On the higher Cheeger problem

TL;DR

This paper introduces higher Cheeger constants for measurable sets, explores their properties, and relates them to spectral minimal partitions, with applications to specific planar domains.

Contribution

It defines higher Cheeger constants, proves existence of minimizers with special properties, and connects these constants to spectral partition problems.

Findings

Existence of minimizers for higher Cheeger constants.

Relation between higher Cheeger constants and spectral minimal partitions.

Application to determine the second Cheeger constant of planar domains.

Abstract

We develop the notion of higher Cheeger constants for a measurable set $\Omega \subset \mathbb{R}^N$. By the $k$-th Cheeger constant we mean the value \[h_k(\Omega) = \inf \max \{h_1(E_1), \dots, h_1(E_k)\},\] where the infimum is taken over all $k$-tuples of mutually disjoint subsets of $\Omega$, and $h_1(E_i)$ is the classical Cheeger constant of $E_i$. We prove the existence of minimizers satisfying additional "adjustment" conditions and study their properties. A relation between $h_k(\Omega)$ and spectral minimal $k$-partitions of $\Omega$ associated with the first eigenvalues of the $p$-Laplacian under homogeneous Dirichlet boundary conditions is stated. The results are applied to determine the second Cheeger constant of some planar domains.

Equations

$$h_k(\Omega )=\inf \max \{h_1(E_1),\dots ,h_1(E_k)\},$$

$$h_1(\Omega ):=\inf \left\{\frac{P(E)}{|E|}:\text{leavevmode}E\subset \Omega ,\text{leavevmode}|E|>0\right\},$$

$$h_k(\Omega ):=\inf \left\{\underset{i=1,\dots ,k}{\max }\frac{P(E_i)}{|E_i|}:\text{leavevmode}{E}_i\subset \Omega ,\text{leavevmode}|E_i|>0\text{leavevmode}\forall i,\text{leavevmode}{E}_i\cap E_j=\varnothing \text{leavevmode}\forall i\ne j\right\},$$

$$h_k(\Omega ):=\inf \left\{\underset{i=1,\dots ,k}{\max }{h}_1(E_i):\text{leavevmode}{E}_i\subset \Omega ,\text{leavevmode}|E_i|>0\text{leavevmode}\forall i,\text{leavevmode}{E}_i\cap E_j=\varnothing \text{leavevmode}\forall i\ne j\right\}.$$

$$\{\begin{array}{rlrl}-{\Delta }_{p}u& =\lambda |u{|}^{p-2}u& & \text{in }\Omega ,\\ u& =0& & \text{on }\partial \Omega .\end{array}$$

$$0<{\lambda }_1(p;\Omega )<{\lambda }_2(p;\Omega )\le \cdots \le {\lambda }_k(p;\Omega )\to +\infty \text{as }k\to +\infty .$$

$$\underset{p\to 1}{\lim }{\lambda }_1(p;\Omega )=h_1(\Omega ).$$

$$\underset{p\to 1}{\lim }{\lambda }_2(p;\Omega )=h_2(\Omega )$$

$$\underset{p\to 1}{\mathrm{limsup}}{\lambda }_k(p;\Omega )\le h_k(\Omega )\text{for }k\ge 3.$$

$$\underset{p\to 1}{\lim }{\lambda }_k(p;\Omega )={\Lambda }_k(\Omega )\text{for }k\in \mathbb{N},$$

$${\mathfrak{L}}_k(p;\Omega ):=\inf \left\{\underset{i=1,\dots ,k}{\max }{\lambda }_1(p;E_i):\text{leavevmode}{E}_i\subset \Omega ,\text{leavevmode}|E_i|>0\text{leavevmode}\forall i,\text{leavevmode}{E}_i\cap E_j=\varnothing \text{leavevmode}\forall i\ne j\right\},$$

$$\underset{p\to 1}{\lim }{\mathfrak{L}}_k(p;\Omega )=h_k(\Omega ).$$

$${\Lambda }_k^{(p)}(\Omega ):=\inf \left\{\sum _{i=1}^k{\lambda }_1(p;E_i):\text{leavevmode}{E}_i\subset \Omega ,\text{leavevmode}|E_i|>0\text{leavevmode}\forall i,\text{leavevmode}{E}_i\cap E_j=\varnothing \text{leavevmode}\forall i\ne j\right\}$$

$$H_k(\Omega ):=\inf \left\{\sum _{i=1}^k{h}_1(E_i):\text{leavevmode}{E}_i\subset \Omega ,\text{leavevmode}|E_i|>0\text{leavevmode}\forall i,\text{leavevmode}{E}_i\cap E_j=\varnothing \text{leavevmode}\forall i\ne j\right\}.$$

$$h_k(\Omega )=\underset{(E_1,\dots ,E_k)\in {\mathcal{E}}_k}{\inf }\underset{i=1,\dots ,k}{\max }\frac{P(E_i)}{|E_i|}\text{leavevmode}\text{and}\text{leavevmode}{h}_k(\Omega )=\underset{(E_1,\dots ,E_k)\in {\mathcal{E}}_k}{\inf }\underset{i=1,\dots ,k}{\max }{h}_1(E_i),$$

$$h_1(E_i)=\frac{P(E_i)}{|E_i|}\text{for all }i\in \{1,\dots ,k\}.$$

$$h_{1}\biggl{(}\Omega\setminus\bigcup_{j\neq i}{E_{j}}\biggr{)}=h_{1}(E_{i})=\frac{P(E_{i})}{|E_{i}|}\quad\text{for all }i\in\{1,\dots,k\},$$

$$\displaystyle h_{2}\biggl{(}\Omega\setminus\bigcup_{j\neq i_{1},i_{2}}{E_{j}}\biggr{)}=h_{2}(E_{i_{1}}\cup E_{i_{2}})$$

$$\text{for all mutually different }{i}_1,i_2\in \{1,\dots ,k\},$$

$$\displaystyle h_{n}\biggl{(}\Omega\setminus\bigcup_{j\neq i_{1},\dots,i_{n}}{E_{j}}\biggr{)}=h_{n}(E_{i_{1}}\cup\cdots\cup E_{i_{n}})$$

$$\text{for all mutually different }{i}_1,\dots ,i_n\in \{1,\dots ,k\},$$

$$J(E)=\frac{P(E)}{|E|}.$$

$$(E_1,\dots ,E_m,E_{m+1},\dots ,E_k)=(E_1,\dots ,E_m,C_{m+1}^{\prime },\dots ,C_k^{\prime }).$$

$$(E_1,\dots ,E_{i-1},E_i,E_{i+1},\dots ,E_k)=(E_1,\dots ,E_{i-1},C_i^{\prime },E_{i+1},\dots ,E_k).$$

$$h_n\left(\Omega \setminus C_{n+1}\right)<\max \{J(C_1),\dots ,J(C_n)\}\le h_{n+1}(\Omega ).$$

$$h_{l}\biggl{(}\Omega\setminus\bigcup_{j\neq i_{1},\dots,i_{l}}C_{j}\biggr{)}<\max\{J(C_{i_{1}}),\dots,J(C_{i_{l}})\}\leq h_{n+1}(\Omega).$$

$$J(E_{i_1}),\dots ,J(E_{i_l})<h_{n+1}(\Omega ),$$

$$h_{n}\biggl{(}\Omega\setminus\bigcup_{j=n+1}^{k}C_{j}\biggr{)}<\max\{J(C_{1}),\dots,J(C_{n})\}\leq h_{k}(\Omega).$$

$$J(C_1^{\prime }),\dots ,J(C_n^{\prime })<h_k(\Omega ),$$

$$(E_1,\dots ,E_m,E_{m+1},\dots ,E_k)=(C_1^{\prime },\dots ,C_m^{\prime },E_{m+1},\dots ,E_k).$$

$$P(E;A)={\mathcal{H}}^{N-1}({\partial }^{\ast }E\cap A)\le {\mathcal{H}}^{N-1}(\partial E\cap A).$$

Full text

[TABLE]

where the infimum is taken over all $k$-tuples of mutually disjoint subsets of $\Omega$, and $h_{1}(E_{i})$ is the classical Cheeger constant of $E_{i}$. We prove the existence of minimizers satisfying additional “adjustment” conditions and study their properties. A relation between $h_{k}(\Omega)$ and spectral minimal $k$-partitions of $\Omega$ associated with the first eigenvalues of the $p$-Laplacian under homogeneous Dirichlet boundary conditions is stated. The results are applied to determine the second Cheeger constant of some planar domains.

1. Introduction

Let $\Omega\subset\mathbb{R}^{N}$ ($N\geq 2$) be a measurable set with positive Lebesgue measure, i.e., $|\Omega|>0$. The Cheeger problem consists in determining the value of the Cheeger constant of $\Omega$, defined as

[TABLE]

and the associated minimizing sets, each of which is called Cheeger set. Here $P(E):=P(E;\mathbb{R}^{N})$ is the distributional perimeter of $E$ with respect to $\mathbb{R}^{N}$ (cf. [15]). For an introduction to the Cheeger problem we refer to the expository articles [27] and [22], and the references therein.

A natural generalization of the above concept is the higher Cheeger problem, where minimization takes place among $k$-tuples of mutually disjoint subsets of $\Omega$. More precisely, the $k$-th Cheeger constant of $\Omega$ is defined as

[TABLE]

or, in an equivalent way, as

[TABLE]

Besides having a geometric interest on its own, the Cheeger problem, as well as its generalization, arises in the study of the asymptotic behaviour of the eigenvalue problem for the $p$-Laplacian as $p$ tends to $1$. Assume that $\Omega$ is a bounded domain. We say that $\lambda\in{\mathbb{R}}$ is an eigenvalue of the $p$-Laplacian if there exists a nontrivial weak solution $u\in W^{1,p}_{0}(\Omega)$, which is called eigenfunction, to the problem

[TABLE]

Here $p\in(1,+\infty)$, and $\Delta_{p}u:=\text{div}(|\nabla u|^{p-2}\nabla u)$. The existence of sequences of eigenvalues of the $p$-Laplacian can be proven by means of minimax principles, see, e.g., [13, 11]. In this paper, we will focus on the sequence $\{\lambda_{k}(p;\Omega)\}_{k\in\mathbb{N}}$ defined using the Krasnoselskii genus [13, § 5], which satisfies

[TABLE]

As for the behaviour of (3) for $p\to 1$, it was proven in [19, Corollary 6 and Remark 7] that

[TABLE]

A few years later, the result was generalized to higher eigenvalues. In [28, Theorem 5.5] it was proven that

[TABLE]

and, more in general,

[TABLE]

We also mention the paper [25], where the authors showed that

[TABLE]

where $\Lambda_{k}(\Omega)$ is the $k$-th Krasnoselskii eigenvalue of the $1$-Laplace operator (cf. [9]).

