow{\mathbb{R}}^+,\,n\in {\mathbb{N}}$, be measurable, such that $\lambda_1(\Omega_n,V_n)$ has multiplicity $1$ for all $n\in {\mathbb{N}}$. Let $\mathfrak{A}((\Omega_n))$ be given by , and let $$\label{extra}
\kappa=\sup\bigg\{\limsup_{n\rightarrow\infty}\frac{\int_{A_{n}}\varphi_{1, \Omega_n,V_n}}{
\|\varphi_{1,\Omega_n,V_n}\|_1}:(A_n)\in\mathfrak{A}((\Omega_n))\bigg\}.$$ We say that $(\varphi_{1,\Omega_n,V_n})$ $\kappa$-[*localises*]{} if $\kappa>0$, $(\varphi_{1,\Omega_n,V_n})$ [*localises*]{} if holds with $\kappa=1$.

\[the8\] Let $(\Omega_n)$ be a sequence of open subsets in ${\mathbb{R}}^m$ with $0 < |\Omega_n| < \infty$, and let $(V_n)$ be a sequence of measurable potentials, $V_n:\Omega_n\rightarrow{\mathbb{R}}^+, n\in {\mathbb{N}}$, be such that $\lambda_1(\Omega_n,V_n)$ has multiplicity $1$. If $(v_{\Omega_n,V_n})$ has vanishing efficiency, $\lim_{n \to \infty}\Phi(\Omega_n,V_n) = 0$, then $(\varphi_{1, \Omega_n,V_n})$ localises. If $(v_{\Omega_n,V_n})$ localises, then $(\varphi_{1,\Omega_n,V_n})$ localises.

This paper is organised as follows. The proofs of Theorems \[the1\], \[the2\], \[the3\], \[the5\], and \[the8\] will be given in Section \[sec2\]. Examples \[exa1\] and \[exa2\] will be analysed in Sections \[sec3\] and \[sec4\] respectively. The proofs of Theorems \[the7\] and \[the9\] will be given in Sections \[sec5\] and \[sec6\] respectively.

Proofs of Theorems \[the1\], \[the2\], \[the3\], \[the5\], and \[the8\] \[sec2\]
================================================================================

To prove Theorem \[the1\] and Theorem \[the2\], we first need to establish some auxiliary results. It is well known that the torsion function can be expressed in terms of the heat kernel of $L$. Let $\Omega \subset {\mathbb{R}}^m$ be open with $0 < |\Omega| < \infty$, and let $V:\Omega\rightarrow {\mathbb{R}}^+$ be measurable. Denote by $p_{\Omega,V}(x,y;t)$, $x\in \Omega, y\in \Omega,t > 0$ the heat kernel of $$\frac{\partial u}{\partial t} = Lu,  \qquad u\in H_0^1(\Omega).$$ The torsion function of $L$ then satisfies $$\label{e24}
v_{\Omega, V}(x) = \int_{\Omega} dy \int_{{\mathbb{R}}^+} dt\, p_{\Omega,V}(x,y;t),\qquad \forall x \in \Omega\ .$$ In case $V=0$ we write $p_{\Omega}$ for $p_{\Omega,V}$.

Recall the Feynman-Kac formula (cf. [@BS]) for non-negative, measurable potentials $V:\Omega\rightarrow {\mathbb{R}}^+,$ $$p_{\Omega,V}(x,y;t) = p_{{\mathbb{R}}^m}(x,y;t) {\mathbb{E}}[e^{- \int_0^t V(\beta(s))ds} \big(\Pi_{s\in[0,t]}{\bf1}_{\Omega}(\beta(s))\big) \, : \,  \beta(0)=x, \beta(t) = y ],$$ where $\beta$ is a Brownian bridge. Hence if $V_1: \Omega \to {\mathbb{R}}^+$ and $V_2: \Omega \to {\mathbb{R}}^+$ are measurable functions with $0 \le V_2 \le  V_1$, then

$$\label{e26}
0\le p_{\Omega,V_1}(x,y;t)\le p_{\Omega,V_2}(x,y;t),\qquad \forall x\in \Omega, \forall y\in \Omega, \forall t>0.$$

Since $V\ge 0$ we then conclude that $$0 \le v_{\Omega, V}(x) \le v_{\Omega}(x)\le \mathfrak{c}_m\lambda_1(\Omega)^{-1}, \qquad  \forall x \in \Omega,$$ with $\mathfrak{c}_m$ given by .

\[lem1\] If $0 < |\Omega|<\infty$, and if $V: \Omega \to {\mathbb{R}}^+$ is measurable, then $$\label{e29}
\lambda_{1}(\Omega, V)^{-1} \le
\|v_{\Omega,V}\|_{\infty}
\le (4+3m\log 2)\lambda_{1}(\Omega, V)^{-1}.$$

Lemma \[lem1\] implies that $$\label{e29a}
\mathfrak{d}_m:=\sup\{\lambda_1(\Omega,V)\|v_{\Omega,V}\|_{\infty}:\Omega\,\textup{ open in }\,{\mathbb{R}}^m,\,0<|\Omega|<\infty,\, V\in \mathcal{V}_{\Omega} \}<\infty,$$ where $\mathcal{V}_{\Omega}$ is the set of measurable functions from $\Omega$ into ${\mathbb{R}}^+$. By choosing $V=0$ in the expression under the supremum in the right-hand side of we see that $\mathfrak{c}_m\le \mathfrak{d}_m$.

Throughout this paper we denote by $B(p;R)=\{x\in{\mathbb{R}}^m:|x-p|<R\}$ the open ball centred at $p$ with radius $R$, $B_R=B(0;R)$, and $$\omega_m=|B_1|.$$

[*Proof of Lemma* ]{} To prove the upper bound in we note that Lemma 1 and its proof in [@vdBC] hold with $\lambda=\lambda_1(\Omega,V)$, and Lemma 2 and its proof in [@vdBC] hold for the semigroup associated with $L$. Finally Lemma 3 and Theorem 1 and their proofs in [@vdBC] hold with $\lambda=\lambda_1(\Omega,V)$. This proves the upper bound in . It remains to prove the lower bound. Let $\Omega_R=\Omega\cap B(0;R)$. Then $0<|\Omega_R|\le \omega_mR^m$. Let $L_R$ be the restriction of $-\Delta+V$ acting in $L^2(\Omega_R)$ with Dirichlet boundary conditions on $\partial \Omega_R$. If we denote $V_R={\bf1}_{B(p;R)}V,$ then $L_R$ is also the operator $-\Delta+V_R$ acting in $L^2(\Omega_R)$ with Dirichlet boundary conditions on $\partial \Omega_R$. Then $L_R$ is self-adjoint, and its spectrum is discrete. Since the first Dirichlet eigenfunction $\varphi_{1,\Omega_R,V_R}$ is non-negative, and $\| \varphi_{1,\Omega_R,V_R} \|_2 =1 $, one has, by the Cauchy-Schwarz inequality, $$\label{e31}
0<\int_{\Omega_R}\varphi_{1,\Omega_R,V_R}\le |\Omega_R|^{1/2}.$$ By self-adjointness $$\begin{aligned}
\label{e32}
\int_{\Omega_R}\varphi_{1,\Omega_R,V_R}
&=\int_{\Omega_R}\varphi_{1,\Omega_R,V_R}L_Rv_{\Omega_R,V_R}=
\int_{\Omega_R}\big(L_R\varphi_{1,\Omega_R,V_R}\big)v_{\Omega_R,V_R}\nonumber \\
&=\lambda_1(\Omega_R,V_R)\int_{\Omega_R}\varphi_{1,\Omega_R,V_R}v_{\Omega_R,V_R}\nonumber \\
& \le \lambda_1(\Omega_{R},V_R)\|v_{\Omega_R,V_R}\|_{\infty}\int_{\Omega_R}\varphi_{1,\Omega_R,V_R}.\end{aligned}$$ By – we conclude that $$\lambda_1(\Omega_R,V_R) \, \|v_{\Omega_R,V_R}\|_{\infty} \ge 1.$$ Since $\|v_{\Omega_R,V_R}\|_{\infty}\le \|v_{\Omega,V}\|_{\infty}$ we have that $$\lambda_1(\Omega_R,V_R) \|v_{\Omega,V}\|_{\infty}\ge 1.$$ The assertion follows since $R\mapsto \lambda_1(\Omega_R,V_R)$ is decreasing to $\lambda_1(\Omega,V)$ as $R\rightarrow\infty$. $\square$

\[lem2\] Let $\Omega\subset {\mathbb{R}}^m$ be open with $0 <  |\Omega | < \infty$. If $V : \Omega \rightarrow {\mathbb{R}}^+$ is measurable with $0 \le c_1 \le V(x) \le c_2 < \infty$, $x \in \Omega$, then $$\begin{aligned}
\label{e35}
 v_{\Omega,c_2}(x) \le
 v_{\Omega,V}(x)\le v_{\Omega,c_1}(x)\ ,\end{aligned}$$ and for any $0 <c < \infty$, $x\in \Omega$, $$\begin{aligned}
\label{e36}
 v_{\Omega,c}(x)\ge 2^{-2(3m+4)c/\lambda_1(\Omega)}\frac{\lambda_1(\Omega)}{8c+\lambda_1(\Omega)}\bigg(\frac{8c}{8c+\lambda_1(\Omega)}\bigg)^{8c/\lambda_1(\Omega)}v_{\Omega}(x).\end{aligned}$$ Hence $$\label{e37}
 \|v_{\Omega,c}\|_1\ge 2^{-2(3m+4)c/\lambda_1(\Omega)}\frac{\lambda_1(\Omega)}{8c+\lambda_1(\Omega)}\bigg(\frac{8c}{8c+\lambda_1(\Omega)}\bigg)^{8c/\lambda_1(\Omega)}\|v_{\Omega}\|_1,$$ and $$\label{e38}
 \|v_{\Omega,c}\|_{\infty}\ge 2^{-2(3m+4)c/\lambda_1(\Omega)}\frac{\lambda_1(\Omega)}{8c+\lambda_1(\Omega)}\bigg(\frac{8c}{8c+\lambda_1(\Omega)}\bigg)^{8c/\lambda_1(\Omega)}\|v_{\Omega}\|_{\infty}.$$ Furthermore the right-hand side of converges to $v_{\Omega}(x)$ as $c\downarrow 0$, and $$\label{e45a}
\int_{\Omega}dy\,p_{\Omega}(x,y;t)\le 2^{3m/4}e^{-t\lambda_1(\Omega)/4},\qquad \forall t>0,\, \forall x\in \Omega.$$

The two inequalities in follow immediately from , and the hypothesis $0\le c_1 \le V \le c_2$. To prove inequality , note that for any $T>0$, and any $c\ge 0$, $$\begin{aligned}
\label{e39}
v_{\Omega,c}(x)&=\int_{\Omega}dy\,\int_{{\mathbb{R}}^+}dt\, p_{\Omega,c}(x,y;t)\nonumber \\ &
=\int_{\Omega}dy\,\int_{{\mathbb{R}}^+}dt\, e^{-ct}p_{\Omega}(x,y;t)\nonumber \\ &
\ge \int_{\Omega}dy\,\int_0^Tdt\, e^{-ct}p_{\Omega}(x,y;t)\nonumber \\ &
\ge e^{-cT}\int_{\Omega}dy\,\int_0^Tdt\, p_{\Omega}(x,y;t)\nonumber \\ &
=e^{-cT}\Big(v_{\Omega}(x)-\int_T^{\infty}dt\,\int_{\Omega}dy\,p_{\Omega}(x,y;t)\Big).\end{aligned}$$ We need to estimate the double integral in the right-hand side of . By the heat semigroup property