The reason of the discrepancy between (4) and (5) is the fact that, while every eigenfunction $u_{2}$ associated to $\lambda_{2}(p;\Omega)$ has exactly two nodal domains (namely, connected components of the set $\{u_{2}\neq 0\}$), an eigenfunction $u_{k}$ associated to $\lambda_{k}(p;\Omega)$ does not have, in general, $k$ nodal domains. Therefore, and also in view of (2), it makes sense to introduce the spectral minimal $k$-partition problem for the $p$-Laplacian as

[TABLE]

see Section 5 below for precise definitions. The existence, regularity and qualitative properties of $\mathfrak{L}_{k}(p;\Omega)$ in the case $p=2$ have been studied intensively nowadays, see, e.g., [5, 10, 17, 18].

One of the main results of this paper is Theorem 5.4, where we prove that $k$-th Cheeger constant $h_{k}(\Omega)$ can be characterized as

[TABLE]

Let us mention that a related spectral partitioning problem was studied in [8]. In that paper, the author investigated the limit as $p\to 1$ of the quantity

[TABLE]

and proved its convergence towards

[TABLE]

Existence and qualitative properties of minimizing $k$-tuples of sets for $H_{k}(\Omega)$ were also comprehensively studied in [8]. On the other hand, the functional $\Lambda_{k}^{(p)}(\Omega)$ in the case $p=2$ also attracted the attention of a significant number of researchers, see, for instance, [5, 10, 7, 4].

Expressions (1) and (2) can be rewritten in shorter forms as

[TABLE]

where $\mathcal{E}_{k}$ is the family of all $k$-tuples of mutually disjoint measurable subsets of $\Omega$ with positive Lebesgue measure. Note that each $E_{i}$ can be assumed to be connected. A natural question is whether the infimum in (7) is actually attained. A first result about the existence of a minimizing $k$-tuple $(E_{1},\dots,E_{k})\in\mathcal{E}_{k}$ for $h_{k}(\Omega)$ is proved in [28, Theorem 3.1]. In what follows, a minimizer of $h_{k}(\Omega)$ will be called a $k$-tuple of multiple Cheeger sets, or, simply, a Cheeger $k$-tuple of $\Omega$. If $k=2$, we also call minimizers $E_{1}$ and $E_{2}$ coupled Cheeger sets. In general, a $k$-tuple of multiple Cheeger sets for $h_{k}(\Omega)$ need not be unique, as the following example shows111The example, as well as Figure 1, has only an illustrative purpose, since the various assertions are not rigorously proven.. Set $k=3$ and let $\Omega$ be a union of a square $K$, a disc $B$, and a negligibly thin channel $T$ which connects $K$ and $B$, see Fig. 1. On the one hand, we can take a radius of $B$ sufficiently small to get $h_{2}(K)<h_{1}(B)$. On the other hand, we can take a radius of $B$ sufficiently large to get $h_{1}(B)<h_{3}(K)$. Thus, $h_{3}(\Omega)=h_{1}(B)$ and we have some freedom to vary sets $E_{2},E_{3}\subset K$ of a minimizer $(E_{1},E_{2},E_{3})\in\mathcal{E}_{3}$ for $h_{3}(\Omega)$ without loss of minimizing property.

The above example also indicates that we cannot expect from an arbitrary taken Cheeger $k$-tuple of $\Omega$ to be regular enough. With this respect, it is reasonable to look for minimizers of $h_{k}(\Omega)$ satisfying some additional adjustment conditions. A first guess could be to require all sets of the Cheeger $k$-tuple to be calibrable (cf. [2]), namely, each $E_{i}$ is a Cheeger set of itself:

[TABLE]

However, this is not completely satisfactory, as can be seen in the example in Fig. 1. A more suitable requirement is defined as follows. We say that a minimizer $(E_{1},\dots,E_{k})\in\mathcal{E}_{k}$ of $h_{k}(\Omega)$ is a $1$-adjusted Cheeger $k$-tuple if

[TABLE]

that is, each $E_{i}$ is a Cheeger set of $\Omega\setminus\bigcup_{j\neq i}{E_{j}}$.

Analogously, we say that a minimizer $(E_{1},\dots,E_{k})\in\mathcal{E}_{k}$ of $h_{k}(\Omega)$ is a $2$-adjusted Cheeger $k$-tuple if it is $1$-adjusted and, additionally,

[TABLE]

that is, each pair $(E_{i_{1}},E_{i_{2}})$ is a $1$-adjusted Cheeger couple of $\Omega\setminus\bigcup_{j\neq i_{1},i_{2}}{E_{j}}$.

Proceeding in this way, we say that a minimizer $(E_{1},\dots,E_{k})\in\mathcal{E}_{k}$ of $h_{k}(\Omega)$ is a $n$-adjusted Cheeger $k$-tuple for $n\in\{2,\dots,k\}$ if it is $(n-1)$-adjusted and, additionally,

[TABLE]

that is, each $n$-tuple $(E_{i_{1}},\dots,E_{i_{n}})$ is a $(n-1)$-adjusted Cheeger $n$-tuple of $\Omega\setminus\bigcup_{j\neq i_{1},\dots,i_{n}}{E_{j}}$.

Equivalently, a minimizer $(E_{1},\dots,E_{k})\in\mathcal{E}_{k}$ of $h_{k}(\Omega)$ is a $n$-adjusted Cheeger $k$-tuple for $n\in\{1,\dots,k\}$ if any $l$-tuple $(E_{i_{1}},\dots,E_{i_{l}})\subset(E_{1},\dots,E_{k})$ with arbitrary $l\in\{1,\dots,n\}$ is a minimizer of $h_{l}\left(\Omega\setminus\bigcup_{j\neq i_{1},\dots,i_{l}}{E_{j}}\right)$. Moreover, it is easy to see from the definitions that any $(k-1)$-adjusted Cheeger $k$-tuple is, in fact, $k$-adjusted.

Let us remark that these adjustment conditions are not necessary for the problem defined in (6), since in this case the Cheeger constant of each component $E_{i}$ of a minimizer $(E_{1},\dots,E_{k})$ contributes to the value of $H_{k}(\Omega)$, while in our problem $h_{k}(\Omega)$ is defined via the maximal Cheeger constant only.

As for the existence of adjusted Cheeger $k$-tuples, our main result is Theorem 2.1, where we prove that, for any $k\in\mathbb{N}$, and for any $n\in\{1,\dots,k\}$, there exists a $n$-adjusted Cheeger $k$-tuple.

The advantage of working with adjusted Cheeger $k$-tuples is the fact that they satisfy the usual regularity properties of perimeter-minimizing sets, which are stated in detail in Theorem 3.8. In particular, each set of any $1$-adjusted Cheeger $k$-tuple can be approximated from the inside by a sequence of smooth sets in a suitable sense (see Proposition 3.10). This result plays a crucial role for the characterization of $h_{k}(\Omega)$ as a limit of spectral minimal partitions, see Theorem 5.4.

Qualitative properties of Cheeger $k$-tuples were also investigated in [28], although with a few imprecisions. In particular, [28, Theorem 3.9] is not correct, since the proof uses the erroneous fact that the perimeter of both Cheeger sets increases under volume-preserving deformations, which is in general not true. We correct that statement by our Proposition 4.7.

The paper is structured as follows. After proving the existence of adjusted Cheeger $k$-tuples for $h_{k}(\Omega)$ by means of inductive arguments, in Section 3 we investigate their regularity properties, and in Section 4 we study further qualitative properties. In Section 5 we consider the spectral minimal partition problem for the $p$-Laplacian and study its limit as $p\to 1$. The last section is devoted to the computation of $h_{2}(\Omega)$ when $\Omega$ is a disc or an annulus.

2. Existence of adjusted Cheeger $k$-tuples

Theorem 2.1**.**

Let $\Omega\subset\mathbb{R}^{N}$ be a bounded measurable set with positive Lebesgue measure. Then, for any $k\in\mathbb{N}$ and $n\in\{1,\dots,k\}$, there exists a $n$-adjusted Cheeger $k$-tuple.

Proof.

Let us denote, for simplicity,

[TABLE]

The proof is combinatorial and based on inductive arguments. Let us outline its scheme:

1. (I)

Show the existence of a $1$-adjusted Cheeger $k$-tuple for any $k\geq 1$. This is the base of induction with respect to the adjustment order. 2. (II)

Fix $n\geq 2$ and suppose that for any $n^{\prime}<n$ there exists a $n^{\prime}$-adjusted Cheeger $k$-tuple for any $k\geq n^{\prime}$. This is the induction hypothesis. Then, prove the existence of a $n$-adjusted Cheeger $k$-tuple for any $k\geq n$. This is the inductive step.

In the steps (I) and (II) we again apply the induction as follows:

1. (1)

Fix $n\geq 1$ and show the existence of a $n$-adjusted Cheeger $(n+1)$-tuple. (Note that the existence of a $n$-adjusted Cheeger $n$-tuple is obvious: if $n=1$, then it is trivial; if $n\geq 2$, then it simply follows from the existence of a $(n-1)$-adjusted Cheeger $n$-tuple, which we know by the induction hypothesis of (II)). This is the base of induction with respect to the index of the higher Cheeger constant. 2. (2)

Fix $k\geq n+1$ and suppose that for any $k^{\prime}\in\{n,\dots,k-1\}$ there exists a $n$-adjusted Cheeger $k^{\prime}$-tuple. This is the induction hypothesis. Then, prove the existence of a $n$-adjusted Cheeger $k$-tuple. This is the inductive step.

Let us emphasize that the induction hypotheses are assumed to be satisfied for arbitrary bounded measurable $\Omega$ with positive Lebesgue measure.

We now turn to details.

(I) First we consider the case of $1$-adjusted Cheeger $k$-tuples.

(1) Let $k=2$, and let $(C_{1},C_{2})$ be a minimizer of $h_{2}(\Omega)$. Without loss of generality, we can assume that $J(C_{1})=h_{2}(\Omega)$. Observe that this does not mean, in general, that $h_{1}(\Omega\setminus C_{2})=h_{1}(C_{1})=J(C_{1})$. We will distinguish two cases:

1. (a)

$J(C_{2})<h_{2}(\Omega)$. In this case, $C_{1}$ must be a Cheeger set of $\Omega\setminus C_{2}$. Then, let $C_{2}^{\prime}$ be a Cheeger set of $\Omega\setminus C_{1}$. It evidently satisfies $J(C_{2}^{\prime})\leq J(C_{2})<h_{2}(\Omega)$. By definition, $C_{1}$ is automatically a Cheeger set for $\Omega\setminus C_{2}^{\prime}$. Hence, $(C_{1},C_{2}^{\prime})$ is a $1$-adjusted Cheeger couple of $\Omega$. 2. (b)

$J(C_{2})=h_{2}(\Omega)$. As before, consider a Cheeger set $C_{2}^{\prime}$ of $\Omega\setminus C_{1}$: if $J(C_{2}^{\prime})<h_{2}(\Omega)$, we fall into the previous case. Otherwise, either $C_{1}$ is a Cheeger set of $\Omega\setminus C_{2}^{\prime}$, and we are done; or this is not the case, and we take $C_{1}^{\prime}$ to be a Cheeger set of $\Omega\setminus C_{2}^{\prime}$. This last set will satisfy $J(C_{1}^{\prime})<h_{2}(\Omega)$, and we fall again into the case (a).

(2) Fix a natural number $k>2$ and suppose that the claim is true for every $k^{\prime}<k$. We will show that it is true also for $k$. Let $(C_{1},\dots,C_{k})$ be a minimizer of $h_{k}(\Omega)$. We proceed as follows. Set $(E_{1},\dots,E_{k})=(C_{1},\dots,C_{k})$.

1. (a)

Without loss of generality, we can suppose that $J(E_{1})=\dots=J(E_{m})=h_{k}(\Omega)$ for some $m\in\{1,\dots,k\}$, and $J(E_{m+1}),\dots,J(E_{k})<h_{k}(\Omega)$ whenever $m<k$. 2. (b)

If $m<k$, then, by the induction hypothesis, there exists a $1$-adjusted Cheeger $(k-m)$-tuple $(C_{m+1}^{\prime},\dots,C_{k}^{\prime})$ corresponding to $h_{k-m}(\Omega\setminus\bigcup_{j=1}^{m}E_{j})$. By definition, we will have $J(C_{i}^{\prime})<h_{k}(\Omega)$ for every $i\in\{m+1,\dots,k\}$. Consider the new $k$-tuple

[TABLE]

It is not hard to see that $E_{i}$ is a Cheeger set of $\Omega\setminus\bigcup_{j\neq i}E_{j}$ for any $i\in\{m+1,\dots,k\}$. 3. (c)

Set $i=1$. 4. (d)

If $i\leq m$ and $E_{i}$ is a Cheeger set of $\Omega\setminus\bigcup_{j\neq i}E_{j}$, then put $i=i+1$ and repeat step (d). 5. (e)

If $i\leq m$ and $E_{i}$ is not a Cheeger set of $\Omega\setminus\bigcup_{j\neq i}E_{j}$, then let $C^{\prime}_{i}$ be such a Cheeger set, and consider the new $k$-tuple

[TABLE]

Observe that $J(E_{i})<h_{k}(\Omega)$ and hence the number of sets $E_{j}$ such that $J(E_{j})=h_{k}(\Omega)$ has been decreased by one unit. Go to step (a).

At the end of this procedure, after a finite number of steps, we will obtain a $1$-adjusted Cheeger $k$-tuple of $\Omega$.

(II) Fix some natural $n\geq 2$ and suppose that for any $n^{\prime}<n$ and $k\geq n^{\prime}$ there exists a $n^{\prime}$-adjusted Cheeger $k$-tuple. Let us show now the existence of a $n$-adjusted Cheeger $k$-tuple for any $k\geq n$. Observe that the existence of a $n$-adjusted Cheeger $n$-tuple trivially follows from the existence of a $(n-1)$-adjusted Cheeger $n$-tuple. Therefore, it is enough to suppose that $k\geq n+1$.

(1) First we show the existence of a $n$-adjusted Cheeger $(n+1)$-tuple. Let $(C_{1},\dots,C_{n+1})$ be a minimizer of $h_{n+1}(\Omega)$. By the induction hypothesis with respect to the adjustment order, we can assume that $(C_{1},\dots,C_{n+1})$ is a $(n-1)$-adjusted Cheeger $(n+1)$-tuple corresponding to $h_{n+1}(\Omega)$. Suppose that $(C_{1},\dots,C_{n+1})$ is not $n$-adjusted. Therefore, there exists a $n$-subtuple of $(C_{1},\dots,C_{n+1})$, say, $(C_{1},\dots,C_{n})$, such that

[TABLE]

Let us substitute $(C_{1},\dots,C_{n})$ by a $n$-adjusted Cheeger $n$-tuple $(C_{1}^{\prime},\dots,C_{n}^{\prime})$ corresponding to $h_{n}(\Omega\setminus C_{n+1})$. Then we get $J(C_{1}^{\prime}),\dots,J(C_{n}^{\prime})<J(C_{n+1})=h_{n+1}(\Omega)$.

Let us show that $(C_{1}^{\prime},\dots,C_{n}^{\prime},C_{n+1})$ is a $n$-adjusted Cheeger $(n+1)$-tuple of $\Omega$. Omit, for simplicity, the superscript ′, and suppose, by contradiction, that there exists a $l$-tuple $(C_{i_{1}},\dots,C_{i_{l}})$ with some $l\in\{1,\dots,n\}$ such that

[TABLE]

Recalling that $(C_{1},\dots,C_{n})$ is a $n$-adjusted Cheeger $n$-tuple of $\Omega\setminus C_{n+1}$, we see that the $l$-tuple $(C_{i_{1}},\dots,C_{i_{l}})$ must necessarily contain $C_{n+1}$. Therefore, if $(E_{i_{1}},\dots,E_{i_{l}})$ is an arbitrary Cheeger $l$-tuple of the set $\Omega\setminus\bigcup_{j\neq i_{1},\dots,i_{l}}C_{j}$, we get a contradiction to the definition of $h_{n+1}(\Omega)$, since

[TABLE]

and $J(C_{i})<h_{n+1}(\Omega)$ for all $i\in\{1,\dots,n+1\}$ satisfying $i\neq i_{j}$ with $j\in\{1,\dots,l\}$.

(2) Fix an arbitrary $k\geq n+1$ and suppose that for any $k^{\prime}\in\{n,\dots,k-1\}$ there exists a $n$-adjusted Cheeger $k^{\prime}$-tuple. Let us prove the existence of a $n$-adjusted Cheeger $k$-tuple.

If $k=n+1$, this is easy to prove, so we can suppose that $k\geq n+2$. Let $(C_{1},\dots,C_{k})$ be a $(n-1)$-adjusted Cheeger $k$-tuple of $\Omega$ which exists by the induction hypothesis with respect to the adjustment order stated in (II). Suppose that $(C_{1},\dots,C_{k})$ is not $n$-adjusted, that is, there exits, say, $(C_{1},\dots,C_{n})$, such that

[TABLE]

Let $(C_{1}^{\prime},\dots,C_{n}^{\prime})$ be a corresponding $n$-adjusted Cheeger $n$-tuple of $\Omega\setminus\bigcup_{j=n+1}^{k}C_{j}$. Note that

[TABLE]

that is, there are at least $n$ elements of $(J(C_{1}^{\prime}),\dots,J(C_{n}^{\prime}),J(C_{n+1}),\dots,J(C_{k}))$ which are strictly smaller than $h_{k}(\Omega)$. We omit, for simplicity, the superscript ′, and proceed as follows. Set $(E_{1},\dots,E_{k})=(C_{1},\dots,C_{k})$.

1. (a)

Without loss of generality, we can suppose that $J(E_{1}),\dots,J(E_{m})<h_{k}(\Omega)$ for some $m\in\{n,\dots,k-1\}$, and $J(E_{m+1})=\dots=J(E_{k})=h_{k}(\Omega)$. 2. (b)

By the induction hypothesis, there exists a $n$-adjusted Cheeger $m$-tuple $(C_{1}^{\prime},\dots,C_{m}^{\prime})$ corresponding to $h_{m}(\Omega\setminus\bigcup_{j=m+1}^{k}E_{j})$. By definition, we will have $J(C_{i}^{\prime})<h_{k}(\Omega)$ for every $i\in\{1,\dots,m\}$. Consider the new $k$-tuple

[TABLE]

It is not hard to see that any $l$-tuple $(E_{i_{1}},\dots,E_{i_{l}})\subset(E_{1},\dots,E_{m})$ is a $l$-adjusted Cheeger $l$-tuple of $\Omega\setminus\bigcup_{j\neq i_{1},\dots,i_{l}}E_{j}$. 3. (c)

Assume that there exists $l\in\{1,\dots,n\}$ and a $l$-tuple $(E_{i_{1}},\dots,E_{i_{l}})\subset(E_{1},\dots,E_{k})$ such that $(E_{i_{1}},\dots,E_{i_{l}})$ is not a minimizer of $h_{l}\left(\Omega\setminus\bigcup_{j\neq i_{1},\dots,i_{l}}E_{j}\right)$. Note that $(E_{i_{1}},\dots,E_{i_{l}})\not\subset(E_{1},\dots,E_{m})$, as it follows from step (b). Therefore, there exists $j\in\{1,\dots,l\}$ such that $E_{i_{j}}\in(E_{m+1},\dots,E_{k})$, i.e., $J(E_{i_{j}})=h_{k}(\Omega)$. Let now $(C_{i_{1}}^{\prime},\dots,C_{i_{l}}^{\prime})$ be any minimizer of $h_{l}\left(\Omega\setminus\bigcup_{j\neq i_{1},\dots,i_{l}}E_{j}\right)$. Then we observe that $J(C_{i_{j}}^{\prime})<h_{k}(\Omega)$. Consider a new $k$-tuple $(E_{1},\dots,E_{k})$ obtained by replacing $(E_{i_{1}},\dots,E_{i_{l}})$ with $(C_{i_{1}}^{\prime},\dots,C_{i_{l}}^{\prime})$. Hence, the number of sets $E_{j}$ such that $J(E_{j})=h_{k}(\Omega)$ has been decreased at least by one unit. Go to step (a).

At the end of this procedure, after a finite number of steps, we will obtain a $n$-adjusted Cheeger $k$-tuple of $\Omega$. ∎

3. Regularity of adjusted Cheeger $k$-tuples

Let $(E_{1},\dots,E_{k})\in\mathcal{E}_{k}$ be a $1$-adjusted Cheeger $k$-tuple of $\Omega$. In this section we prove some regularity properties of $E_{i}$ by adapting the results obtained by Caroccia in [8]. Throughout this section we assume that $\Omega$ is a bounded open set.

We denote by $\mathcal{H}^{N-1}$ the $(N-1)$-dimensional Hausdorff measure. Given two Borel sets $E$ and $F$, we will write $E\approx F$ whenever $\mathcal{H}^{N-1}(E\bigtriangleup F)=0$, where $E\bigtriangleup F$ stands for the symmetric difference between $E$ and $F$. Given a Borel set $A$ and a set of finite perimeter $E$, we denote by $P(E;A)$ the perimeter of $E$ measured with respect to $A$. Recall also that, for the sake of simplicity, we write $P(E):=P(E;{\mathbb{R}}^{N})$. We denote by $\partial^{*}E$ the reduced boundary of $E$ (see [26, p. 167]). We recall that

[TABLE]

As an immediate consequence, if we denote the complement of a set $A$ in ${\mathbb{R}}^{N}$ by $A^{c}$, we have the basic relation

[TABLE]

For $\alpha\in[0,1]$, we denote by $E^{(\alpha)}$ the set of points of Lebesgue density $s$, namely,

[TABLE]

The set $E^{(1)}$ is called essential interior of $E$, the set $E^{(0)}$ is the essential exterior of $E$, and $\partial^{e}E={\mathbb{R}}^{N}\setminus(E^{(0)}\cup E^{(1)})$ is the essential boundary of $E$. By Federer’s Theorem (cf. [26, Theorem 16.2]), we have

[TABLE]

This implies in particular that, for every set of finite perimeter $E$,

[TABLE]

Lemma 3.1**.**

Let $E$ and $F$ be sets of finite perimeter and $A$ be a Borel set. Then

[TABLE]

Proof.