$$p_{\Omega}(x,y;t)=\int_{\Omega}dz\,p_{\Omega}(x,z;t/2)p_{\Omega}(z,y;t/2)$$ we have $$\begin{aligned}
\label{e40}
\int_{\Omega}dy\, & p_{\Omega}(x,y;t) =\int_{\Omega}dy\,\int_{\Omega}dz\,p_{\Omega}(x,z;t/2)p_{\Omega}(z,y;t/2)\nonumber \\ &
= \int_{\Omega}dz\,p_{\Omega}(x,z;t/2)\int_{\Omega}dy\,\big(p_{\Omega}(z,y;t/2)\big)^{1/2}\big(p_{\Omega}(z,y;t/2)\big)^{1/2}.\end{aligned}$$ We estimate $p_{\Omega}(z,y;t/2)$ in terms of $\lambda_1(\Omega)$. Recall that $p_{\Omega}(x,y;t)$ can be expressed in terms of the eigenfunctions $\varphi_{j, \Omega}(x)$ by $$\label{e41}
p_{\Omega}(x,y;t) = \sum_{j =1}^{\infty} e^{- t\lambda_j(\Omega)} \varphi_{j, \Omega}(x)  \varphi_{j, \Omega}(y).$$ Moreover for any $\beta\in [0,1)$, $$\label{beta}
p_{\Omega}(x,x;t)\le e^{-\beta t\lambda_1(\Omega)}p_{\Omega}(x,x;(1-\beta)t).$$ The heat kernel for ${\mathbb{R}}^m$ is given by $$\label{e42}
p_{{\mathbb{R}}^m}(x,y;t)=(4\pi t)^{-m/2}e^{-|x-y|^2/(4t)},$$ and satisfies $$\label{e43}
\int_{{\mathbb{R}}^m}dy\,p_{{\mathbb{R}}^m}(x,y;t)=1,\qquad  \forall x\in {\mathbb{R}}^m,\, \forall t>0.$$ By the heat semigroup property, the Cauchy-Schwarz inequality, for $\beta=\frac12$, domain monotonicity of the Dirichlet heat kernel, and we have $$\begin{aligned}
\label{e44}
p_{\Omega}(z,y;t/2)&=\int_{\Omega}dw\,p_{\Omega}(z,w;t/4)p_{\Omega}(w,y;t/4)\nonumber \\ &
\le \bigg(\int_{\Omega}dw\,\big(p_{\Omega}(z,w;t/4)\big)^2\bigg)^{1/2}\bigg(\int_{\Omega}dw\,\big(p_{\Omega}(w,y;t/4)\big)^2\bigg)^{1/2}\nonumber \\ &
=\big(p_{\Omega}(z,z;t/2)\big)^{1/2}\big(p_{\Omega}(y,y;t/2)\big)^{1/2}\nonumber \\ &
\le e^{-t\lambda_1(\Omega)/4}\big(p_{\Omega}(z,z;t/4)\big)^{1/2}\big(p_{\Omega}(y,y;t/4)\big)^{1/2}\nonumber \\ &
\le e^{-t\lambda_1(\Omega)/4}\big(p_{{\mathbb{R}}^m}(z,z;t/4)\big)^{1/2}\big(p_{{\mathbb{R}}^m}(y,y;t/4)\big)^{1/2}\nonumber \\ &
=e^{-t\lambda_1(\Omega)/4}(\pi t)^{-m/2}.\end{aligned}$$ Substituting into we find $$\begin{aligned}
\label{e45}
\int_{\Omega}dy&\,p_{\Omega}(x,y;t)\le e^{-t\lambda_1(\Omega)/8}(\pi t)^{-m/4}\int_{\Omega}dz\,p_{\Omega}(x,z;t/2)\int_{\Omega}dy\,\big(p_{\Omega}(z,y;t/2)\big)^{1/2}\nonumber \\ &
\le e^{-t\lambda_1(\Omega)/8}(\pi t)^{-m/4}\int_{\Omega}dz\,p_{\Omega}(x,z;t/2)\int_{{\mathbb{R}}^m}dy\,\big(p_{{\mathbb{R}}^m}(z,y;t/2)\big)^{1/2}\nonumber \\ &
=2^{3m/4}e^{-t\lambda_1(\Omega)/8}\int_{\Omega}dz\,p_{\Omega}(x,z;t/2).\end{aligned}$$ Integrating both sides of with respect to $t$ over the set $[T,\infty)$ gives $$\begin{aligned}
\label{e46}
\int_T^{\infty}dt\,\int_{\Omega}dy\,p_{\Omega}(x,y;t)&\le 2^{3m/4}e^{-T\lambda_1(\Omega)/8}\int_T^{\infty}dt\,\int_{\Omega}dz\,p_{\Omega}(x,z;t/2)\nonumber \\ &
\le 2^{3m/4}e^{-T\lambda_1(\Omega)/8}\int_{{\mathbb{R}}^+}dt\,\int_{\Omega}dz\,p_{\Omega}(x,z;t/2)\nonumber \\ &
=2^{(4+3m)/4}e^{-T\lambda_1(\Omega)/8}v_{\Omega}(x).\end{aligned}$$ By combining and we find $$\label{e47}
v_{\Omega,c}(x)\ge e^{-cT}\big(1-2^{(4+3m)/4}e^{-T\lambda_1(\Omega)/8}\big)v_{\Omega}(x).$$ Choosing $T$ as to maximise the right-hand side of gives that $$\label{e48}
T=\frac{8}{\lambda_1(\Omega)}\log\Big(2^{(4+3m)/4}\Big(1+\frac{\lambda_1(\Omega)}{8c}\Big)\Big).$$ Inequality then follows by and . The inequalities and follow immediately from .

To prove we have by , $$\label{e45b}
p_{\Omega}(x,y;t)\le e^{-t\lambda_1(\Omega)/2}(2\pi t)^{-m/2}.$$ By the domain monotonicity, the non-negativity of the Dirichlet heat kernel, and , $$\begin{aligned}
\int_{\Omega}dy\,p_{\Omega}(x,y;t)&\le e^{-t\lambda_1(\Omega)/4}(2\pi t)^{-m/4}\int_{\Omega}dy\,\big(p_{\Omega}(x,y;t)\big)^{1/2}\nonumber \\ &
\le e^{-t\lambda_1(\Omega)/4}(2\pi t)^{-m/4}\int_{{\mathbb{R}}^m}dy\,\big(p_{{\mathbb{R}}^m}(x,y;t)\big)^{1/2}\nonumber \\ &
=2^{3m/4}e^{-t\lambda_1(\Omega)/4},\end{aligned}$$ which proves .

[*Proof of Theorem* ]{} Since $0\le V\le c$ we have by , $$\label{e56}
\| v_{\Omega,V} \|_1  \le \| v_{\Omega} \|_1,$$ and $$\label{e57}
\|v_{\Omega,V}\|_{\infty}\ge \|v_{\Omega,c}\|_{\infty}.$$ The upper bound follows from , , and .

Similarly, by , we have $$\label{e58}
\|v_{\Omega,V}\|_{\infty}\le \|v_{\Omega}\|_{\infty}.$$ and $$\begin{aligned}
\label{e59}
\|v_{\Omega,V}\|_1\ge\|v_{\Omega,c}\|_1.\end{aligned}$$ The lower bound follows from , , and . $\square$

[*Proof of Theorem* ]{} By using the inequality $$\big(1+\theta^{-1}\big)^{\theta}\le e, \qquad \forall\theta>0,$$ with $\theta=8\Vert V_n\Vert_{\infty}/{\lambda_1(\Omega_n)}$, we obtain by , $$\Phi(\Omega_n,V_n)\le e2^{2(3m+4)\eta}(1+8\eta)\Phi(\Omega_n),$$ and by , $$\Phi(\Omega_n,V_n)\ge e^{-1}2^{-2(3m+4)\eta}(1+8\eta)^{-1}\Phi(\Omega_n).$$ This proves Theorem \[the2\]. $\square$

[*Proof of Theorem .*]{} To verify we have by Lemma 4 in [@MvdBS], $$\int_{\Omega}dy\, p_{\Omega}(x,y;t)\ge 1-2^{(m+2)/2}e^{-d_{\Omega}(x)^2/(8t)}.$$ Hence we find by that $$\label{e53}
\int_{\Omega}dx\,\int_{\Omega}dy\, p_{\Omega}(x,y;t)\ge |\Omega|-2^{(m+2)/2}\int_{\Omega}dx\,e^{-d_{\Omega}(x)^2/(8t)}.$$ Multiplying both sides of by $e^{-ct}$ and integrating with respect to $t$ gives that $$\begin{aligned}
\label{e54}
\| v_{\Omega,c} \|_1&\ge \frac{|\Omega|}{c}-2^{(m+2)/2}\int_{\Omega}dx\,\int_{{\mathbb{R}}^+}dt\, e^{-ct-d_{\Omega}(x)^2/(8t)}\nonumber \\ &
\ge \frac{|\Omega|}{c}-2^{(m+2)/2}\int_{\Omega}dx\,\int_{{\mathbb{R}}^+}dt\, e^{-ct/2}\sup\{e^{-ct/2-d_{\Omega}(x)^2/(8t)}:t>0\}\nonumber \\ &
=\frac{|\Omega|}{c}-\frac{2^{(m+4)/2}}{c}\int_{\Omega}dx\,e^{-c^{1/2}d_{\Omega}(x)/2}.\end{aligned}$$ On the other hand using , $$\begin{aligned}
\label{e55}
\| v_{\Omega,c} \|_\infty &=\sup_{x\in\Omega}\int_{{\mathbb{R}}^+}dt\, e^{-ct}\int_{\Omega}dy\, p_{\Omega}(x,y;t)\nonumber \\ &
\le\sup_{x\in\Omega}\int_{{\mathbb{R}}^+}dt\, e^{-ct}\int_{{\mathbb{R}}^m}dy\, p_{{\mathbb{R}}^m}(x,y;t)\nonumber \\ &
=\frac{1}{c}.\end{aligned}$$ The second inequality in follows from and . Finally, follows from , and Lebesgue’s dominated convergence theorem, while follows from of Theorem \[the1\]. $\square$

[*Proof of Theorem* ]{} Let $\frac12<\kappa<1$ be arbitrary, and let $(A_{n})\in\mathfrak{A}((\Omega)_n)$ be such that for all $n$ sufficiently large, $\Vert v_{\Omega_n,V_n}\Vert_1\le (2\kappa-1)^{-1}\int_{A_{n}} v_{\Omega_n,V_n}$, and $|A_n|/|\Omega_n|<1-\kappa$. Then, for all such $n$, $$\begin{aligned}
\Vert v_{\Omega_n,V_n}\Vert_1&\le (2\kappa-1)^{-1}\int_{A_{n}}v_{\Omega_n,V_n}\nonumber \\ &\le (2\kappa-1)^{-1}\Vert v_{\Omega_n,V_n}\Vert_{\infty}|A_{n}|\nonumber \\ &\le \frac{1-\kappa}{2\kappa-1}\Vert v_{\Omega_n,V_n}\Vert_{\infty}|\Omega_n|.\end{aligned}$$ Hence for all $n$ sufficiently large, $\Phi(\Omega_n,V_n)\le \frac{1-\kappa}{2\kappa-1}$. Since $\frac12<\kappa<1$ was arbitrary, $\lim_{n\rightarrow\infty}\Phi(\Omega_n,V_n)=0.$ $\square$

Theorem \[the5\] implies that if $(\Omega_n,V_n)$ satisfies the $\eta$ condition of Theorem \[the2\], and if either $(v_{\Omega_n})$ or $(v_{\Omega_n,V_n})$ have non-vanishing efficiencies then both $(v_{\Omega_n})$ and $(v_{\Omega_n,V_n})$ are not localising.

[*Proof of Theorem* ]{} Using and we obtain $$v_{\Omega,V}(x)=\sum_{j =1}^{\infty} \lambda_j(\Omega,V)^{-1} \bigg(\int_{\Omega}\varphi_{j, \Omega,V}\bigg)\varphi_{j, \Omega,V}(x).$$ Integrating with respect to $x$ over $\Omega$ gives $$\begin{aligned}
\int_{\Omega}v_{\Omega,V}&=\sum_{j =1}^{\infty} \lambda_j(\Omega,V)^{-1} \bigg(\int_{\Omega}\varphi_{j, \Omega,V}\bigg)^2\nonumber \\ &
\ge \lambda_1(\Omega,V)^{-1} \bigg(\int_{\Omega}\varphi_{1, \Omega,V}\bigg)^2.\end{aligned}$$ Multiplying both sides with $\lambda_1(\Omega,V)$, and using the definition of $\mathfrak{d}_m$ in gives $$\mathfrak{d}_m\frac{\|v_{\Omega,V}\|_1}{\|v_{\Omega,V}\|_{\infty}}\ge \bigg(\int_{\Omega}\varphi_{1, \Omega,V}\bigg)^2.$$ This implies that $$\mathfrak{d}_m\Phi(\Omega,V)\ge \frac{1}{|\Omega|}\bigg(\int_{\Omega}\varphi_{1, \Omega,V}\bigg)^2.$$ If $\lim_{n\rightarrow\infty}\Phi(\Omega_n,V_n)=0$, then it follows that $\lim_{n\rightarrow\infty}\frac{1}{|\Omega_n|}\big(\int_{\Omega_n}\varphi_{1, \Omega_n,V_n}\big)^2=0$. This implies localisation by Lemma 3 in [@vdBBDPG]. Since by Theorem \[the5\] $(v_{\Omega_n,V_n})$ has vanishing efficiency if it localises, the last assertion in Theorem \[the8\] follows. $\square$