Define the set

[TABLE]

By [26, Theorem 16.3] we have

[TABLE]

where we used relation (8). ∎

Lemma 3.2**.**

Let $E$, $F$ and $L$ be sets of finite perimeter and $|E\cap F|=0$. Then

[TABLE]

Proof.

By [26, Theorem 16.3] we have

[TABLE]

Using the fact that $\{\nu_{L}=\nu_{E}\}$ and $\{\nu_{L}=\nu_{F}\}$ are subsets of $\partial^{*}L\cap\partial^{*}E$ and $\partial^{*}L\cap\partial^{*}F$ respectively, and that $D^{(1)}\cap\partial^{*}D=\emptyset$ for every set of finite perimeter $D$, we obtain

[TABLE]

Since $|E\cap F|=0$, it holds $E^{(1)}\subset F^{(0)}$ and $F^{(1)}\subset E^{(0)}$, and therefore

[TABLE]

by Federer’s Theorem. Finally, if $x\in\partial^{*}E\cap\partial^{*}F$, then it is a point of density $\frac{1}{2}$ for both $E$ and $F$. Since $|E\cap F|=0$, it follows that $\nu_{E}(x)=-\nu_{F}(x)$ (see [26, Exercice 12.9]), and therefore

[TABLE]

Hence we obtain the claim. ∎

Definition 3.3**.**

A set $M\subset\Omega$ has distributional mean curvature bounded from above by $g\in L^{1}_{loc}(\Omega)$ in $\Omega$ if there exists $r_{0}>0$ such that, for every $B_{r}\subset\subset\Omega$ with $r\in(0,r_{0})$, and for every $L\subset M$ with $M\setminus L\subset\subset B_{r}$, it holds

[TABLE]

Definition 3.4**.**

Let $\Lambda$, $r_{0}>0$. We say that a set of finite perimeter $E$ is $(\Lambda,r_{0})$-perimeter minimizing in $\Omega$ if, for every $B_{r}\subset\subset\Omega$ with $r\in(0,r_{0})$, and for every set of finite perimeter $F$ with $E\bigtriangleup F\subset\subset B_{r}$, it holds

[TABLE]

Lemma 3.5**.**

Let $(E_{1},\dots,E_{k})\in\mathcal{E}_{k}$ be a $1$-adjusted Cheeger $k$-tuple of $\Omega$, and define

[TABLE]

Then, each $M_{i}$ has distributional mean curvature bounded from above by $h_{k}(\Omega)$, namely, for every $L\subset M_{i}$ with $M_{i}\setminus L\subset\subset B_{r}$, where $B_{r}$ is a ball of radius $r<\frac{1}{h_{k}(\Omega)}$, it holds

[TABLE]

Proof.

Set $r_{0}=\frac{1}{h_{k}(\Omega)}$. Let $B_{r}\subset\subset\Omega$ be a ball of radius $r\in(0,r_{0})$, and let $L\subset M_{i}$ be a set of finite perimeter such that $M_{i}\setminus L\subset\subset B_{r}$. Define $F_{j}:=E_{j}\cap L$. By our choice of $r_{0}$, it holds $|F_{j}|>0$ for every $j\neq i$; if this was not the case, then there would exist a set $E_{j}$ such that, up to negligible sets, $E_{j}\subset M_{i}\setminus L\subset\subset B_{r}$, and therefore

[TABLE]

a contradiction. Since $F_{j}\subset\Omega\setminus M_{j}$ for every $j\neq i$, and $E_{j}$ is a Cheeger set of $\Omega\setminus M_{j}$ (by $1$-adjustment assumption), it holds

[TABLE]

and therefore

[TABLE]

which implies

[TABLE]

Using [8, Lemma 3.3], we obtain

[TABLE]

Moreover, applying again [8, Lemma 3.3], we get

[TABLE]

where we used Lemma 3.2 and the fact that $M_{i}\cap L=L$. The above inequalities finally give

[TABLE]

∎

Remark 3.6**.**

Reasoning in a similar way, one can show that each Cheeger set $E_{i}$, $i\in\{1,\dots,k\}$, has distributional mean curvature bounded from above by $h_{k}(\Omega)$ in ${\mathbb{R}}^{N}$.

Proposition 3.7**.**

Let $(E_{1},\dots,E_{k})\in\mathcal{E}_{k}$ be a $1$-adjusted Cheeger $k$-tuple of $\Omega$. Then, each $E_{i}$ is $(\Lambda,r_{0})$-perimeter minimizing in $\Omega$ for $\Lambda=h_{k}(\Omega)$ and $r_{0}=\frac{1}{h_{k}(\Omega)}$.

Proof.

As in Lemma 3.5, define

[TABLE]

Let $r_{0}=\frac{1}{h_{k}(\Omega)}$. Let $B_{r}\subset\subset\Omega$ be a ball of radius $r\in(0,r_{0})$, and let $F_{i}$ be such that $F_{i}\bigtriangleup E_{i}\subset\subset B_{r}$. Define $D_{i}:=F_{i}\setminus M_{i}$ and observe that, by the definition of a $1$-adjusted Cheeger $k$-tuple,

[TABLE]

Therefore, noting that $E_{i}=D_{i}=F_{i}$ in $B_{r}^{c}$, we get

[TABLE]

and hence

[TABLE]

Observing that $F_{i}\cap M_{i}\subset F_{i}\setminus E_{i}$, we obtain

[TABLE]

and hence

[TABLE]

On the other hand, using the fact that $M_{i}\setminus F_{i}\subset M_{i}$ and $M_{i}\setminus(M_{i}\setminus F_{i})=M_{i}\cap F_{i}\subset\subset B_{r}$, by Lemma 3.5 we get

[TABLE]

Therefore, by Lemma 3.1,

[TABLE]

Finally, combining (9) and (10), we obtain

[TABLE]

which proves that $E_{i}$ is $(\Lambda,r_{0})$-perimeter minimizing in $\Omega$ with $\Lambda=h_{k}(\Omega)$ and $r_{0}=\frac{1}{h_{k}(\Omega)}$. ∎

Theorem 3.8**.**

Let $(E_{1},\dots,E_{k})\in\mathcal{E}_{k}$ be a $1$-adjusted Cheeger $k$-tuple of $\Omega$. Then, for each $i\in\{1,\dots,k\}$, the following assertions hold:

1. (i)

$\partial^{*}E_{i}\cap\Omega$* is of class $C^{1,\gamma}$ for every $\gamma\in(0,\frac{1}{2})$.* 2. (ii)

The set $(\partial E_{i}\setminus\partial^{*}E_{i})\cap\Omega$ has Hausdorff dimension at most $N-8$. 3. (iii)

If $N\leq 7$, then $\partial E_{i}\cap\Omega$ is of class $C^{1,\gamma}$ for every $\gamma\in(0,\frac{1}{2})$. 4. (iv)

Suppose that $\mathcal{H}^{N-1}(\partial\Omega)<+\infty$. Then there exists an open set $\widetilde{E}_{i}$ such that $|E_{i}\bigtriangleup\widetilde{E}_{i}|=0$. Moreover, $(\widetilde{E}_{1},\dots,\widetilde{E}_{k})$ is a $1$-adjusted Cheeger $k$-tuple. 5. (v)

Suppose that $\Omega$ has finite perimeter. Then $\partial E_{i}\cap\Omega$ can meet $\partial^{*}\Omega$ only in a tangential way, that is, if $x\in\partial^{*}\Omega\cap\partial E_{i}$, then $x\in\partial^{*}E_{i}$, and $\nu_{\Omega}(x)=\nu_{E_{i}}(x)$.

Proof.

The proof of (i) and (ii) follows from the fact that each $E_{i}$ is $(\Lambda,r_{0})$-perimeter minimizing in $\Omega$, and from classical regularity results, cf. [26, Theorems 21.8 and 28.1]. Assertion (iii) easily follows from (i) and (ii). Let us now prove (iv). Since $\mathcal{H}^{N-1}(\partial\Omega)<+\infty$, by (ii) we have that the topological boundary of each $E_{i}$ has Hausdorff dimension $N-1$. If we define $\widetilde{E}_{i}:=E_{i}\setminus\partial E_{i}$, then $\widetilde{E}_{i}$ is an open set such that $P(\widetilde{E}_{i})=P(E_{i})$ and $|\widetilde{E}_{i}|=|E_{i}|$. Therefore, $(\widetilde{E}_{1},\dots,\widetilde{E}_{k})$ is a $1$-adjusted Cheeger $k$-tuple of $\Omega$. Finally, assertion (v) can be proven as in [24, Appendix A]. ∎

Remark 3.9**.**

If $\Omega$ has a boundary of class $C^{1}$, and $N\leq 7$, then Theorem 3.8 implies that the boundary of each $1$-adjusted Cheeger set $E_{i}$ is of class $C^{1}$ as well. Concerning the boundary regularity, we also refer to [14, Theorem 3].

Proposition 3.10**.**

Suppose that $\mathcal{H}^{N-1}(\partial\Omega\setminus\partial^{*}\Omega)=0$. Let $(E_{1},\dots,E_{k})\in\mathcal{E}_{k}$ be a $1$-adjusted Cheeger $k$-tuple of $\Omega$ such that every $E_{i}$ is open. Then, for every $i\in\{1,\dots,k\}$, there exists a sequence $\{E_{i,m}\}_{m\in\mathbb{N}}$ such that:

1. (i)

$E_{i,m}\subset\subset E_{i}$* for every $m\in\mathbb{N}$;* 2. (ii)

$\partial E_{i,m}$* is smooth for every $m\in\mathbb{N}$;* 3. (iii)

$E_{i,m}\to E_{i}$* in $L^{1}(\Omega)$ as $m\to+\infty$;* 4. (iv)

$P(E_{i,m})\to P(E_{i})$* as $m\to+\infty$.*

Proof.

By assertion (ii) of Theorem 3.8, each $E_{i}$ satisfies $\mathcal{H}^{N-1}((\partial E_{i}\cap\Omega)\setminus\partial^{*}E_{i})=0$. Moreover, by assertion (v) of Theorem 3.8, $\partial^{*}\Omega\cap\partial E_{i}\subset\partial^{*}E_{i}$. Therefore

[TABLE]

Therefore, we can apply the approximation result of [29, Theorem 1.1] to obtain the desired claims. ∎

4. Properties of Cheeger $k$-tuples

In this section we prove some qualitative properties of Cheeger $k$-tuples $(E_{1},\dots,E_{k})\in\mathcal{E}_{k}$ of $\Omega$. Throughout this section we assume that $\Omega$ is a bounded open set. First we introduce several notations.

Definition 4.1**.**

The free boundary of $E_{i}$ is

[TABLE]

Definition 4.2**.**

The contact surface between $E_{i}$ and $E_{j}$ (for $i\neq j$) is

[TABLE]

Definition 4.3**.**

The boundary surface of $E_{i}$ is the contact surface between $E_{i}$ and $\partial\Omega$, that is,

[TABLE]

If $(E_{1},\dots,E_{k})$ is a $1$-adjusted Cheeger $k$-tuple of $\Omega$, then, in view of Theorem 3.8, the following decomposition takes place:

[TABLE]

We will denote by $\partial_{f}^{*}E_{i}$ and $\partial^{*}(E_{i}E_{j})$ the reduced part of $\partial_{f}E_{i}$ and $\partial(E_{i}E_{j})$, respectively.