 Example \[exa1\] {#sec3}
=================

In this section we analyse a one parameter family of Schrödinger operators in one dimension, where $(\kappa,\delta)$-localisation for certain parameter choices occurs.

\[exa1\] If $\Omega=(-1, 1)$, and $$V_\nu(x) = \nu^4 1_{(-1, -\frac{1}{\nu})}(x) +
\nu^4 1_{(\frac{1}{\nu}, 1)}(x)
\, , \quad x \in (-1, 1)\, ,$$ where $\nu>1$ is a real parameter, then the torsion function $v_\nu := v_{\Omega, V_\nu}$ satisfies $$\label{fundamental estimate}
\frac{\int_{1/\nu}^1 v_\nu}{\int_{0}^{1/\nu} v_\nu}
\le \frac{6}{\nu}\, .$$ As a consequence, for any $0 < \kappa < 1,$ $0 < \delta < 1$ there exists $\nu_{\kappa, \delta} >1$ so that for any $\nu \ge \nu_{\kappa, \delta}$, $A_\nu := [-\frac{1}{\nu}, \frac{1}{\nu}]$ satisfies, $$\frac{|A_\nu|}{|\Omega|} \le  \delta\, , \qquad
\frac{\int_{A_\nu} v_\nu}{\| v_\nu \|_1} \ge \kappa.$$

Since the potential $V_\nu$ is even, so is the torsion function $v_\nu$ and hence it suffices to determine $v_\nu$ on $[0, 1]$. On the interval $[ 0,\frac{1}{\nu}]$, the function $v_\nu$ is of the form $$v_{\nu, 1}(x) := - \frac{1}{2}x^2 + \gamma_\nu\, ,
\quad 0 \le x \le \frac{1}{\nu}\, ,$$ whereas on the interval $[\frac{1}{\nu}, 1]$, we make the following Ansatz $$v_{\nu, 2}(x) := \frac{1}{\nu^4}
- \alpha_\nu e^{\nu^2 x}
+ \beta_\nu e^{-\nu^2 x}\, ,
\quad \frac{1}{\nu}  \le x \le 1\, .$$ It is straightforward to verify that $(-\Delta + V_{\nu}) v_{\nu, 1} = 1$ on $[0, \frac{1}{\nu}]$ and $(-\Delta + V_{\nu}) v_{\nu, 2} = 1$ on $[\frac{1}{\nu}, 1]$. The constants $\gamma_\nu,$ $\alpha_\nu$, and $\beta_\nu$ are determined by the boundary condition $v_{\nu, 2}(1) = 0$ and the matching conditions $$v_{\nu, 1}(\nu^{-1}) = v_{\nu, 2}(\nu^{-1})\, ,\quad
v_{\nu, 1}'(\nu^{-1}) = v_{\nu, 2}'(\nu^{-1})\, ,$$ where $'$ denotes the derivative with respect to the variable $x$. The boundary condition $v_{\nu, 2}(1) = 0$, or explicitly, $$\frac{1}{\nu^4} - \alpha_\nu e^{\nu^2} + \beta_\nu e^{-\nu^2} =0\, ,$$ yields $$\label{first formula beta nu}
\beta_\nu = \alpha_\nu e^{2\nu^2} -
\frac{1}{\nu^4} e^{\nu^2}\,$$ and the matching condition $v_{\nu, 1}'(\frac{1}{\nu}) =
v_{\nu, 2}'(\frac{1}{\nu})$ reads $$\frac{1}{\nu} =
\nu^2 \alpha_\nu e^{\nu} + \nu^2 \beta_\nu e^{-\nu} \, ,$$ leading to $$\alpha_\nu = \frac{1}{\nu^4 e^{\nu^2}}
\frac{1 + \nu e^{-\nu^2 +\nu}}{1 + e^{-2\nu^2 + 2\nu}}\,.$$ Substituting this value for $\alpha_\nu$ into the formula for $\beta_{\nu}$ one obtains $$\beta_\nu =
\frac{e^\nu}{\nu^3}
\frac{1 - \frac{1}{\nu} e^{-\nu^2 +\nu}}{1 + e^{-2\nu^2 + 2\nu}}\,.$$ It remains to determine $\gamma_\nu$ from the matching condition $v_{\nu, 1}(\frac{1}{\nu}) = v_{\nu, 2}(\frac{1}{\nu})$. One has $$- \frac{1}{2 \nu^2} + \gamma_\nu =
\frac{1}{\nu^4}  - \alpha_\nu e^{\nu}
+ \beta_\nu e^{-\nu}\, ,$$ which gives the following formula for $\gamma_\nu$ $$\gamma_\nu = \frac{1}{2 \nu^2} + \frac{1}{\nu^4}
+ \frac{1}{\nu^3} \frac{1 - \frac{1}{\nu} e^{-\nu^2 +\nu}}
{1 + e^{-2\nu^2 + 2\nu}}
- \frac{1}{\nu^4 e^{\nu^2 - \nu}}
\frac{1 + \nu e^{-\nu^2 +\nu}}{1 + e^{-2\nu^2 + 2\nu}} \, .$$ Now let us turn to the estimate of the quotient, $\frac{\int_{1/\nu}^1 v_\nu}{\int_{0}^{1/\nu} v_\nu}$. We estimate the two integrals separately. First we derive a lower bound for $\int_{0}^{1/\nu} v_\nu$. To this end note that $$\int_{0}^{1/\nu} v_\nu =
\bigg( -\frac{1}{6}x^3 + \gamma_\nu x \bigg)\bigg|_0^{1/\nu}
= \frac{\gamma_\nu}{\nu} - \frac{1}{6\nu^3}$$ yielding $$\int_{0}^{1/\nu} v_\nu =
\frac{1}{3 \nu^3}
+ \frac{1}{\nu^4} \frac{1 - \frac{1}{\nu} e^{-\nu^2 +\nu}}
{1 + e^{-2\nu^2 + 2\nu}}
+ \frac{1}{\nu^5}\Big( 1 -  \frac{1}{ e^{\nu^2 - \nu}}
\frac{1 + \nu e^{-\nu^2 +\nu}}{1 + e^{-2\nu^2 + 2\nu}} \Big).$$ Note that for any $\nu > 1$, one has $1 - \frac{1}{\nu} e^{-\nu^2 +\nu} > 0,$ and $$\frac{1}{ e^{\nu^2 - \nu}}
\frac{1 + \nu e^{-\nu^2 +\nu}}{1 + e^{-2\nu^2 + 2\nu}}
\le 1\, ,$$ implying that $$\label{estimate integral in denominator}
\int_{0}^{1/\nu} v_\nu  \ge \frac{1}{3\nu^3} \, , \quad
\forall \nu > 1\, .$$ Next we derive an upper bound for $\int_{1/\nu}^1 v_\nu$. One computes $$\begin{aligned}
\label{t1}
\int_{1/\nu}^1 v_\nu &=
\frac{1}{\nu^4} (1 - \frac{1}{\nu})
- \frac{\alpha_\nu}{\nu^2} e^{\nu^2 x}|_{1/\nu}^1
- \frac{\beta_\nu}{\nu^2} e^{-\nu^2 x}|_{1/\nu}^1 \nonumber \\
& \le
\frac{1}{\nu^4}
+ \frac{\alpha_\nu}{\nu^2} e^{\nu}
+ \frac{\beta_\nu}{\nu^2} e^{-\nu} \nonumber \\
& \le \frac{1}{\nu^4}  ( 1+ R_\nu ),\,\end{aligned}$$ where the remainder $R_\nu$ is given, and bounded, by $$\begin{aligned}
\label{t2}
R_\nu &= \frac{1}{\nu^2 e^{\nu^2 - \nu}}
\frac{1 + \nu e^{-\nu^2}}{1 + e^{-2\nu^2 + 2\nu}}
+ \frac{1}{\nu}
\frac{1 - \frac{1}{\nu} e^{-\nu^2 +\nu}}{1 + e^{-2\nu^2 + 2\nu}}\nonumber \\ &
 =\frac{1+e^{-2\nu^2 + \nu}}{\nu\big(1+e^{-2\nu^2 + 2\nu}\big)}\nonumber \\ &\le \frac{1}{\nu},\qquad \forall\nu>1.\end{aligned}$$ By and , $$\label{estimate integral in nominator}
\int_{1/\nu}^1 v_\nu \le \frac{1}{\nu^4}+\frac{1}{\nu^5},\qquad \forall\nu\ge 1.$$ Combining and we obtain the claimed estimate , $$\frac{\int_{1/\nu}^1 v_\nu}{\int_{0}^{1/\nu} v_\nu}
\le \frac{3}{\nu}+\frac{3}{\nu^2}\le \frac{6}{\nu},\, \forall\nu\ge 1.$$ The remaining statements of the theorem now follow in a straightforward way from since $$\frac{\int_{A_\nu} v_\nu}{\| v_\nu \|_1} = 1 - \frac{\int_{1/\nu}^1 v_\nu}
{\int_{0}^{1/\nu} v_\nu + \int_{1/\nu}^1 v_\nu}\ge 1-\frac{\int_{1/\nu}^1 v_\nu}{\int_{0}^{1/\nu} v_\nu}.$$

 Example \[exa2\] {#sec4}
=================

In this section we analyse localisation for a sequence of open, bounded sets $(\Omega_{\alpha,n}), n\in{\mathbb{N}}$ in ${\mathbb{R}}^2$, where $\Omega_{\alpha,n}$ is the disjoint union of $n$ rectangles of size $n^{-1}\times 1$, and $1$ rectangle of size $cn^{-\alpha}\times 1$. The constant $c$ is strictly positive, and $0<\alpha<1$.

\[exa2\] Let $$N_{\alpha,c}=\min\{n\in{\mathbb{N}}:n^{\alpha}>c,\, n^{1-\alpha}>c^{-1}\}.$$

-   If $n\ge N_{\alpha,c}$, then $$\label{e16}
    \frac{8}{45c^2(1+c)}\big(n^{2\alpha-2}+c^3n^{-\alpha}\big)\le\Phi(\Omega_{\alpha,n})\le \frac{4}{3c^2}\big(n^{2\alpha-2}+c^3n^{-\alpha}\big).$$

-   If $\frac23<\alpha<1$, then $(v_{\Omega_{\alpha,n}})$ is not localising, $\kappa=0.$

-   If $0<\alpha<\frac23$, then $(v_{\Omega_{\alpha,n}})$ is localising, $\kappa=1.$

-   If $\alpha=\frac23$, then $(v_{\Omega_{\alpha,n}})$ is $\kappa_c$-localising with $$\label{e13}
    \kappa_c=\frac{c^3}{1+c^3},$$ and the supremum defining $\kappa_c$ in is a maximum.

First we establish the following estimate for the torsion function of a rectangle in ${\mathbb{R}}^2$.

Let $R_{a,b}=(-a/2,a/2)\times(-b/2,b/2)$ be a rectangle with side lengths $a$ and $b$ respectively, and let $x=(x_1,x_2)\in R_{a,b}$. $$\label{e49a}
v_{R_{a,b}}(x_1,x_2)\le v_{(-a/2,a/2)}(x_1)=\frac{a^2-4x_1^2}{8},\qquad \forall (x_1,x_2)\in R_{a,b}.$$

By for $V=0$, separation of variables, and monotonicity and non-negativity of Dirichlet heat kernels we have $$\begin{aligned}
\label{e49b}
v_{R_{a,b}}(x_1,x_2)&=\int_0^{\infty}dt\,\int_{(-a/2,a/2)}dy_1\,p_{(-a/2,a/2)}(x_1,y_1;t)\nonumber \\ &
\hspace{20mm}\times\int_{(-b/2,b/2)}dy_2\,p_{(-b/2,b/2)}(x_2,y_2;t)\nonumber \\ &
\le\int_0^{\infty}dt\,\int_{(-a/2,a/2)}dy_1\,p_{(-a/2,a/2)}(x_1,y_1;t)\int_{{\mathbb{R}}}dy_2\,p_{{\mathbb{R}}}(x_2,y_2;t)\nonumber \\ &
=\int_0^{\infty}dt\,\int_{(-a/2,a/2)}dy_1\,p_{(-a/2,a/2)}(x_1,y_1;t)\nonumber \\ &
=v_{(-a/2,a/2)}(x_1)\nonumber \\&
=\frac{a^2-4x_1^2}{8},\end{aligned}$$ where we have used formula , $p_{{\mathbb{R}}}(x_2,y_2;t)=(4\pi t)^{-1/2}e^{-|x_2-y_2|^2/(4t)}$, identity , and the fact that the last equality in is the one-dimensional version of the torsion function for a ball $B_a\subset {\mathbb{R}}^m$ with radius $a$, centred at $0$. One verifies that $$\label{e49c}
v_{B_a}(x)=\frac{a^2-|x|^2}{2m},$$ satisfies for $\Omega=B_a$, [@PSZ].

[*Proof of Example* ]{} (i) We have $$\label{e49}
1\le |\Omega_{\alpha,n}|\le 1+c.$$ By , $$\|v_{\Omega_{\alpha,n}}\|_{\infty}\le \frac18\max\bigg\{\frac{1}{n^2},\frac{c^2}{n^{2\alpha}}\bigg\}.$$ Since $n\ge N_{\alpha,c}$ we obtain $$\label{e70a}
\|v_{\Omega_{\alpha,n}}\|_{\infty}\le \frac{c^2}{8n^{2\alpha}}.$$ On the other hand since $n\ge N_{\alpha,c}$, $\Omega_{\alpha,n}$ contains a ball with radius $\frac12 cn^{-\alpha}$. The maximum of the torsion function is, by domain monotonicity, bounded from below by the maximum for that ball. Since by with $m=2$, $\|v_{B_a}\|_{\infty}=\frac{a^2}{4}$ we have $$\label{e71}
\|v_{\Omega_{\alpha,n}}\|_{\infty}\ge \frac{c^2}{16n^{2\alpha}}.$$ It was shown in Proposition 3.1 in [@MvdBBV] that if $a\le b$, then $$\label{e72}
\frac{a^3b}{12}-\frac{11a^4}{180}\le \|v_{R_{a,b}}\|_1\le \frac{a^3b}{12}.$$ Since all rectangles in $\Omega_{\alpha,n}$ are disjoint $$\label{f1}
\|v_{\Omega_{\alpha,n}}\|_1\le \frac{1}{12n^2}+\frac{c^3n^{-3\alpha}}{12},$$ and $$\begin{aligned}
\label{f2}
\|v_{\Omega_{\alpha,n}}\|_1&\ge \frac{1}{12n^2}+\frac{c^3n^{-3\alpha}}{12}-\frac{11}{180n^3}-\frac{11c^4n^{-4\alpha}}{180}\nonumber \\ &
\ge \frac{1}{45}\big(n^{-2}+ c^3n^{-3\alpha}\big),\qquad n\ge N_{\alpha,c}.\end{aligned}$$ The upper bound in follows from the lower bound in , , and the upper bound in . The lower bound in follows the upper bound in , , and the lower bound in .

\(ii) Let $A_n$ be a measurable subset of $\Omega_{\alpha,n}$. Denote by $S_{\alpha,n}$ the rectangle of size $cn^{-\alpha}\times 1$. We have $$\begin{aligned}
\label{f3}
\int_{A_n}v_{\Omega_{\alpha,n}}&=\int_{A_n\cap S_{\alpha,n}}v_{\Omega_{\alpha,n}}+\int_{A_n\setminus S_{\alpha,n}}v_{\Omega_{\alpha,n}}\nonumber \\ &
\le\frac{c^3n^{-3\alpha}}{12}+|A_n|\max\{v_{\Omega_{\alpha,n}}:x\in \Omega_{\alpha,n}\setminus S_{\alpha,n}\}\nonumber \\ &
=\frac{c^3n^{-3\alpha}}{12}+\frac{|A_n|}{8n^2}.\end{aligned}$$ By , $\|v_{\Omega_{\alpha,n}}\|_1\ge \frac{1}{45n^{2}}$ so that $$\label{f4}
\frac{\int_{A_n}v_{\Omega_{\alpha,n}}}{\|v_{\Omega_{\alpha,n}}\|_1}\le\frac{15c^3n^{2-3\alpha}}{4}+\frac{45|A_n|}{8}.$$ Since $\alpha>\frac23$ we have that the right-hand side of tends to $0$ as $n\rightarrow\infty$ by the hypothesis on $|A_n|$. That hypothesis also shows that $$\lim_{n\rightarrow\infty}\frac{|A_n|}{|\Omega_{\alpha,n}|} \le \lim_{n\rightarrow\infty}|A_n|=0.$$ This proves that $\kappa=0$.

\(iii) We will show that for $n\rightarrow\infty$ all of the torsion mass localises at $S_{\alpha,n}$. This is equivalent to showing that $$\lim_{n\rightarrow\infty}\frac{\int_{\Omega_{\alpha,n}\setminus S_{\alpha,n}}v_{\Omega_{\alpha,n}}}{\int_{\Omega_{\alpha,n}}v_{\Omega_{\alpha,n}}}=0.$$ This follows from $$\int_{\Omega_{\alpha,n}\setminus S_{\alpha,n}}v_{\Omega_{\alpha,n}}\le \frac{1}{12n^2},$$ the estimate , $$\int_{\Omega_{\alpha,n}}v_{\Omega_{\alpha,n}}\ge \frac{c^3n^{-3\alpha}}{45},$$ and the hypotheses on $\alpha$ and on $n$. This proves assertion (iii).

\(iv) To obtain a lower bound for $\kappa$ in with $V=0,\, \Omega_n=\Omega_{\frac23,n}$ we choose $A_n=S_{\frac23,n}$. Then $ \lim_{n\rightarrow\infty} | S_{\frac23,n} | = 0$, and it follows by that $$\lim_{n\rightarrow\infty}n^2\int_{S_{\frac23,n}}v_{S_{\frac23,n}}=\frac{c^3}{12},$$ and $$\lim_{n\rightarrow\infty}n^2\int_{\Omega_{\frac23,n}\setminus S_{\frac23,n}}v_{\Omega_{\frac23,n}}=\lim_{n\rightarrow\infty}n^2\int_{\Omega_{\frac23,n}\setminus S_{\frac23,n}}v_{\Omega_{\frac23,n}\setminus S_{\frac23,n}}=\frac{1}{12}.$$ This implies $$\label{h8}
 \kappa_c\ge \frac{c^3}{1+c^3}.$$ To prove the reverse inequality we let $(A_n)_n$ be an arbitrary sequence of measurable sets with $A_n\subset \Omega_{\frac23,n},\,n\in{\mathbb{N}}$ and $\lim_{n\rightarrow\infty}|A_n|=0$. Then, by , $$\label{h6}
\int_{A_n}v_{\Omega_{\frac23,n}}\le
\frac{1}{n^2}\bigg(\frac{c^3}{12}+\frac{|A_n|}{8}\bigg).$$ By the first inequality in , $$\label{h7}
\|v_{\Omega_{\frac23,n}}\|_1\ge \frac{n^{-2}}{12}+\frac{c^3n^{-2}}{12}-\frac{11n^{-3}}{180}-\frac{11c^4n^{-8/3}}{180}.$$ By and , $$\label{h9}
\limsup_{n\rightarrow\infty}\frac{\int_{A_n}v_{\Omega_{\frac23,n}}}{\|v_{\Omega_{\frac23,n}}\|_1}\le \frac{c^3}{1+c^3}.$$ Taking the supremum in the left-hand side of over all sequences $(A_n)$ in yields, $$ \kappa_c\le \frac{c^3}{1+c^3}.$$ This, together with gives . Furthermore, the supremum in is a maximum at $A_n=S_{\frac23,n}$. This maximiser is unique modulo sequences of measurable sets $B_n \subset A_n,\,n\in {\mathbb{N}}$ with measure $|B_n|=0,\,n\in {\mathbb{N}}$, and sequences of measurable sets $C_n \subset \Omega_{\frac23,n}\setminus A_n,\,n\in {\mathbb{N}}$, with $\lim_{n\rightarrow 0}|C_n|=0$. This completes the proof of assertion (iv). $\square$

Proof of Theorem \[the7\]  {#sec5}
==========================

[*Proof of Theorem .*]{} The proof follows a method from [@HLP], which can be summarised as follows: construct a capacitary measure $\mu$ so that the efficiency of the eigenfunction of the first eigenvalue of $- \Delta + \mu$ almost equals $1$ and then approximate the capacitary measure in the sense of $\gamma$-convergence by a sequence of domains. [ Such an]{} approximation is possible in view of the Dal Maso-Mosco density result. (It could possibly be done locally using a Cioranescu-Murat periodic structure.) We refer the reader to [@BB05 Chapter 4] for the notions of $\gamma$-convergence, capacitary measures and approximation by sequences of domains.

Let ${\overline}B_r$ denote the closure of $B_r$. For any $0 < \varepsilon < 1$, we consider the Dirichlet-Neumann eigenvalue problem on the annulus $A_{\varepsilon}=B_1{\setminus}{\overline}B_{1-{\varepsilon}}$. Denote by $\lambda_{\varepsilon}$ the first eigenvalue and by $u_{\varepsilon}$ a corresponding eigenfunction,

$$\begin{cases}
-\Delta u_{\varepsilon}=\lambda_{\varepsilon}u_{\varepsilon},&\text{in }A_{\varepsilon},\\
{ u_{\varepsilon}}=0,&\text{on }\partial B_{1},\\
\frac{\partial { u_{\varepsilon}}}{\partial \nu}= 0,&\text{on }\partial B_{1-{\varepsilon}}.
\end{cases}$$

where $\nu$ denotes the inward-pointing normal on the sphere $\partial B_{1-{\varepsilon}}$. One can show that ${\lambda}_{\varepsilon}$ is simple and strictly positive and that $u_{\varepsilon}$ is radially symmetric and has a constant sign, say positive. In particular, the restriction of $u_{\varepsilon}$ to $\partial B_{1-{\varepsilon}}$ equals a positive constant, $c_{\varepsilon}> 0$. We continuously extend $u_{\varepsilon}$ inside $B_{1-{\varepsilon}}$ by $c_{\varepsilon}$ and denote the resulting function, defined on ${\overline}B_1$, by $v_{\varepsilon}$.

Since the normal derivatives of $v_{\varepsilon}$ on both sides of $\partial B_{1-{\varepsilon}}$ vanish, $\Delta  v_{\varepsilon}$ is an $L^2$-function. More precisely, one has $$-\Delta v_{\varepsilon}= {\lambda}_{\varepsilon}u_{\varepsilon}1_{A_{\varepsilon}}, \quad \mbox  {in } {\mathcal D}'(B_1),$$ where ${\mathcal D}'(B_1)$ denotes the space of distributions on $B_1$. By adding on both sides the $L^2$ function $ {\lambda}_{\varepsilon}v_{\varepsilon}1_{{\overline}B_{1-{\varepsilon}}}$ we get\
$$-\Delta v_{\varepsilon}+  {\lambda}_{\varepsilon}v_{\varepsilon}1_{{\overline}B_{1-{\varepsilon}}}= {\lambda}_{\varepsilon}v_{\varepsilon},\quad \mbox  {in } {\mathcal D}'(B_1).$$ We view $\mu :=  {\lambda}_{\varepsilon}1_{{\overline}B_{1-{\varepsilon}}}$ as a capacitary measure on $B_1$. Then formally, we get $$-\Delta v_{\varepsilon}+ \mu  v_{\varepsilon}= {\lambda}_{\varepsilon}v_{\varepsilon},\quad \mbox  {in } {\mathcal D}'(B_1).$$ Combined with the fact that $v_{\varepsilon}>0$, this means that ${\lambda}_{\varepsilon}$ is the first Dirichlet eigenvalue of $- \Delta + \mu$. We assume from now on that $v_{\varepsilon}$ is normalized by $\|v_{\varepsilon}\|_{L^2(B_1)}=1$. Since the measure $\mu$ is finite with support in $B_1$, the first eigenvalue $\lambda_{\varepsilon}$ is simple.