The following results are a consequence of [14, Theorem 2].

Proposition 4.4**.**

Let $(E_{1},\dots,E_{k})\in\mathcal{E}_{k}$ be a $1$-adjusted Cheeger $k$-tuple of $\Omega$. Then the mean curvature of $\partial_{f}^{*}E_{i}$ (measured from the inside of $E_{i}$) is a constant equal to $\frac{h_{1}(E_{i})}{N-1}$ for every $i\in\{1,\dots,k\}$.

Proposition 4.5**.**

Let $(E_{1},\dots,E_{k})\in\mathcal{E}_{k}$ be a $2$-adjusted Cheeger $k$-tuple of $\Omega$. Then the mean curvature of $\partial^{*}(E_{i}E_{j})$ (for $i\neq j$) is constant.

Remark 4.6**.**

Note that the result of Proposition 4.5 can be false for $1$-adjusted Cheeger $k$-tuples, see Fig. 1, c).

Hereinafter, for a $2$-adjusted Cheeger $k$-tuple $(E_{1},\dots,E_{k})$ of $\Omega$, we denote by $c_{ij}$ the mean curvature of $\partial^{*}(E_{i}E_{j})$ measured from the inside of $E_{i}$.

Proposition 4.7**.**

Let $E_{i}$ and $E_{j}$ be any elements of a $2$-adjusted Cheeger $k$-tuple $(E_{1},\dots,E_{k})$ of $\Omega$ such that $\partial^{*}(E_{i}E_{j})\neq\emptyset$. Assume that $c_{ij}\geq 0$. Then $h_{1}(E_{i})\geq h_{1}(E_{j})$ and $h_{1}(E_{i})\geq c_{ij}$. Moreover, if $h_{1}(E_{i})>h_{1}(E_{j})$, then $h_{1}(E_{i})=c_{ij}$.

Proof.

Let us show first that $h_{1}(E_{i})\geq c_{ij}$. If $c_{ij}=0$, then the result is obvious. Assume that $c_{ij}>0$. Suppose, by contradiction, that $h_{1}(E_{i})<c_{ij}$. Take any $x\in\partial^{*}(E_{i}E_{j})$. By Theorem 3.8 (i), there is a neighborhood of $x$ where $\partial^{*}(E_{i}E_{j})$ can be described as a graph of a function $u\in C^{1,\gamma}(\omega)$, where $\omega$ is an open subset of $\mathbb{R}^{N-1}$. Let us take any $\varphi\in C_{0}^{\infty}(\omega)\setminus\{0\}$ such that $\varphi\geq 0$, and let us perturb $\partial^{*}(E_{i}E_{j})$ by $-\varepsilon\varphi$ for $\varepsilon>0$ small enough, so that we obtain two new sets $E_{i}^{\varepsilon}$ and $E_{j}^{\varepsilon}$. The quotient between the perimeter and the volume of $E_{i}^{\varepsilon}$ satisfies the relation

[TABLE]

Since $\partial^{*}(E_{i}E_{j})$ has a constant mean curvature $c_{ij}$, we have

[TABLE]

Therefore, recalling that $\frac{P(E_{i})}{|E_{i}|}=h_{1}(E_{i})<c_{ij}$, we easily deduce that

[TABLE]

On the other hand, applying the same perturbation $-\varepsilon\varphi$ to $E_{j}$, we see that $P(E_{j}^{\varepsilon})$ decreases and $|E_{j}^{\varepsilon}|$ increases, which implies that

[TABLE]

However, (11) and (12) contradict the fact that $(E_{i},E_{j})$ is a minimizer of $h_{2}\left(\Omega\setminus\bigcup_{m\neq i,j}E_{m}\right)$, since $(E_{1},\dots,E_{k})$ is a $2$-adjusted Cheeger $k$-tuple of $\Omega$. Therefore, $h_{1}(E_{i})\geq c_{ij}$.

Let us show now that $h_{1}(E_{i})\geq h_{1}(E_{j})$. Suppose, by contradiction, that $h_{1}(E_{i})<h_{1}(E_{j})$. Then $h_{2}\left(\Omega\setminus\bigcup_{m\neq i,j}E_{m}\right)=h_{1}(E_{j})$. However, applying the perturbation $-\varepsilon\varphi$ as above, we see from (12) that

[TABLE]

for sufficiently small $\varepsilon>0$. A contradiction.

Let us show finally that if $h_{1}(E_{i})>h_{1}(E_{j})$, then $h_{1}(E_{i})=c_{ij}$. Suppose, by contradiction, that $h_{1}(E_{i})>c_{ij}$. Using the perturbation argument as above, but with a positive perturbation $\varepsilon\varphi$ we get

[TABLE]

for sufficiently small $\varepsilon>0$. However, we again get a contradiction as in (13), since $(E_{1},\dots,E_{k})$ is a $2$-adjusted Cheeger $k$-tuple of $\Omega$. ∎

5. Spectral minimal partitions

In this section we show that $h_{k}(\Omega)$ can be characterized as a limit of the energy $\mathfrak{L}(p;\Omega)$ of the spectral minimal partition of $\Omega$ with respect to the $p$-Laplacian as $p\to 1$. Along the whole section, we will assume that $\Omega\subset\mathbb{R}^{N}$ is a bounded open set.

Let $E$ be a measurable subset of $\Omega$. We define the first eigenvalue of the $p$-Laplacian on $E$ as

[TABLE]

and we put $\lambda_{1}(p;E)=+\infty$ whenever the admissible set of functions is empty. Since the constraint $u=0$ a.e. on $\Omega\setminus E$ is weakly compact, we get the existence of a minimizer (i.e., eigenfunction) of $\lambda_{1}(p;E)$. Note that if $E$ is open and $\partial E$ is continuous, then $\lambda_{1}(p;E)$ is the usual first eigenvalue of the $p$-Laplacian on $E$, cf. [1, Theorem 5.29].

Let us define $\mathfrak{L}(p;\Omega)$ as

[TABLE]

A $k$-tuple $(E_{1},\dots,E_{k})\in\mathcal{E}_{k}$ which delivers a minimum to $\mathfrak{L}_{k}(p;\Omega)$ is called a $k$-th spectral minimal partition of $\Omega$. We start with the existence result for $\mathfrak{L}(p;\Omega)$.

Proposition 5.1**.**

$\mathfrak{L}_{k}(p;\Omega)$* is attained for any $k\in\mathbb{N}$ and $p>1$, that is, there exists a $k$-th spectral minimal partition of $\Omega$.*

Proof.

First we introduce the following auxiliary minimization problem:

[TABLE]

where the functional $\mathcal{L}:(W_{0}^{1,p}(\Omega))^{k}\to\mathbb{R}$ is defined by

[TABLE]

We claim that for any $p>1$ and $k\in\mathbb{N}$ there exists a minimizer $(\varphi_{1},\dots,\varphi_{k})\in(W_{0}^{1,p}(\Omega))^{k}$ for $\widetilde{\mathfrak{L}}_{k}(p;\Omega)$, where each $\varphi_{i}\not\equiv 0$ in $\Omega$. Let $(\varphi_{1}^{n},\dots,\varphi_{k}^{n})\in(W_{0}^{1,p}(\Omega))^{k}$, $n\in\mathbb{N}$, be a minimizing sequence for $\widetilde{\mathfrak{L}}_{k}(p;\Omega)$. Let $C>0$ be such that $\mathcal{L}(\varphi_{1}^{n},\dots,\varphi_{k}^{n})<C$ for all $n\in\mathbb{N}$. Due to the [math]-homogeneity of $\mathcal{L}$, we can assume that $\|\varphi_{i}^{n}\|_{L^{p}(\Omega)}=1$ for each $i\in\{1,\dots,k\}$ and $n\in\mathbb{N}$. Therefore, we obtain that $\|\nabla\varphi_{i}^{n}\|_{L^{p}(\Omega)}<C$ for each $i\in\{1,\dots,k\}$ and $n\in\mathbb{N}$. Hence, by a diagonal argument, we can find a subsequence $(\varphi_{1}^{n_{j}},\dots,\varphi_{k}^{n_{j}})$, $j\in\mathbb{N}$, and a vector $(\varphi_{1},\dots,\varphi_{k})\in(W_{0}^{1,p}(\Omega))^{k}$ such that each $\varphi_{i}^{n_{j}}\to\varphi_{i}$ weakly in $W_{0}^{1,p}(\Omega)$, strongly in $L^{p}(\Omega)$, and almost everywhere in $\Omega$ as $j\to+\infty$. Thus, since $\|\varphi_{i}^{n}\|_{L^{p}(\Omega)}=1$, we conclude that $\|\varphi_{i}\|_{L^{p}(\Omega)}=1$, that is, $\varphi_{i}\not\equiv 0$ in $\Omega$. Moreover, due to a.e.-convergence, we deduce that $\varphi_{i}\cdot\varphi_{j}=0$ a.e. on $\Omega$ for $i\neq j$. Therefore, $(\varphi_{1},\dots,\varphi_{k})$ is an admissible vector for $\widetilde{\mathfrak{L}}_{k}(p;\Omega)$. Finally, considering $m\in\{1,\dots,k\}$ such that

[TABLE]

and noting that

[TABLE]

we conclude that $(\varphi_{1},\dots,\varphi_{k})$ is a minimizer of $\widetilde{\mathfrak{L}}_{k}(p;\Omega)$.

Evidently, we can assume that each $\varphi_{i}\geq 0$ a.e. on $\Omega$. Let us denote $E_{i}=\{x\in\Omega:\varphi_{i}(x)>0\}$ for $i\in\{1,\dots,k\}$. Obviously, each $|E_{i}|>0$. Moreover, since $\varphi_{i}\cdot\varphi_{j}=0$ a.e. on $\Omega$ for $i\neq j$, we get $E_{i}\cap E_{j}=\emptyset$ a.e. on $\Omega$. Therefore, we can assume that $(E_{1},\dots,E_{k})\in\mathcal{E}_{k}$ and hence $\mathfrak{L}_{k}(p;\Omega)\leq\widetilde{\mathfrak{L}}_{k}(p;\Omega)$. The opposite inequality $\widetilde{\mathfrak{L}}_{k}(p;\Omega)\leq\mathfrak{L}_{k}(p;\Omega)$ is obvious by the definitions. Thus, we conclude that $\mathfrak{L}_{k}(p;\Omega)=\widetilde{\mathfrak{L}}_{k}(p;\Omega)$ and a minimizer of $\widetilde{\mathfrak{L}}_{k}(p;\Omega)$ defines a $k$-th spectral minimal partition of $\Omega$. ∎

Remark 5.2**.**

Evidently, $\mathfrak{L}_{1}(p;\Omega)=\lambda_{1}(p;\Omega)$. Moreover, using the variational characterization of $\lambda_{2}(p;\Omega)$ (see, e.g., [13] or [11]), it is not hard to obtain that $\mathfrak{L}_{2}(p;\Omega)=\lambda_{2}(p;\Omega)$.