In view of the Dal Maso-Mosco density result [@DMM87], there exists a sequence ${\Omega}_n{\subseteq}B_1$, $n \ge 1$, such that ${\Omega}_n $ $\gamma$-converges to $\mu$. We can assume that the boundary $\partial {\Omega}_n$ of ${\Omega}_n$ is smooth. Furthermore, we can also assume that ${\Omega}_n$ is connected. For otherwise we keep from ${\Omega}_n$ only a connected component $\Omega_n'$ with the largest $L^1-$norm of its torsion function among all the connected components of $\Omega_n$. Then $ \Omega_n \setminus \Omega_n'$ $\gamma$-converges to the empty set as $n \to \infty$ and the removal of $ \Omega_n \setminus \Omega_n'$ from $\Omega_n$ does not affect the first Dirichlet eigenvalue of $-\Delta$ on $\Omega_n$, while $|\Omega_n'| \le |\Omega_n|$. It is possible to explicitly construct such a sequence $(\Omega_n)$ in the spirit of Cioranescu-Murat [@CM], but such a construction is not needed for the rest of the proof. The only fact we need to keep in mind is that ${\Omega}_n {\subseteq}B_1$, so that $|{\Omega}_n| \le |B_1|$. Denote by $\lambda_1(\Omega_n)$ the first Dirichlet eigenvalue of $-\Delta$ on $\Omega_n$ and by $u_n$ the $L^2-$normalized, positive eigenfunction corresponding to $\lambda_1(\Omega_n)$. We extend $u_n$ to $ B_1$ by setting it to $0$ on $B_1 \setminus {\overline}\Omega_n$ and by a slight abuse of notation, denote this extension again by $u_n$. The $\gamma$-convergence of $(\Omega_n)$, together with the compact embedding of $H^1_0(B_1)$ in $L^1(B_1)$, imply that (i) ${\lambda}_1({\Omega}_n) \rightarrow {\lambda}_{\varepsilon}$, and (ii) $u_n\rightharpoonup v_{\varepsilon}$ weakly in $H^1_0(B_1)$ and strongly in $L^1(B_1)$. Assuming that $$\label{L infty}
\lim_{n \to \infty }\|u_n\|_\infty = \|v_{\varepsilon}\|_\infty.$$ we get $$\liminf_{n \rightarrow +\infty} \frac{\int_{{\Omega}_n} u_n dx}{\|u_n\|_\infty |{\Omega}_n|} \ge  \frac{\int_{B_1} v_{\varepsilon}dx}{\|v_{\varepsilon}\|_\infty |B_1|}.$$ Since the right-hand side is arbitrarily close to $1$ when ${\varepsilon}\downarrow 0$, the proof of Theorem \[the7\] is then completed by a diagonal selection procedure.

It remains to show . This kind of assertion is known to be true in a general setup. In essence it is a consequence of the subharmonicity of the eigenfunctions $u_n$, $n \ge 1$. (For a similar result for the torsion function see [@HLP Theorem 2.2].) For the sake of completeness, we give below a proof. It slightly differs from the one in [@HLP Theorem 2.2].

First note that since $u_n \to v_{\varepsilon}$ strongly in $L^1$, it follows that $$\liminf _{n \to +\infty} \|u_n\|_\infty  \ge  \|v_{\varepsilon}\|_\infty .$$ To prove that $ \limsup _{n \to +\infty} \|u_n\|_\infty  \le  \|v_{\varepsilon}\|_\infty $ we argue as follows. Being convergent, the sequence $({\lambda}_1(\Omega_n))$ is bounded, and so is $\| u_n \|_\infty$. Choose $M>0$ so that for any $n \in {\mathbb{N}}$ $${\lambda}_1(\Omega_n) u_n(x)  \le M, \quad \forall \ x \in B_1,$$ and therefore $$-\Delta u_n(x)  \le M ,  \qquad \forall \ x \in B_1 .$$ Let $x_n \in B_1$ be a maximum point for $u_n$. By taking, if necessary, a subsequence we may assume that $x_n \to x^*$. Furthermore, $$-\Delta u_n \le M= M \Delta \frac{|x-x_n|^2}{2m},  \quad \mbox  {in } {\mathcal D}'( {\mathbb{R}}^m),$$ or $$-\Delta \Big (u_n + M  \frac{|x-x_n|^2}{2m}\Big)\le 0, \quad \mbox  {in } {\mathcal D}'( {\mathbb{R}}^m).$$ By the subharmonicity of the function $x \mapsto u_n(x) + M  \frac{|x-x_n|^2}{2m}$ around $x_n$, it then follows that for $\delta >0$ sufficiently small, $$\|u_n \|_\infty = u_n(x_n) \le \frac{\int _{B(x_n; \delta)} \big( u_n(x) + M  \frac{|x-x_n|^2}{2m}\big) }{|B(x_n; \delta)|}.$$ Taking the limit $n \to +\infty$ we obtain $$\limsup _{n \to +\infty} \|u_n\|_\infty \le \frac{\int _{B(x^*; \delta)}\Big(v_{\varepsilon}(x) + M  \frac{|x-x^*|^2}{2m}\Big)}{|B(x^*;\delta)|},$$ or $$\limsup _{n \to +\infty} \|u_n\|_\infty \le \frac{\|v_{\varepsilon}\|_\infty |B(x^*; \delta)| + M  \frac{\delta^2}{2m} |B(x^*; \delta)|}{|B(x^*; \delta)|}
= \|v_{\varepsilon}\|_\infty + M  \frac{\delta^2}{2m}.$$ Letting $\delta \downarrow 0$, completes the proof. $\square$

\[rem0\] If $\Omega$ is an open connected subset of ${\mathbb{R}}^m$ with $m \ge 2$ and $0<|\Omega|<\infty$, then

1.  $$\label{p1}
    \Phi(\Omega)\ge \frac{E(\Omega)}{1+k_m\lambda_1(\Omega)^{m/4}|\Omega|^{1/2}\big(1-E(\Omega)\big)^{1/2}},\hspace{14mm} m=2,3,$$

2.  $$\label{p2}
    \Phi(\Omega)\ge \frac{E(\Omega)}{1+k_m\lambda_1(\Omega)|\Omega|^{2/m}\big(1-E(\Omega)\big)^{1/(m-1)}},\qquad m\ge 4,$$

where $$\label{p0}
k_m=
\begin{cases}
2(8\pi)^{-m/4}\Gamma((4-m)/4),\hspace {38mm} m=2,3,\\
\pi^{-1}(m-2)^{-1}m^{-1/(m-1)}\big(\Gamma((m+2)/2)\big)^{2/m},\qquad m\ge 4.
\end{cases}$$

Putting $V=0$ in one obtains by using , $$\begin{aligned}
\label{p3}
v_{\Omega}(x)&\ge \int_{\Omega} dy \int_{{\mathbb{R}}^+} dt\, p_{\Omega}(x,y;t)\frac{\varphi_{1,\Omega}(y)}{\|\varphi_{1, \Omega}\|_{\infty}}\nonumber \\ &
=\lambda_{1}(\Omega)^{-1}\frac{\varphi_{1,\Omega}(x)}{\|\varphi_{1, \Omega}\|_{\infty}}.\end{aligned}$$ Integrating both sides of yields, $$\label{p4}
\|v_{\Omega}\|_1\ge \lambda_{1}(\Omega)^{-1}\frac{\|\varphi_{1,\Omega}\|_1}{\|\varphi_{1, \Omega}\|_{\infty}}.$$ To obtain the stated lower bound for $\Phi(\Omega)$ it remains to find an upper bound for $\|v_{\Omega}\|_{\infty}$.

First consider case (i). By , the Cauchy-Schwarz inequality, and the heat semigroup property, one sees that $$\begin{aligned}
\label{p5}
v_{\Omega}(x)-\lambda_{1}&(\Omega)^{-1}\frac{\varphi_{1,\Omega}(x)}{\|\varphi_{1, \Omega}\|_{\infty}}
=\int_{\Omega} dy \int_{{\mathbb{R}}^+} dt\, p_{\Omega}(x,y;t)\bigg(1-\frac{\varphi_{1,\Omega}(y)}{\|\varphi_{1, \Omega}\|_{\infty}}\bigg)\nonumber \\ &
\le \int_{{\mathbb{R}}^+} dt\, \bigg(\int_{\Omega} dy(p_{\Omega}(x,y;t))^2\bigg)^{1/2}\bigg(\int_{\Omega} dy\bigg(1-\frac{\varphi_{1,\Omega}(y)}{\|\varphi_{1, \Omega}\|_{\infty}}\bigg)\bigg)^{1/2}\nonumber \\ &
=|\Omega|^{1/2}\int_{{\mathbb{R}}^+} dt\,p_{\Omega}(x,x;2t)^{1/2}\big(1-E(\Omega)\big)^{1/2}.\end{aligned}$$ Choosing $\beta=\frac12$ in gives by domain monotonicity of the Dirichlet heat kernel, and $$\label{p6}
p_{\Omega}(x,x;2t)\le e^{-t\lambda_1(\Omega)}(4\pi t)^{-m/2}.$$ Substitution of into the right-hand side of , evaluating the resulting integral with respect to $t$, and taking the supremum over all $x\in\Omega$ gives $$\label{p7}
\|v_{\Omega}\|_{\infty}-\lambda_{1}(\Omega)^{-1}\le 2(8\pi)^{-m/4}|\Omega|^{1/2}\Gamma((4-m)/4)\lambda_1(\Omega)^{-1 + m/4}\big(1-E(\Omega)\big)^{1/2}.$$ Inequality follows from , with the values for $k_2$ and $k_3$ given in .

Next consider case (ii). By the first equality in we have by domain monotonicity of the Dirichlet heat kernel $$\begin{aligned}
v_{\Omega}(x)-\lambda_{1}&(\Omega)^{-1}\frac{\varphi_{1,\Omega}(x)}{\|\varphi_{1, \Omega}\|_{\infty}}
\le \int_{\Omega} dy \int_{{\mathbb{R}}^+} dt\, p_{{\mathbb{R}}^m}(x,y;t)\bigg(1-\frac{\varphi_{1,\Omega}(y)}{\|\varphi_{1, \Omega}\|_{\infty}}\bigg) \nonumber \\ &
=c_m\int_{\Omega} dy |x-y|^{2-m}\bigg(1-\frac{\varphi_{1,\Omega}(y)}{\|\varphi_{1, \Omega}\|_{\infty}}\bigg),\end{aligned}$$ where $$c_m=\frac{\Gamma((m-2)/2)}{4\pi^{m/2}}.$$ By Hölder’s inequality with exponents $p=\frac{m-1}{m-2}$ and $q=m-1$ we have $$\begin{aligned}
v_{\Omega}(x)-&\lambda_{1}(\Omega)^{-1}\frac{\varphi_{1,\Omega}(x)}{\|\varphi_{1,\Omega}\|_{\infty}}\nonumber \\ &
\le c_m\bigg(\int_{\Omega}\frac{dy}{|x-y|^{m-1}}\bigg)^{(m-2)/(m-1)}\bigg( \int_\Omega dy \big(1-\frac{\varphi_{1,\Omega}(y)}{\|\varphi_{1, \Omega}\|_{\infty}} )\bigg)^{1/(m-1)}\nonumber \\ &
\le c_m\bigg(\int_{\Omega^*}\frac{dy}{|y|^{m-1}}\bigg)^{(m-2)/(m-1)}\bigg(\int_\Omega dy \big(1-\frac{\varphi_{1,\Omega}(y)}{\|\varphi_{1, \Omega}\|_{\infty}} \big)\bigg)^{1/(m-1)}\nonumber \\ &
=c_m \big(m\omega_m R^*\big)^{(m-2)/(m-1)}|\Omega|^{1/(m-1)}\big(1-E(\Omega)\big)^{1/(m-1)},\end{aligned}$$ where we have used Schwarz symmetrisation with $\Omega^*=B_{R^*}$, and $\omega_m (R^*)^m=|\Omega|.$

Taking the supremum over all $x\in \Omega$ and using the formulae for $c_m$ and $R^*$ gives, $$\begin{aligned}
\|v_{\Omega}\|_{\infty}\le\lambda_{1}(\Omega)^{-1}&+\pi^{-1}(m-2)^{-1}m^{-1/(m-1)}\big(\Gamma((m+2)/2)\big)^{2/m}\nonumber \\ &\times|\Omega|^{2/m}\big(1-E(\Omega)\big)^{1/(m-1)}.\end{aligned}$$ This, together with , implies the assertion for $m\ge 4$.

We see from the proof of Remark \[rem0\](ii) that the case $m=3$ could also have been included. However, that would have given $\lambda_1(\Omega)|\Omega|^{2/3}$ in the denominator. By Theorem \[the9\] (iii) we have, for $m=3$, that $\lambda_1(\Omega)|\Omega|^{2/3}\gg1$ if $E(\Omega)$ is close to $1$. Then $\lambda_1(\Omega)^{3/4}|\Omega|^{1/2}\ll\lambda_1(\Omega)|\Omega|^{2/3},$ and so gives a better bound in that case. However, bounds and do not imply that if $\Phi(\Omega)$ is close to $1$ then $E(\Omega)$ is close to $1$ since, by Theorem \[the9\] (iii) $\lambda_1(\Omega)|\Omega|^{2/m}$, becomes large.