The result of Proposition 5.1 can be refined in the following way. Let us recall several definitions. Under $p$-capacity of a measurable set $E\subset\Omega$ we mean

[TABLE]

A subset $E$ of $\Omega$ is called $p$-quasi-open if for every $\varepsilon>0$ there exists an open subset $E_{\varepsilon}$ of $\Omega$, such that ${\mathrm{cap}_{p}}(E_{\varepsilon}\bigtriangleup E)<\varepsilon$. If some abstract property $G(x)$ is satisfied for all $x\in\Omega$ except, possibly, for elements of a set $Z\subset\Omega$ with ${\mathrm{cap}_{p}}(Z)=0$, we say that $G(x)$ is satisfied $p$-quasi-everywhere on $\Omega$, or $p$-q.e. on $\Omega$, for short. Note that ${\mathrm{cap}_{p}}(Z)=0$ implies $\mathcal{H}^{N-1}(Z)=0$ and hence $|Z|=0$, cf. [12, Section 4.7.2, Theorem 4].

Consider now the subclass $\mathcal{E}_{k}^{*}\subset\mathcal{E}_{k}$ of $k$-tuples $(E_{1},\dots,E_{k})$ where each $E_{i}$ is $p$-quasi-open, and define the quantity

[TABLE]

Lemma 5.3**.**

$\mathfrak{L}_{k}^{*}(p;\Omega)=\mathfrak{L}_{k}(p;\Omega)$* for any $k\in\mathbb{N}$ and $p>1$, that is, there exists a $p$-quasi-open $k$-th spectral minimal partition of $\Omega$.*

Proof.

We prove that $\mathfrak{L}_{k}^{*}(p;\Omega)=\widetilde{\mathfrak{L}}_{k}(p;\Omega)$ where $\widetilde{\mathfrak{L}}_{k}(p;\Omega)$ is defined in Proposition 5.1. Let $(\varphi_{1},\dots\varphi_{k})$ by a minimizer of $\widetilde{\mathfrak{L}}_{k}(p;\Omega)$. We can assume that each $\varphi_{i}\geq 0$ a.e. on $\Omega$. Moreover, we can identify each $\varphi_{i}$ with its $p$-quasi-continuous representative (cf. [30, Section 3.3, Theorem 3.3.3]), that is, we can assume that for every $\varepsilon>0$ there exists a continuous function $\varphi_{i}^{\varepsilon}$ such that ${\mathrm{cap}_{p}}(\{\varphi_{i}\neq\varphi_{i}^{\varepsilon}\})<\varepsilon$. Therefore, denoting $E_{i}=\{x\in\Omega:\varphi_{i}(x)>0\}$, we see that each $E_{i}$ is a $p$-quasi-open subset of $\Omega$. Moreover, since $\varphi_{i}\cdot\varphi_{j}=0$ a.e. on open $\Omega$ for $i\neq j$, we derive from [16] that $\varphi_{i}\cdot\varphi_{j}=0$ $p$-q.e. on $\Omega$ and hence $E_{i}\cap E_{j}=\emptyset$ $p$-q.e. on $\Omega$. Therefore, we can assume that $(E_{1},\dots,E_{k})\in\mathcal{E}_{k}^{*}$ and hence $\mathfrak{L}_{k}^{*}(p;\Omega)\leq\widetilde{\mathfrak{L}}_{k}(p;\Omega)$. The opposite inequality $\widetilde{\mathfrak{L}}_{k}(p;\Omega)\leq\mathfrak{L}_{k}^{*}(p;\Omega)$ is trivial. Thus, we proved that $\mathfrak{L}_{k}^{*}(p;\Omega)=\mathfrak{L}_{k}(p;\Omega)$. ∎

Now we prove the main result of this section.

Theorem 5.4**.**

$h_{k}(\Omega)=\lim\limits_{p\to 1}\mathfrak{L}_{k}(p;\Omega)$* for any $k\in\mathbb{N}$.*

Proof.

We follow the strategy of [28, Section 5]. Let us fix an arbitrary $k\in\mathbb{N}$. We show first that

[TABLE]

Let $(E_{1},\dots,E_{k})\in\mathcal{E}_{k}$ be a $1$-adjusted Cheeger $k$-tuple of $\Omega$ obtained by Theorem 2.1. By Proposition 3.10, we can approximate each $E_{i}$ by a sequence of sets of finite perimeter $\{E_{i,m}\}_{m\in\mathbb{N}}$ with smooth boundary, compactly contained in $E_{i}$, such that

[TABLE]

For $\varepsilon>0$, define

[TABLE]

Let $m\in\mathbb{N}$ be fixed. If $\varepsilon>0$ is sufficiently small in order to have $E_{i,m}^{\varepsilon}\subset\subset E_{i}$ for each $i\in\{1,\dots,k\}$, we define functions $\{v_{i,m}\}_{i=1}^{k}\subset W_{0}^{1,p}(\Omega)$ such that $v_{i,m}=1$ on $E_{i,m}$, $v_{i,m}=0$ on $\Omega\setminus E_{i,m}^{\varepsilon}$ and $|\nabla v_{i,m}|=\varepsilon^{-1}$ on $E_{i,m}^{\varepsilon}\setminus E_{i,m}$. Then we have

[TABLE]

Noting that $|{E}_{i,m}^{\varepsilon}\setminus E_{i,m}|=\varepsilon P(E_{i,m})+o(\varepsilon)$, where $o(\varepsilon)/\varepsilon\to 0$ as $\varepsilon\to 0$ (cf. [3, Corollary 1]), and that the volume of the sets $E_{i,m}$ is uniformly bounded from below, we conclude that

[TABLE]

Therefore, for each $m\in\mathbb{N}$ we have

[TABLE]

Letting first $\varepsilon\to 0$, and then $m\to+\infty$, we derive (14).

Let us show now that

[TABLE]

From Proposition 5.1 we know that for any $p>1$ there exists a spectral minimal partition $(E_{1},\dots,E_{k})\in\mathcal{E}_{k}$ for $\Omega$, that is, $\mathfrak{L}_{k}(p;\Omega)=\max\limits_{i=1,\dots,k}\lambda_{1}(p;E_{i})$. Suppose for a moment that the following lower estimate is valid for each $i\in\{1,\dots,k\}$:

[TABLE]

Then for any $i\in\{1,\dots,k\}$ we get

[TABLE]

and hence

[TABLE]

where the second inequality is obtained by the definition (7). Letting now $p\to 1$, we get the desired lower bound (15). Combining it with (14), we conclude finally that

[TABLE]

Let us now prove the estimate (16). Note that (16) is valid for bounded open sets, see [21, Appendix]. However, in general, our $E_{i}$’s are only measurable (or $p$-quasi-open by Lemma 5.3). We will detalize the proof of [21] in order to cover our case. Let $v\in W_{0}^{1,p}(\Omega)\setminus\{0\}$ be a minimizer of $\lambda_{1}(p;E_{i})$. Denoting $\Phi(v):=|v|^{p-1}v$, we get

[TABLE]

which implies that $\Phi(v)\in W_{0}^{1,1}(\Omega)$. Since $v\equiv 0$ a.e. on $\Omega\setminus E_{i}$ and we can assume that $v\geq 0$ a.e. on $\Omega$, we have $\Phi(v)\equiv 0$ a.e. on $\Omega\setminus E_{i}$ and $\Phi(v)\geq 0$ a.e. on $\Omega$. Therefore, denoting

[TABLE]

and arguing as in the proof of [20, Corollary 2.2.3], we see that $P(F_{t};\Omega)=P(F_{t};\mathbb{R}^{N})=:P(F_{t})$ for all $t>0$. Consequently, using the co-area formula (cf. [20, Theorem 2.2.1 and Corollary 2.2.1]) with the layer-cake representation, and noting that $F_{t}\subset E_{i}$ for $t>0$, we derive

[TABLE]

Finally, applying the inequality (17), we get

[TABLE]

and hence (16) follows. ∎

6. Applications

6.1. Second Cheeger constant for a ring

Let $\Omega$ be a concentric ring in $\mathbb{R}^{2}$. We can assume, without loss of generality, that $\Omega=B_{1}\setminus\overline{B_{R}}$, where $B_{1}$ and $B_{R}$ are open concentric discs with radii $1$ and $R$, respectively, and $R\in(0,1)$. We start with a discussion of a configuration for coupled Cheeger sets in $\Omega$ which is empirically optimal. We will call it reference configuration. Then we prove its optimality rigorously.

6.1.1. Reference configuration

Let $\Omega^{\prime}:=\Omega\cap\{(x,y)\in\mathbb{R}^{2}:y>0\}$ be the upper half-ring corresponding to $\Omega$, see Fig. 3. Let us compute $h_{1}(\Omega^{\prime})$. For this end, we consider a set $\mathcal{O}_{r}$ obtained by rolling a disc with fixed radius $r\in\left(0,\frac{1-R}{2}\right]$ inside of $\Omega^{\prime}$, i.e.,

[TABLE]

where $[\Omega^{\prime}]^{r}$ is the inner parallel set to $\Omega^{\prime}$ at distance $r$, that is,

[TABLE]

Denote the quotient perimeter/area of $\mathcal{O}_{r}$ by $\mathcal{F}(r)$, i.e.,

[TABLE]

where for the numerator we have

[TABLE]

and for the denominator we have

[TABLE]

and

[TABLE]

Lemma 6.1**.**

$h_{1}(\Omega^{\prime})=\min\limits_{r\in\left[0,\frac{1-R}{2}\right]}\mathcal{F}(r)$. Moreover, a minimizer of $h_{1}(\Omega^{\prime})$ is unique and given by $\mathcal{O}_{r_{0}}$ with some $r_{0}\in\left(0,\frac{1-R}{2}\right)$.

Proof.

Evidently, each $\mathcal{O}_{r}$ with $r\in\left[0,\frac{1-R}{2}\right]$ is an admissible set for the minimization problem $h_{1}(\Omega^{\prime})$, that is,

[TABLE]

To show that $h_{1}(\Omega^{\prime})=\mathcal{F}(r_{0})$, let us recall the following definition from [23]. It is said that $\Omega^{\prime}$ has no necks of radius $r$ if for any two balls $B_{r}(x_{0}),B_{r}(x_{1})\subset\Omega^{\prime}$ there exists a continuous curve $\gamma:[0,1]\to\Omega^{\prime}$ such that

[TABLE]

It is not hard to observe that $\Omega^{\prime}$ satisfies this property for all $r\in\left[0,\frac{1-R}{2}\right]$. Moreover, this property is also satisfied for all $r>\frac{1-R}{2}$ since $\Omega^{\prime}$ contains no ball of such radius $r$. Applying now [23, Theorem 1.4], we conclude that the Cheeger set of $\Omega^{\prime}$ is given by $\mathcal{O}_{r_{0}}$, that is, $h_{1}(\Omega^{\prime})=\mathcal{F}(r_{0})$. Moreover, $\mathcal{O}_{r_{0}}$ is a unique minimizer of $h_{1}(\Omega^{\prime})$.

Let us show that $r_{0}\in\left(0,\frac{1-R}{2}\right)$. The case $r_{0}=0$ is impossible, since otherwise we get a contradiction to [23, Theorem 1.4]. On the other hand, if $r_{0}=\frac{1-R}{2}$, then we must have $h_{1}(\Omega^{\prime})=\frac{2}{1-R}$, cf. [22, Proposition 3.5 (iv)] or Proposition 4.4. However, we see that

[TABLE]

which is impossible. ∎

6.1.2. Optimality

Let us show that the reference configuration is indeed a minimizer of $h_{2}(\Omega)$ and it is unique up to rotation. Denote by $(E_{1},E_{2})\in\mathcal{E}_{2}$ any $1$-adjusted Cheeger couple of $\Omega$, that is, $(E_{1},E_{2})$ is a minimizer of $h_{2}(\Omega)$ and

[TABLE]

Note that $(E_{1},E_{2})$ exists by Theorem 2.1. Moreover, evidently, $(E_{1},E_{2})$ is $2$-adjusted. We assume that each $E_{i}$ has only one connected component. Throughout the proof we will frequently use the following properties of $E_{1}$ and $E_{2}$ which follow from Theorem 3.8 and Propositions 4.4 and 4.5:

• •

$\partial E_{i}\subset\overline{\Omega}$ is $C^{1}$-smooth.

• •

The mean curvatures of the free boundaries $\partial_{f}E_{1}$ and $\partial_{f}E_{2}$ are constant. We will denote them as $c_{1}$ and $c_{2}$, respectively.

• •

The mean curvature of the contact surface $\partial(E_{1}E_{2})$ is constant. We will denote it as $c_{12}$.

• •

$\partial E_{i}$ consists of arcs of circles and $\partial E_{i}=\partial_{f}E_{i}\cup\partial_{b}E_{i}\cup\partial(E_{1}E_{2})$.

• •

If $\partial(E_{1}E_{2})$ is not closed, then each end-point of $\partial(E_{1}E_{2})$ lies inside $\Omega$ and gives rise to exactly two arcs of free boundaries $\partial_{f}E_{i}$.

• •

Each end-point of $\partial_{f}E_{i}$ lies either on $\partial B_{1}$ or on $\partial B_{R}$ or coincides with an end-point of $\partial(E_{1}E_{2})$.

We will say that a part of $\partial E_{i}$ has nonempty interior if this part is not empty and not discrete.

The proof will be performed through the following steps:

1. (i)

Show that $\partial(E_{1}E_{2})$ has nonempty interior. 2. (ii)

Show that $\partial E_{1}\cap\partial B_{R}$ or $\partial E_{2}\cap\partial B_{R}$ has nonempty interior. 3. (iii)

Show that $\partial(E_{1}E_{2})$ does not have closed connected components. 4. (iv)

Show that $c_{1}=c_{2}$. 5. (v)

Show that $c_{12}=0$. 6. (vi)

Show the optimality of the reference configuration.

(i) We start with proving that the contact surface between $E_{1}$ and $E_{2}$ has nonempty interior. Suppose the claim was false. Thus, we have

[TABLE]

where the closure is taken with respect to the relative topology. Note that $\partial_{f}E_{1}$ and $\partial_{f}E_{2}$ have nonempty interiors. Indeed, if we suppose that, say, the interior of $\partial_{f}E_{1}$ is empty, then $\partial E_{1}=\partial_{b}E_{1}$. This readily yields $\partial E_{1}=\partial\Omega$ and $E_{1}=\Omega$, and hence $E_{2}=\emptyset$, which is impossible. By the same reasoning, $\partial_{f}E_{2}$ also has nonempty interior.

Let us show that $\partial_{f}E_{1}$ does not have closed connected components. Suppose, by contradiction, that there is a closed connected component $S$ of $\partial_{f}E_{1}$. Then, recalling that the curvature of $\partial_{f}E_{1}$ is a positive constant $c_{1}$, we see that $S$ is a circle. Denoting by $B_{0}$ a ball such that $\partial B_{0}=S$, we obtain $E_{1}\subset B_{0}$. There are two possible positions for $B_{0}$: either $B_{0}\subset\Omega$ or $B_{R}\subset B_{0}$. If $B_{0}\subset\Omega$, then we get a contradiction since such a configuration is not optimal. Indeed, a maximal ball which can be inscribed in $\Omega$ has radius $\frac{1-R}{2}$. Moreover, $\Omega$ contains at least two such balls. Therefore, it is not hard to see that in the best configuration with $E_{1}\subset B_{0}$ it will hold $E_{1}=B_{0}$ and $\text{radius}(B_{0})=\frac{1-R}{2}$, that is,

[TABLE]

However, taking two half-rings as a pair of admissible sets for $h_{2}(\Omega)$, we easily obtain a contradiction:

[TABLE]

Suppose now that $B_{R}\subset B_{0}$. Then it is not hard to see that $E_{2}\subset B_{1}\setminus B_{0}$. Therefore, in view of (19), we see that $E_{2}$ must be a ball, and the contradiction follows as in (20). Thus, $\partial_{f}E_{1}$ does not have closed connected components. Analogously, the same conclusion holds for $\partial_{f}E_{2}$.

Fix now any connected component of $\partial_{f}E_{1}$. Since this component is not closed and it is an arc of a circle of radius $\frac{1}{c_{1}}$, and $\partial(E_{1}E_{2})$ has empty interior by the assumption, we see that this component must touch both $\partial B_{1}$ and $\partial B_{R}$. Thus, either $\frac{1}{c_{1}}=\frac{1-R}{2}$ or $\frac{1}{c_{1}}=\frac{1+R}{2}$. However, the latter case is impossible due to the $C^{1}$-smoothness of $\partial E_{1}$. Therefore, we conclude that $h_{1}(E_{1})=c_{1}=\frac{2}{1-R}$. Let us show that in this case the best configuration for $E_{1}$ and $E_{2}$ will be given by $\mathcal{O}_{\frac{1-R}{2}}$, see (18). For this end, we study the behaviour of $\partial_{f}E_{1}$ and $\partial_{f}E_{2}$, see Fig. 5. Let an arc $PQ\subset\partial_{f}E_{1}$ be such that $P\in\partial B_{1}$ and $Q\in\partial B_{R}$. If there exists an arc $PP_{1}\subset\partial_{f}E_{1}$ such that $P_{1}\in\Omega\setminus PQ$, or an arc $QQ_{1}\subset\partial_{f}E_{1}$ such that $Q_{1}\in\Omega\setminus PQ$, then, recalling that $\partial(E_{1}E_{2})$ has empty interior, this arcs can be prolonged such that $P_{1}=Q$ or $Q_{1}=P$, and we obtain a contradiction as in (20). Hence, there are arcs $PR\subset\partial E_{1}\cap\partial B_{1}$ and $QS\subset\partial E_{1}\cap\partial B_{R}$. Moreover, noting that the same holds true for $\partial E_{2}$, we conclude that $PR\not\equiv\partial B_{1}$ and $QS\not\equiv\partial B_{R}$, and we can take $R$ and $S$ such that $RS\subset\partial_{f}E_{1}$. Thus, we see that $E_{1}$ and $E_{2}$ have the shapes as depicted in Fig. 5, and at least one of $E_{i}$ is contained in a half-ring of $\Omega$. It is not hard to show that the best configuration for each $E_{i}$ will be given, up to rotation, by $\mathcal{O}_{\frac{1-R}{2}}$ (see (18)). However, $\mathcal{O}_{\frac{1-R}{2}}$ is not optimal, as it follows from Lemma 6.1. Therefore, we conclude that $\partial(E_{1}E_{2})$ has nonempty interior.

(ii) Assume, without loss of generality, that $c_{12}\geq 0$. Let us show that $\partial E_{1}\cap\partial B_{R}$ has nonempty interior. Suppose, by contradiction, that the interior of $\partial E_{1}\cap\partial B_{R}$ is empty. Since the curvature of $\partial_{f}E_{1}$ is $c_{1}>0$ and the curvature of $\partial E_{1}\cap\partial B_{1}$ equals $1$ (whenever these parts have nonempty interiors), and the curvature of $\partial(E_{1}E_{2})$ is $c_{12}\geq 0$, we conclude that $E_{1}$ has a piecewise nonnegative curvature. Combining this fact with the $C^{1}$-smoothness of $\partial E_{1}$, we conclude that $E_{1}$ is convex. Evidently, the largest convex set which can be inscribed in $\Omega$ is a circular segment as on Fig. 3. Therefore, $E_{1}$ must be contained inside such segment and, consequently, strictly inside a half-ring. But this fact contradicts Lemma 6.1.

Note that if $c_{12}=0$, then the above arguments can be applied to both $E_{1}$ and $E_{2}$ to conclude that $\partial E_{1}\cap\partial B_{R}$ and $\partial E_{2}\cap\partial B_{R}$ have nonempty interiors.

(iii) Let us now prove that the contact surface between $E_{1}$ and $E_{2}$ does not have closed connected components. Suppose the assertion is false. Since the curvature of $\partial(E_{1}E_{2})$ is a constant $c_{12}$, there exists a connected component $S$ of $\partial(E_{1}E_{2})$ which is a circle. Let $B_{0}$ be a ball such that $\partial B_{0}=S$. We again have two possibilities: either $B_{0}\subset\Omega$ or $B_{R}\subset B_{0}$. If $B_{0}\subset\Omega$, then we get a contradiction as in (20). Suppose now that $B_{R}\subset B_{0}$. Since each $E_{i}$ has only one connected component, we can assume, without loss of generality, that $E_{1}\subset B_{0}\setminus\overline{B_{R}}$ and $E_{2}\subset B_{1}\setminus\overline{B_{0}}$. Let us show that, in fact, $E_{1}=B_{0}\setminus\overline{B_{R}}$ and $E_{2}=B_{1}\setminus\overline{B_{0}}$. Suppose, by contradiction, that, say, $E_{2}$ has a free boundary. Since $\partial B_{0}\subset\partial(E_{1}E_{2})$ and $\partial E_{2}$ is $C^{1}$-smooth, we see that there is no arc of $\partial_{f}E_{2}$ with an end-point on $\partial B_{0}$. Hence, the only possibility is that $\partial_{f}E_{2}$ is a circle $\partial B_{s}$ such that $\partial B_{s}\cap\Omega$ is of angle $2\pi$, which contradicts [24, Lemma 2.11]. Similarly, we can conclude that $E_{1}=B_{0}\setminus\overline{B_{R}}$. Varying the radius of $B_{0}$, it is easy to see that the best configuration is achieved when $h_{1}(E_{1})=h_{1}(E_{2})$. Denoting the radius of $B_{0}$ as $s$ and noting that

[TABLE]

we obtain that $s=\frac{1+R}{2}$, and hence $h_{2}(\Omega)=\frac{4}{1-R}$. The contradiction then follows as in (20). Thus, $\partial(E_{1}E_{2})$ does not have closed connected components.