Proof of Theorem \[the9\]  {#sec6}
==========================

[*Proof of Theorem* ]{}. By re-scaling both $u$ and $\Omega$, $$\label{p24}
\frac{u(tx)}{\|u\|_{\infty}}, \qquad x \in t^{-1} \Omega ,$$ it suffices to consider the case where $\|u\|_{\infty}=1$, and $|\Omega|=\omega_m$. Inequality then reads $$\label{p25}
\bigg( 1 - \fint_{\Omega}u\bigg)\int_{\Omega}|\nabla u|^2\ge \frac{4m^2}{(m+2)^2}\omega_m,$$ with $|\Omega|=\omega_m, \,  \|u\|_\infty = 1,$ and $\fint_{\Omega}u\ge \frac{2}{m+2}.$ Note that replacing $u$ by its positive part $u^{+}$ decreases the left-hand side of , and furthermore, $u^{+}\le 1,\,\fint_{\Omega} u^+ \ge \frac{2}{m+2}$. So it suffices to prove that for any $m \ge 2,$ $$(1-\theta) F(\theta) \ge \frac{4m^2}{(m+2)^2}\omega_m \, , \qquad  \forall \, \theta\in [2/(m+2),1),$$ where $$F(\theta):= \inf\big\{\int_{\Omega}|\nabla u|^2: \, u\in H_0^{1}(\Omega), 0 \le u \le 1,\fint_{\Omega}u=\theta \big\} \, .$$

We make some preliminary observations. By Schwarz rearrangement we may consider the infimum in the definition of $F$ over the collection $H_0^{*1}(B_1)$ of all radially symmetric, decreasing functions $u$ in $H_0^1(B_1)$ since this rearrangement decreases the energy and leaves the other constraints unchanged. So, $$F(\theta)  \ge  F^*(\theta), \quad 2/(m+2) \le \theta < 1,$$ where $$\label{minimization 2}
 F^*(\theta) =   \inf\big\{\int_{B_1}|\nabla u|^2 : \, u \in H_0^{*1}(B_1), 0 \le u \le 1,\,\fint_{B_1}u=\theta \big\}.$$ First note that $$\inf\big\{\int_{B_1}|\nabla u|^2 : \, u \in H_0^{*1}(B_1), \,  \fint_{B_1}u=\theta \big\},$$ admits a minimiser. By the Langrange multiplier theorem, it is given by the radially symmetric, decreasing solution of $$- \Delta u = c\, , \qquad u(1) = 0\, , \qquad \fint_{B_1} u = \theta,$$ where here, and in the sequel, by a slight abuse of notation, we write $u(r)$ instead of $u(x)$. By a straightforward computation one sees that for any $2/(m+2) \le \theta < 1$ $$\label{unconstrained}
u(r) = 2^{-1}(m+2)\theta(1 - r^2).$$ In particular, $\|u \|_\infty = u(0) = c \ge 1$. Note that we could have written $0\le u,\, \| u\|_\infty =1$ instead of $0\le u\le 1$ in the right-hand side of .

For any $2/(m+2) \le \theta <1,$ $ F^*(\theta)$ admits a minimiser. Since the Dirichlet energy is strictly convex, it is unique, and we denote it by $u_\theta$. Let $$f(\theta) = (1- \theta)F^*(\theta).$$ Since $F(\theta) \ge F^*(\theta)$, it suffices to show that $$\label{to be proved}
f(\theta) \ge \frac{4m^2}{(m+2)^2}\omega_m\, , \qquad \ \forall  \theta \in [2/(m+2) , 1)\, .$$ If $\theta =  \frac{2}{m+2}$, then by , $u_{2/(m+2)} (r) = 1- r^2$. Hence $$\label{minimum}
 f(2/(m+2)) = \Big(1- \frac{2}{m+2}\Big) \int_{B_1} (-2r)^2 mr^{m-1} \omega_m dr = \frac{ 4m^2}{(m+2)^2} \omega_m.$$ Hence is satisfied for $\theta =  \frac{2}{m+2}$.

The remaining part of the proof consists of four parts. In part (a) we show that holds for any $m\ge 2$ and $\theta\in [\frac{2}{m+2},\theta_m^*]$ where $$\label{p27}
\theta_m^* :=\frac{m+\big(m^2+8m\big)^{1/2}}{2(m+2)}.$$ We note that $2/(m+2) <  2/3 < \theta_m^* < 1$ for any $m \ge 2$. In part (b) we prove that holds for any $m\ge 4$ and $\theta\in [\theta_m^*,1)$. In part (c) we show that holds for $m=2,3$, using the Euler-Lagrange equation of a variational problem, related to an obstacle problem. See below. Finally, in part (d) we verify that [*equality*]{} in holds if and only if $\Omega$ is a ball and $u$ is a multiple of the torsion function for that ball. This completes the proof of Theorem \[the9\] (i).

\(a) Assume that $\frac{2}{m+2} \le \theta \le \theta_m^*.$ In a first step we prove that $\theta\mapsto f(\theta)$ is increasing on $[\frac{2}{m+2},\frac23]$ which by implies that holds for $\theta$ in the latter interval. Given $\frac{2}{m+2} < \theta < 1$, choose any $\varepsilon>0$ with the property $$\frac{\theta}{1+\varepsilon}>\frac{2}{m+2}.$$ Let $\overline{u}_\theta$ be the minimiser of the obstacle problem $$\inf\{\int_{B_1}|\nabla u|^2: \, \, u\in  H_0^{*1}(B_1), \, 0\le u \le 1+\varepsilon, \, \fint_{B_1} u=\theta\}.$$ Then, by inclusion of the class of admissible test functions, $$\label{p52}
\int_{B_1}|\nabla\overline {u}_\theta|^2\le \int_{B_1}|\nabla u_{\theta}|^2,$$ where $u_{\theta}$ is the minimiser of $F^*(\theta)$. Furthermore $(1+\varepsilon)^{-1}\overline{u}_\theta$ is the minimizer $u_{\theta/(1+\varepsilon)}$ for $F^*( \theta/(1 + \varepsilon))$, since $0 \le (1+\varepsilon)^{-1}\overline{u}_\theta \le 1$, and $\fint_{B_1}\frac{\overline{u}_\theta}{1+\varepsilon}=\frac{\theta}{1+\varepsilon}$. We wish to prove that $$\label{p53}
f(\theta/(1 + \varepsilon)) = \Big(1- \frac{\theta}{ 1 + \varepsilon}\Big) \int_{B_1}\frac{|\nabla \overline{ u}_\theta |}{ (1 + \varepsilon)^2} \le
(1-\theta) \int_{B_1}|\nabla u_{\theta}|^2 = f(\theta),$$ By , inequality holds if $$\frac{1}{(1+\varepsilon)^2}\bigg(1-\frac{\theta}{1+\varepsilon}\bigg)\le(1-\theta),$$ or, after simplifying, $$\label{p56}
1\le (3+3\varepsilon +\varepsilon^2)(1-\theta).$$ The latter inequality clearly holds for any $\varepsilon \ge 0$ if $ \frac{2}{m+2} \le \theta\le \frac23$. Thus we have proved that $f(\theta)$ is increasing on the interval $[\frac{2}{m+2},\frac23]$.

By the same argument we now prove that $f(\theta) \ge f(2/(m+2))= \frac{ 4m^2}{(m+2)^2}\omega_m $ also for $\theta \in ( \frac23, \theta_m^*]$. (However, we do not prove that $f$ is increasing in the interval $( \frac23, \theta_m^*]$.) Given $\theta \in ( \frac23, \theta_m^*]$, we want to find $\varepsilon>0$ so that $$\label{p57}
 \frac{2}{m+2}  \le  \frac{\theta}{1+\varepsilon}\le \frac 23 ,\qquad 1\le \big( 3+3\varepsilon+\varepsilon^2\big) (1-\theta).$$ By and , one then infers that $$f(\theta) \ge f(\theta/(1+\varepsilon)) \ge f(2/(m+2)).$$ To find $\varepsilon > 0$ so that holds, we solve $$\label{eq for epsilon}
1= 3(1-\theta)+3\varepsilon(1-\theta)+\varepsilon^2(1-\theta)$$ for $\varepsilon$ and verify that for the given range of $\theta,$ one has $\frac{2}{m+2}  \le  \frac{\theta}{1+\varepsilon}\le \frac 23$. The positive solution of is given by $$\varepsilon=-\frac32+ \frac12\bigg(\frac{1+3\theta}{1-\theta}\bigg)^{1/2}>0.$$ The requirement $ \frac{2}{m+2} \le \frac{\theta}{1+\varepsilon}$ yields, by a direct computation, $$\theta\le \theta_m^*.$$ Using $\theta \ge \frac 23$, one sees that $ \frac{\theta}{1+\varepsilon} \le \frac 23$.

\(b) In this part we assume that $m \ge 4$ and $\theta_m^* \le \theta < 1.$ Let $ 0 \le  r_0 < 1$ and let $u \in H_0^{*1}(B_1)$ with $0 \le u \le 1$ and $\theta = \fint_{B_1} u.$ Since $u$ is radially symmetric and decreasing, $$\omega_m^{-1}\int_{B_1\setminus\overline{B}_{r_0}}u \le \, \omega_m^{-1}u(r_0) \, |B_1\setminus\overline{B}_{r_0}|=u(r_0)(1-r_0^m).$$ Since, $0 \le u \le 1$ and $\theta = \fint_{B_1} u,$ we conclude that $$ r_0^m\ge \omega_m^{-1}\int_{B_{r_0}}u = \omega_m^{-1}\int_{B_1}u - \omega_m^{-1}\int_{B_1 \setminus  {\overline}B_{r_0}}u
 \ge \theta-u(r_0)(1-r_0^m).$$ It then follows that $$\label{p31}
u(r_0)\ge  \eta_m(\theta, r_0) := \frac{ \theta-r_0^m}{1-r_0^m}  \ge 0 \, , \qquad  \forall \, r_0\in [0,\theta^{1/m}].$$ By inclusion of the admissible test functions one has $$\begin{aligned}
\label{p32}
f(\theta)\ge (1-\theta)\inf\big\{\int_{B_1}|\nabla u|^2 : \, u \in H_0^{*1}(B_1), \, u(r_0)\ge \eta_m(\theta, r_0) \big\}.\end{aligned}$$ The infimum in the right-hand side of is attained and its minimiser $u^*$ is given by $$u^*(r)=
\begin{cases}
\eta_m(\theta, r_0),\hspace {18mm} 0\le r\le r_0,\\
\frac{1-r^{2-m}}{1-r_0^{2-m}}\eta_m(\theta, r_0),\qquad r_0\le r\le 1.
\end{cases}$$ A straightforward calculation gives $$\label{p34}
\int_{B_1}|\nabla u^*|^2=m(m-2)\omega_m\frac{\eta_m^2(\theta, r_0)}{r_0^{2-m}-1}\, .$$ We now choose $$\label{p35}
r_0=\theta^{c/m},$$ where $c\ge 1$ is to be determined later. This choice satisfies the constraint $0 \le r_0 \le \theta^{1/m}$ in . By and , $$\label{p36}
f(\theta)\ge m(m-2)\omega_m\bigg(\frac{\theta-\theta^c}{1-\theta^c}\bigg)^2\frac{1-\theta}{\theta^{c(2-m)/m}-1}.$$ By L’Hôpital’s rule, $$\label{p371}
\liminf_{\theta\uparrow 1}f(\theta) \,  \ge  \, m^2 \omega_m\frac{(c-1)^2}{c^3}.$$ The right-hand side of is maximised for $c=3$. This choice yields, $$\liminf_{\theta\uparrow 1}f(\theta)\ge \frac{4m^2}{27} \omega_m.$$ Note that $$\label{p40}
\frac{4m^2}{27} \omega_m \ge \frac{4m^2}{(m+2)^2}\omega_m.$$ if and only if $m\ge 4$. This is why the proof of for $m=2,3$ has been deferred to part (c). By we have for $c=3$, $$\begin{aligned}
\label{p38}
f(\theta)&\ge m(m-2)\omega_m\bigg(\frac{\theta(1+\theta)}{1+\theta+\theta^2}\bigg)^2\frac{1-\theta}{\theta^{3(2-m)/m}-1}\nonumber \\ &
\ge m(m-2)\omega_m\bigg(\frac{\theta(1+\theta)}{1+\theta+\theta^2}\bigg)^2\frac{1-\theta}{\theta^{-3}-1}\nonumber \\ &
=m(m-2)\omega_m\frac{\theta^5(1+\theta)^2}{(1+\theta+\theta^2)^3}.\end{aligned}$$

To prove for $\theta\in [\theta_m^*,1)$ and $m \ge 4$, it suffices, by , to show that $$\label{p41}
\frac{\theta^5(1+\theta)^2}{(1+\theta+\theta^2)^3}\ge \frac{4m}{(m-2)(m+2)^2},\qquad \forall \, \theta\in [\theta_m^*,1).$$ First observe that the left-hand side of is a product of non-negative increasing functions, $\theta\mapsto\frac{\theta(1+\theta)}{1+\theta+\theta^2}$ and $\theta\mapsto\frac{\theta^3}{1+\theta+\theta^2}$, and so is increasing. So if holds for $\theta=\theta_m^*$ then it holds on the interval $[\theta_m^*,1)$. Furthermore by , $$\theta_m^*=1-\frac{8}{(m+2)(m+4+(m^2+ 8m)^{1/2})}.$$ Hence $(\theta_m^*)_m$ is a strictly increasing sequence. Since the right-hand side of is decreasing in $m$ we conclude that if holds for $m=m_1$ then it holds for all $m\ge m_1$. It is straightforward to verify $$\theta_6^*> \frac{15}{16},$$ and that holds for $\theta=\frac{15}{16}$, and $m=6$. Hence it follows that holds for all $\theta \in [\theta_m^*, 1)$ with $m\ge 6$.

To complete the proof of part (b) it remains to treat the cases $m=4$ and $m=5$. We first consider the case $m=4$. One computes $$\theta_4^*=\frac{1+\sqrt 3}{3},$$ and by the first inequality in , one gets $$\label{p45}
f(\theta)\ge 8\omega_4\bigg(\frac{\theta(1+\theta)}{1+\theta+\theta^2}\bigg)^2\bigg(\frac{1-\theta}{\theta^{-3/2}-1}\bigg).$$ Since both $\theta\mapsto \frac{\theta(1+\theta)}{1+\theta+\theta^2}$ and $\theta\mapsto \frac{1-\theta}{\theta^{-3/2}-1}$ are non-negative increasing functions on the interval $[0, 1)$ the right-hand side of is increasing in $\theta$. Note that the right-hand side of equals $16\omega_4/9$. Hence by it suffices to verify that $$\label{p46}
\bigg(\frac{\theta^*_4(1+\theta^*_4)}{1+\theta^*_4+\theta{^*_4}^2}\bigg)^2\bigg(\frac{1-\theta^*_4}{\theta{^*_4}^{-3/2}-1}\bigg)\ge \frac{2}{9}.$$ Numerical evaluation of the left-hand side of yields $$\bigg(\frac{\theta^*_4(1+\theta^*_4)}{1+\theta^*_4+\theta{^*_4}^2}\bigg)^2\bigg(\frac{1-\theta^*_4}{\theta{^*_4}^{-3/2}-1}\bigg)\ge .238,$$ which implies .

Finally we consider the case $m=5$. One computes that $$\label{p48}
\theta_5^*\ge \frac{13}{14},$$ and by the first inequality in , $$f(\theta)\ge 15\omega_5\bigg(\frac{\theta(1+\theta)}{1+\theta+\theta^2}\bigg)^2\frac{1-\theta}{\theta^{-9/5}-1}.$$ Since both $\theta\mapsto \frac{\theta(1+\theta)}{1+\theta+\theta^2}$ and $\theta\mapsto \frac{1-\theta}{\theta^{-9/5}-1}$ are non-negative increasing functions on the interval $[0,1)$, so is the right-hand side of and it remains, by , to verify that $$\label{p49}
\bigg(\frac{\theta^*_5(1+\theta^*_5)}{1+\theta^*_5+\theta{^*_5}^2}\bigg)^2\bigg(\frac{1-\theta^*_5}{\theta{^*_5}^{-9/5}-1}\bigg)\ge \frac{20}{147}.$$ By , the left-hand side of is bounded from below by $.206$ while the right hand side of is bounded from above by $.137$. This completes the proof of part (b).

\(c) In this part we treat the cases $m=2$ and $m=3$. We begin with some preliminary considerations. We note that the minimisation problem is related to a volume constraint obstacle problem in $B_1$: we claim that there exist $c>0$ and $0 \le l < 1$, depending on $\theta$, so that $u_\theta$ satisfies the following system of equations, $$\label{np1}
\begin{cases}
-\Delta u=c,&\text{in }B_1\setminus {\overline}B_l,\\
u=1,&\text{on } {\overline}B_l,\\
u= 0,&\text{on }\partial B_1,\\
\frac {\partial u}{\partial \nu} =0, &\text{on }\partial B_l,
\end{cases}$$ where $\nu$ denotes the inward pointing normal on the sphere $\partial B_l$. Indeed, since $u_\theta$ is radially symmetric, decreasing and since $u_\theta(0) = 1$ (cf. ) and $u_\theta(1) = 0$, there exists a maximal number $0 \le l \equiv l(\theta) < 1$ so that $u_\theta (r) =1$ for $0 \le r \le l$. By the Lagrange multiplier theorem, there exists a constant $c >0$ so that $-\Delta u_\theta  =c$ on $B_1 {\setminus}{\overline}B_l$ in the sense of distributions. It then follows from [@GE Theorem 2] that $u_\theta$ is $C^{1, \alpha}$ on ${\overline}B_1$ which implies that $\frac {\partial u_\theta}{\partial \nu} =0 $ on $\partial B_l$. These observations establish . Note that both $c$ and $l$ are uniquely determined by $\theta$.

We claim that the map $$\mathfrak{b} :  [\frac{2}{m+2}, 1) \to  [0,1) , \,    \theta \mapsto l(\theta) \, ,$$ is an increasing bijection. To prove the latter assertion, we construct for any given $0 \le l < 1$ a unique radially symmetric, decreasing solution $u(\cdot ; l)$ of and show that $\theta \equiv \theta(l) := \fint_{B_1} u(\cdot; l)$, satisfies $2/(m+2) \le \theta <1$ with $\theta(0) = 2/(m+2)$ and $\lim_{l \to 1} \theta(l)= 1$. First we note that $c$ is uniquely determined by $l$ since for any given $0 \le l < 1$, the solution of is given by a formula. To obtain it, note that the general radially symmetric solution of $- \Delta u = c$ on the annulus $B_1 \setminus {\overline}B_l$ is of the form $$u(r)=
\begin{cases}
-c\frac{r^2}{4}+a \ln(r) +b \qquad \quad  \mbox{ if } m=2 \, ,\\
-c \frac{r^2}{2m} -\frac{a}{(m-2)r^{m-2}} +b \quad \mbox{ if } m\ge3
\end{cases}$$ for some real constants $a,$ $b,$ $c$. The condition $\frac {\partial u}{\partial \nu}(l) =0 $ implies that $a= \frac{l^m c}{m}$ so that the boundary condition $u(1)=0$ reads $$\label{np2}
u(r)=
\begin{cases}
\frac{c}{4}( 1 - r^2) + \frac{l^2 c}{2}  \ln(r)  \qquad \qquad \quad \quad \mbox{ if } m=2,\\
 \frac{c}{2m}(1-r^2) + \frac{l^m c}{m(m-2)} (1-\frac{1}{r^{m-2}} ) \ \   \mbox{ if } m\ge3.
\end{cases}$$ The value of $c$ is now obtained by the requirement $u(l)=1$, $$\label{np3}
c=
\begin{cases}
 \frac{1}{\frac{1-l^2}{4}+ \frac{l^2}{2}\ln(l)} \qquad \quad \quad  \mbox{ if } m=2,\\
\frac{1}{\frac{1}{2m}+ \frac{l^m}{m(m-2)}-\frac{l^2}{2(m-2)}} \quad \mbox{ if } m\ge3.
\end{cases}$$ One verifies in a straightforward way that the resulting function $u \equiv u(\cdot; l)$ is decreasing for $l \le r \le 1$, that $\theta(0) = 2/(m+2),$ and that $l \mapsto c \equiv c(l)$ is a continuous, strictly increasing function of $0 \le l < 1$. We claim that $l \mapsto \theta(l)$ is also strictly increasing. To verify that this is indeed the case, one could explicitly compute $\theta$ in terms of $l$, but the formula is rather complicated. Instead we prove the claim by using the maximum principle. By contradiction, suppose there exist $0 \le l_2 < l_1 < 1$ with $ \theta_2:=\theta(l_2) > \theta_1:= \theta(l_1)$. By the considerations above, $ c_2:= c(l_2) < c_1:= c(l_1).$ Hence $-\Delta(u_1 - u_2) = c_1 - c_2 > 0$ on $B_1 \setminus B_{l_1}$ where $u_j:= u(\cdot, l_j)$ for $j=1,2$. Since $\theta_1 <\theta_2$, there exist $l_1 < r_1 < r_2 <1$ so that $(u_1 - u_2)(r) < 0$ for any $r_1 < r < r_2$, contradicting the maximum principle.

From the formula of $u(\cdot; l)$ one infers that $\theta(l)$ is a continuous function of $l$ and that and $\lim_{l \uparrow 1} \theta(l)= 1$. Hence for any $0 \le l < 1$, $u(\cdot; l)$ coincides with $u_\theta$ where $\theta \equiv \theta(l) = \fint_{B_1}  u(\cdot; l)$. Altogether we have shown that $ \mathfrak{b}$ is a continuous, increasing bijection.

Define $g : [0, 1) \to {\mathbb{R}}$ by $$\label{formula g}
g(l) :=f(\theta(l)) = (1 - \theta(l)) \int_{B_1} | \nabla u_{\theta(l)}|^2 \, , \qquad \theta(l) := \mathfrak{b}^{-1}(l) \, .$$ In view of it then suffices to show that $g$ is increasing on $[0, 1)$.

We first consider the case $m=2$. Integrating by parts, one obtains from $$\int _{B_1} |\nabla u_{\theta(l)}|^2 = c(l) 2\pi \int_l^1u_{\theta(l)}(r) r dr,$$ and $$\theta(l)= l^2+ 2 \int_l^1u_{\theta(l)}(r) r dr.$$ Using and , one infers from that $$g(l)= 2\pi \frac{(\frac{1}{16}-\frac{l^2}{4}+ \frac{3l^4}{16}-\frac{l^4 \ln (l)}{4})(\frac{1}{8}-\frac{l^4}{8}+\frac{l^2 \ln(l)}{2} )}{(\frac 14-\frac{l^2}4+\frac{l^2}{2}\ln (l))^3},$$ and a straightforward computation yields $$g'(l)= \pi  \frac{l(l^2-1)(-l^2+l^2\ln(l)+\ln(l)+1)(-5l^4+4l^4\ln(l)+4l^2+8l^2\ln(l)+1)}{(1-l^2+2l^2\ln(l))^4}.$$ By inspection one verifies that $g'(l) >0$ on $(0,1)$.

Without any additional effort we may consider the general case $m\ge 3$, and follow the line of arguments above. Integrating by parts, one has $$\int_{B_1} |\nabla u_{\theta(l)}|^2  = c(l) m \omega_m \int_l^1u_{\theta(l)}(r) r^{m-1} dr$$ and one computes that $$\theta(l) = l^m+ m \int_l^1u_{\theta(l)}(r) r^{m-1} dr.$$ Using formula for $m \ge 3$, one infers $$\int_l^1u_{\theta(l)}(r)r^{m-1} dr= c(l) \, \Big (\frac{l^m(l^2-1)}{2m(m-2)} +\frac {l^{m+2}-1}{2m(m+2)}+ \frac{l^m(1-l^m)}{m^2(m-2)}+\frac{1-l^m}{2m^2}\Big)$$ so that by $$\label{np4}
g(l)= c(l) m\omega_m \Big (1-l^m-m\int_l^1u_{\theta(l)}(r) r^{m-1} dr\Big)\int_l^1u_{\theta(l)}(r) r^{m-1} dr.$$

In the case $m=3$, one gets in this way $$g(l)=\frac{24\pi}{25}\frac{(5(1-l^3)+10l^3(1-l^3)-15l^3(1-l^2)-3(1-l^5))(1-l^5+5(l^3-l^2))}{(2l^3+1-3l^2)^3},$$ and a lengthy computation leads to the formula $$g'(l)=\frac{24\pi}{25} \frac{2 l (20 l^4 + 67 l^3 + 84 l^2 + 46 l + 8)}{(2 l + 1)^4}.$$ Clearly, $g'(l) >0$ on $(0,1)$ for $m=3$.[^1]

\(d) In this last part we prove that equality in holds if and only if $\Omega$ is a ball and $u$ is a multiple of the torsion function for that ball. Clearly, if $\Omega$ is a ball and $u$ is a multiple of the torsion function for that ball, then holds (see ). Conversely, assume that equality holds in . As in , we can re-scale the measure of $\Omega$ and the $L^\infty$-norm of $u$. Equality in implies that $u$ has the same Dirichlet integral as its Schwarz rearrangement $u^*$, $$\label{Dirichlet}
\int_{\Omega}|\nabla u|^2 dx= \int_{B_1}|\nabla u^*|^2 dx\, ,$$ and its Schwarz rearrangement is the solution of the obstacle problem on the ball $B_1$ – see . In view of the strict monotonicity of $f$ on $[\frac{2}{m+2}, \frac 23)$ (cf. part (a)) and the (strict) inequalities obtained above $\frac 23$ (cf. parts (a)-(c)), this implies that $\theta =\frac{2}{m+2}$, which corresponds to $l=0$ and to $c=2m$ (cf. , ). It means that $u^*$ is a multiple of the torsion function on $B_1$ (cf. ).

In order to justify that $u$ has to be equal to $u^*$, recall that holds and that $u^*$, being a multiple of the torsion function on $B_1$, has a critical set of zero measure. Equality between $u$ and $u^*$, up to a translation, comes from the classical result of Brothers and Ziemer [@BZ88 Theorem 1.1]. $\square$

[*Proof of Theorem* ]{} The key ingredient into the proof is inequality . First note that since for any $t > 0$, $\frac{1}{t^2}v_{\Omega}(tx)$ is the torsion function of $\frac{1}{t}\Omega$. Choosing $t = (|\Omega| /\omega_m)^{1/m}$, one infers that it suffices to prove estimate in the case $|\Omega| = \omega_m$.

We apply (i) to $$u(x)=\frac{v_{\Omega}(x)}{\|v_{\Omega}\|_{\infty}},\qquad x\in \Omega.$$ Observe that $\fint_{\Omega} u = \Phi(\Omega).$ First we consider the case where $\fint_{\Omega} u\ge \frac{2}{m+2}$. Then by $$\frac{1}{\|v_{\Omega}\|_{\infty}^2}\big(1-\Phi(\Omega)\big)\int_{\Omega}|\nabla v_{\Omega}|^2\ge \frac{4m^2}{(m+2)^2}\omega_m\, .$$ Since $- \Delta v_\Omega = 1,$ $$\frac{1}{\omega_m\|v_{\Omega}\|_{\infty}}\int_{\Omega} | \nabla v_{\Omega}|^2 =
\frac{1}{\omega_m\|v_{\Omega}\|_{\infty}}\int_{\Omega} v_{\Omega}=\Phi(\Omega).$$ Since $\Phi(\Omega) \le 1$, we find that $$\begin{aligned}
\|v_{\Omega}\|_{\infty}&\le \frac{(m+2)^2}{4m^2}\Phi(\Omega)\big(1-\Phi(\Omega)\big)\nonumber \\
& \le \frac{(m+2)^2}{4m^2}\big(1-\Phi(\Omega)\big),\end{aligned}$$ which gives . Next consider the case $\fint_{\Omega} u\le \frac{2}{m+2}$. Since by Saint Venant’s principle $\|v_\Omega\|_\infty \le \|v_{B_1}\|_\infty$ and since $ \|v_{B_1}\|_\infty = 1/2m$ we find that $$\begin{aligned}
1-\Phi(\Omega)&\ge \frac{m}{m+2}=\frac{2m^2}{m+2}.\frac{1}{2m}\nonumber \\ &
\ge \frac{2m^2}{m+2}\|v_{\Omega}\|_{\infty}.\end{aligned}$$ Note that $\frac{2m^2}{m+2} \ge \frac{4m^2}{(m+2)^2}$ and hence the estimate also holds in this case.\
Inequality follows from and . $\square$

[*Proof of Theorem* ]{} The key ingredient in the proof is inequality . Since $\Omega$ is connected, $\lambda_1(\Omega)$ has multiplicity $1$ and hence both, $\varphi_{1,\Omega}$ and $E(\Omega)$, are well defined. First note that since for any $t > 0$, $t^{m/2}\varphi_{1, \Omega}(tx)$ is the positive $L^2-$normalized Dirichlet eigenfunction of $\frac{1}{t}\Omega$, choosing $t = (|\Omega| /\omega_m)^{1/m}$, one infers that it suffices to prove estimate in the case $|\Omega| = \omega_m$. By (26) in [@vdB], one has $$\|\varphi_{1,\Omega}\|_{\infty}\le \bigg(\frac{e}{2\pi m}\bigg)^{m/4}\lambda_1(\Omega)^{m/4}.$$ Hence $\varphi_{1,\Omega}\in L^{\infty}(\Omega)$. We apply (i) to $$u(x)=\frac{\varphi_{1,\Omega}(x)}{\|\varphi_{1,\Omega}\|_{\infty}},\qquad x\in \Omega.$$ First we consider the case where $\fint_{\Omega} u\ge \frac{2}{m+2}$. Then by $$\big(1-\fint_{\Omega}u\big) \int_{\Omega}|\nabla u|^2 \ge \frac{4m^2}{(m+2)^2}\omega_m .$$ Since $\int_{\Omega}|\nabla u|^2=\lambda_1(\Omega)\int_{\Omega}u^2\le \lambda_1(\Omega)\omega_m$, and $\fint_{\Omega} u=E(\Omega)$ we obtain, $$\label{p67}
\lambda_1(\Omega)\ge \frac{4m^2}{(m+2)^2}\big(1-E(\Omega)\big)^{-1}.$$ Now let us consider the case where $\fint_{\Omega} u\le \frac{2}{m+2}$. Then $1-E(\Omega) \ge \frac{m}{m+2}$, and hence by Faber Krahn, $$\label{p68}
\lambda_1(\Omega)\ge \lambda_1(B_1) \frac{m}{m+2}\big(1-E(\Omega)\big)^{-1}.$$ Combining and gives $$\label{p69}
\lambda_1(\Omega)\ge \min\bigg\{\frac{4m^2}{(m+2)^2},\frac{m}{m+2}\lambda_1(B_1)\bigg\}\big(1-E(\Omega)\big)^{-1}.$$ To finish the proof we recall that $$\label{p70}
\lambda_1(B_1)=j^2_{(m-2)/2}.$$ By the results of [@LL], we have that $$\label{p71}
j^2_{(m-2)/2}\ge \frac{m(m+8)}{4}.$$ Hence by , , , $$\lambda_1(\Omega)\ge \frac{4m^2}{(m+2)^2}\big(1-E(\Omega)\big)^{-1} ,$$ which is the inequality in the case $|\Omega | = \omega_m$. $\square$

\[rem1\] By an elementary computation, using the expression for $g$ in terms of $l$ from , one can show that $$\forall m\ge 2, \quad \lim_{\theta \uparrow 1} f(\theta) = \frac{4}{9}m^2 \omega_m.$$

\[rem2\] If $\Omega$ is any open set in ${\mathbb{R}}^m$ with $0<|\Omega|<\infty$, then $$\label{p17}
\lambda_1(\Omega)\ge \frac{\nu_m\mu_m}{\nu_m+\mu_m}\big(1-\Phi(\Omega)\big)^{-1}|\Omega|^{-2/m},$$ and $$\label{p18}
\|v_{\Omega}\|_{\infty}\le \mathfrak{c}_m\frac{\nu_m+\mu_m}{\nu_m\mu_m}\big(1-\Phi(\Omega)\big)|\Omega|^{2/m},$$ where $$\nu_m=\frac{2m\omega_m^{2/m}}{m+2},$$ $$\mu_m =  \lambda_1(B_1)|B_1|^{2/m} =j_{(m-2)/2}^2\omega_m^{2/m},$$ and where $\mathfrak{c}_m$ is defined in .

The proof of and is elementary but the constants are not as good as in Theorem \[the9\](ii).

[*Proof of Remark .*]{} In [@vdBFNT Theorem 1.1] it was shown that $$\label{p19}
\frac{\|v_{\Omega}\|_1\lambda_1(\Omega)}{|\Omega|}\le 1-\frac{\nu_m\|v_{\Omega}\|_1}{|\Omega|^{(m+2)/m}}.$$ Since $1 \le \lambda_1(\Omega) \|v_\Omega\|_\infty$ by it then follows from that $$\label{p20}
\|v_{\Omega}\|_1\le \lambda_1(\Omega)\|v_{\Omega}\|_{\infty}\bigg(\frac{\lambda_1(\Omega)}{|\Omega|}+\frac{\nu_m}{|\Omega|^{(m+2)/m}}\bigg)^{-1}.$$ Faber Krahn asserts that $$\label{p21}
\lambda_1(\Omega)|\Omega|^{2/m}\ge \lambda_1(B_1)|B_1|^{2/m}=\mu_m.$$ By and , $$\begin{aligned}
\Phi(\Omega)&\le \frac{\lambda_1(\Omega)}{\lambda_1(\Omega)+\nu_m|\Omega|^{-2/m}}\nonumber \\ &
=1-\frac{\nu_m}{\lambda_1(\Omega)|\Omega|^{2/m}+\nu_m}\nonumber \\ &
\le 1-\frac{\nu_m}{\lambda_1(\Omega)|\Omega|^{2/m}\big(1+\frac{\nu_m}{\mu_m}\big)}.\end{aligned}$$ This implies the assertion . Inequality follows from and . $\square$

Below we show that $\lambda_1(\Omega)$ cannot be bounded from above in terms of $(1-E(\Omega))^{-1}|\Omega|^{-2/m}$ nor of $(1-\Phi(\Omega))^{-1}|\Omega|^{-2/m}$.

\[rem3\] We have $$\label{ea1}
\sup\{\lambda_1(\Omega)(1-E(\Omega))|\Omega|^{2/m}:\Omega\,\textup{open, convex},\,0<|\Omega|<\infty\}=+\infty,$$ and $$\label{ea2}
\sup\{\lambda_1(\Omega)(1-\Phi(\Omega))|\Omega|^{2/m}:\Omega\,\textup{open, convex},\,0<|\Omega|<\infty\}=+\infty.$$

Let $\Omega_n=\{x_1^2+...+x_{m-1}^2<1, -\frac{n}{2}<x_m<\frac{n}{2}\}$. Then $\lambda_1(\Omega_n)\ge (m-1)\pi^2.$ A straightforward calculation shows that for an interval of length $L,\, L>0,\,
E((0,L))=\frac{2}{\pi}$. By separation of variables $E(\Omega_n)=\frac{2^m}{\pi^m}$. We conclude that the supremum in is bounded from below by $(m-1)\big(1-\frac{2^m}{\pi^m}\big)\pi^2n^{2/m}.$ Letting $n\rightarrow\infty$ concludes the proof of .

$\Phi(\Omega_n)$ cannot be computed in closed form by separation of variables. For large $n$ however, we have approximate separation of the $x_m$ variable. Similar to , $\|v_{\Omega_n}\|_1\le n\|v_{(0,1)^{m-1}}\|_1$, and $\lim_{n\rightarrow\infty}\|v_{\Omega_n}\|_{\infty}=\|v_{(0,1)^{m-1}}\|_{\infty}$. Hence for any $\varepsilon\in(0,1)$, there exists $N_{\varepsilon}\in {\mathbb{N}}$ such that for all $n\ge N_{\varepsilon}$, $\|v_{\Omega_n}\|_{\infty}\ge(1-\varepsilon)\|v_{(0,1)^{m-1}}\|_{\infty}.$ It follows that $\Phi(\Omega_n)\le (1-\varepsilon)^{-1}\Phi((0,1)^{m-1})$. We choose $\varepsilon=\frac12(1-\Phi((0,1)^{m-1}))$, and obtain that for all $n$ sufficiently large, $$1-\Phi(\Omega_n)\ge \frac{1-\Phi((0,1)^{m-1})}{1+\Phi((0,1)^{m-1})}.$$ We conclude that for all $n$ sufficiently large, the supremum in is bounded from below by $$\frac{(m-1)(1-\Phi((0,1)^{m-1}))\pi^2n^{2/m}}{(1+\Phi((0,1)^{m-1}))}.$$ Letting $n\rightarrow\infty$ concludes the proof of .

[**Acknowledgments.**]{} MvdB and TK acknowledge support by the Leverhulme Trust through Emeritus Fellowship EM-2018-011-9, and the Swiss National Science Foundation respectively. DB was supported by the LabEx PERSYVAL-Lab GeoSpec (ANR-11-LABX-0025-01) and ANR SHAPO (ANR-18-CE40-0013).

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[^1]: For general $m \ge 4$, the formula for $g'$ can be computed to be a quotient of two polynomials with degrees depending on $m$. We believe that $g'(l)$ is strictly positive for every $l$ on $(0,1)$, but a direct proof, covering all dimensions $m\ge 5$, based on the formula of $g'$ seems out of reach. For $m=4$ the quotient of the polynomials simplifies, and gives $g(l)=\omega_4\Big(\frac {16}{3}l^2+\frac{16}{9}\Big ).$ We see that for $m=4$, $g(1)=\frac{64\omega_4}{9}$ agrees with the value $f(1)$ given in Remark \[rem1\]. We also have that $g(0)=\frac{16\omega_4}{3}$ agrees with the value $f(1/3)$ from Theorem \[the9\](i). Indeed for $m=4$ and $\theta=\frac13$ we have equality in . Note that for $m=4$, $g(l)$ is increasing on $(0,1)$.