(iv) As a consequence of step (iii), we see that the free boundaries of $E_{1}$ and $E_{2}$ are not empty. Let us prove that $c_{1}=c_{2}$. Suppose, without loss of generality, that $c_{12}\geq 0$. Then we get from step (ii) that $\partial E_{1}\cap\partial B_{R}$ has nonempty interior. We know from Proposition 4.7 that $c_{1}\geq c_{2}$ and $c_{1}\geq c_{12}$, and if $c_{1}>c_{2}$, then $c_{1}=c_{12}$. We prove that $c_{1}>c_{12}$, which will imply that the case $c_{1}>c_{2}$ is impossible, and hence $c_{1}=c_{2}$. Suppose, contrary to our claim, that $c_{1}=c_{12}\geq c_{2}$. If $\partial B_{R}\not\subset\partial E_{1}$, then we argue as in step (i) to deduce that the best configuration for $E_{1}$ and $E_{2}$ in this case is given by $\mathcal{O}_{\frac{1-R}{2}}$, which is impossible. Therefore, $\partial B_{R}\subset\partial E_{1}$. Then, the $C^{1}$-smoothness of $\partial E_{1}$ implies that no arc of $\partial_{f}E_{1}$ can have an end-point on $\partial B_{R}$. Hence, since $c_{1}=c_{12}$, we conclude that there exists a circle $\partial B_{0}\subset\partial E_{1}$. Suppose that $B_{0}\subset\Omega$. Since $E_{1}$ has only one connected component and $c_{1}=c_{12}>0$, we see that $E_{1}\subset B_{0}$. However, it contradicts the fact that $\partial B_{0}\subset\partial E_{1}$. Suppose now that $B_{R}\subset B_{0}$. Since, again, $E_{1}$ has only one connected component and $c_{1}=c_{12}>0$, we deduce that $E_{1}\subset B_{0}\setminus\overline{B_{R}}$ and $E_{2}\subset B_{1}\setminus\overline{B_{0}}$. Moreover, it is easy to see that $E_{1}=B_{0}\setminus\overline{B_{R}}$. Denoting by $s$ the radius of $B_{0}$ and recalling that $\partial_{f}E_{1}\cap\partial B_{0}$ has nonempty interior, we get the following contradiction:

[TABLE]

Therefore, $c_{1}\neq c_{12}$. Finally, applying Proposition 4.7, we conclude that $c_{1}=c_{2}$.

(v) Let us show now that $c_{12}=0$. Suppose, by contradiction and without loss of generality, that $c_{12}>0$. To prove the claim, we will study the behaviour of $\partial E_{1}$ and $\partial E_{2}$.

From step (ii) we know that $\partial E_{1}\cap\partial B_{R}$ has nonempty interior. Take an arc $PQ\subset\partial E_{1}\cap\partial B_{R}$ and points $F,G\in\partial(E_{1}E_{2})$ such that the arc $QF\subset\partial_{f}E_{1}$ and $PG\subset\partial_{f}E_{1}$, see Fig. 5. Note that such $F$ and $G$ exist, since otherwise, without loss of generality, either $F\in\partial B_{R}$ or $F\in\partial B_{1}$. The first case is impossible, since then $F=Q$ and the contradiction follows, for instance, as in (20). The second case is impossible, since then the best configuration will be given by $\mathcal{O}_{\frac{1-R}{2}}$ which is not optimal, see step (i) and Lemma 6.1. Consider maximal arcs $FK,GM\subset\partial(E_{1}E_{2})$.

1. Suppose first that $FK\equiv GM$, i.e., $F=M$ and $K=G$, see Fig. 7. Consider arcs $FX_{1},GX_{2}\subset\partial_{f}E_{2}$. There are several possible positions of $X_{1}$ and $X_{2}$:

• (a)

Assume first that $X_{1},X_{2}\in\partial B_{1}$. Let $GF$ be the arc of a circle $\partial B_{0}$. Let us show that $B_{0}$ is concentric with $B_{1}$ and $B_{R}$. Suppose the claim is false. Noting that the arcs $PG$ and $GX_{2}$ cannot be of angle greater than $\pi$ [24, Lemma 2.11], we easily get a contradiction, see Fig. 7. Therefore, $B_{0}$ is concentric with $B_{1}$ and $B_{R}$, and $E_{1}\subset B_{0}\setminus\overline{B_{R}}$ and $E_{2}\subset B_{1}\setminus\overline{B_{0}}$. However, it is known that rings $B_{0}\setminus\overline{B_{R}}$ and $B_{1}\setminus\overline{B_{0}}$ are calibrable [6], that is, they are Cheeger sets of themselves. This implies that $E_{1}=B_{0}\setminus\overline{B_{R}}$ and hence $\partial(E_{1}E_{2})=\partial B_{0}$, i.e., $\partial(E_{1}E_{2})$ is closed. A contradiction to step (iii).

• (b)

Assume now that $X_{1}=G$ and $X_{2}=F$. That is, the arc $GX_{2}\subset\partial_{f}E_{2}$ touches $PG$ and $QF$ at corresponding points. Since the arc $FG\subset\partial(E_{1}E_{2})$ also touches $PG$ and $QF$ at the same points as $GX_{2}$, we readily get a contradiction to the fact that $c_{2},c_{12}>0$.

• (c)

Other positions for $X_{1}$ and $X_{2}$ are impossible by evident reasons.

2. Suppose now that $FK\not\equiv GM$, i.e., $F\neq M$ and $K\neq G$, see Fig. 5. Then, consider arcs $KL_{1},MN_{1}\subset\partial_{f}E_{1}$ and $KL_{2},MN_{2}\subset\partial_{f}E_{2}$. Consider also a segment $KR$ tangent to the arc $FK$ at the point $K$ such that $R\in\partial B_{1}$, and, analogously, a segment $MS$ tangent to the arc $GM$ at $M$ such that $S\in\partial B_{1}$. Recall that the curvature of $\partial E_{1}$ is piecewise nonnegative except of the part $\partial E_{1}\cap\partial B_{R}$. Therefore, it is not hard to see that if $KR$ and $GM$ intersect at a point $T$ inside $\Omega$, then $E_{1}$ is contained inside a set bounded by a closed path

[TABLE]

Analogously, if $KR$ and $GM$ do not intersect inside $\Omega$ (as it is depicted on Fig. 5), then $E_{1}$ is contained inside a set bounded by a closed path

[TABLE]

Let us denote a set bounded by paths (21) or (22) as $\mathcal{Z}_{1}$. That is, $E_{1}\subset\mathcal{Z}_{1}$. This implies that $L_{2},N_{2}\in\partial B_{1}$. Indeed, if, say, $L_{2}\in\partial(E_{1}E_{2})$, then the only possibility is $L_{2}=M$. However, recalling that $c_{12}>0$, it is not hard to observe that the arcs $FX_{1},GX_{2}\subset\partial_{f}E_{2}$ will intersect $\partial B_{R}$ transversally (see, for example, Fig. 9), which is impossible in view of the $C^{1}$-smoothness of $\partial E_{2}$. The case $L_{2}\in\partial B_{R}$ is obviously impossible, too. Thus, we conclude that $L_{2}\in\partial B_{1}$. By the same arguments, $N_{2}\in\partial B_{1}$.

These facts readily imply that $E_{2}$ is located inside a set $\mathcal{Z}_{2}$ bounded by a closed path

[TABLE]

3. Suppose that $L_{1}=M$. Consider arcs $FX_{1},GX_{2}\subset\partial_{f}E_{2}$. There are two cases for $X_{1}$:

1. (a)

Assume first that $X_{1}\neq G$ (and hence $X_{2}\neq F$). Therefore, since $E_{1}\subset\mathcal{Z}_{1}$, we see that $X_{1},X_{2}\not\in\partial(E_{1}E_{2})$. If $X_{1}\in\partial B_{1}$, then we get at least two connected components of $E_{2}$, which is impossible. Analogously, $X_{2}\not\in\partial B_{1}$. Therefore, we must have $X_{1},X_{2}\in\partial B_{R}$, see Fig. 7. If the angle $\angle FOG$ measured from $E_{1}$ is less than or equal to $\pi$, then $E_{1}$ is contained strictly inside a half-ring, which contradicts Lemma 6.1. Thus, $\angle FOG>\pi$. However, it is not hard to see that, in this case, $FK$ and $GM$ cannot be connected by $KL_{1}=KM\subset\partial_{f}E_{1}$ in such a way that $\partial E_{1}$ is $C^{1}$-smooth, since $c_{1}\geq c_{12}$. A contradiction. 2. (b)

Assume now that $X_{1}=G$ (and hence $X_{2}=F$). Then we see that $\angle FOG>\pi$ and, as above, we see that this case is impossible.

4. Suppose now that $L_{1}\neq M$. Then we deduce that $L_{1}\in\partial B_{1}$ and $N_{1}\in\partial B_{1}$. Indeed, since $L_{1}\neq M$, we have $L_{1}\not\in\mathcal{Z}_{2}$ and hence $L_{1}\not\in\partial(E_{1}E_{2})$. If $L_{1}\in\partial B_{R}$, then $L_{1}=P$ and $FK=GM$, which is impossible as shown in case 1) above. Thus, $L_{1}\in\partial B_{1}$ and, by the same arguments, $N_{1}\in\partial B_{1}$.

5. At the end, we have $L_{1},L_{2},N_{1},N_{2}\in\partial B_{1}$. However, recalling that $c_{12}>0$ by the assumption, and $c_{1}=c_{2}$ by step (iv), it is not hard to see that an arc $FX_{1}\subset\partial_{f}E_{2}$ will intersect with $\partial B_{R}$ not transversely, see Fig. 9. A contradiction to the $C^{1}$-smoothness of $\partial E_{2}$. Thus, $c_{12}=0$.

(vi) Let us finish the proof by showing that the reference configuration is optimal. Since $c_{12}=0$ by step (v), we see from step (ii) that the arcs $P_{1}Q_{1}:=\partial E_{1}\cap\partial B_{R}$ and $P_{2}Q_{2}:=\partial E_{2}\cap\partial B_{R}$ are nonempty and they are the only parts of $\partial E_{1}$ and $\partial E_{2}$ with strictly negative curvature.

Take, without loss of generality, an arc $P_{1}Q_{1}$, and let $F\in\partial(E_{1}E_{2})$ be such that the arc $Q_{1}F\subset\partial_{f}E_{1}$. We want to prove that the arc $FK\subset\partial(E_{1}E_{2})$ is “perpendicular” to $\partial B_{R}$. Let $X_{1}$ be the corresponding end point of the arc $FX_{1}\subset\partial_{f}E_{2}$. Then there are three possibilities for a position of $X_{1}$:

1. (a)

$X_{1}\in\partial B_{R}$. In this case $FK$ is “perpendicular” to $\partial B_{R}$. 2. (b)

$X_{1}\in\partial(E_{1}E_{2})$. Due to piecewise nonnegativity of the curvature of $\partial E_{2}$ (except of the part $\partial E_{2}\cap\partial B_{R}$), we see that $E_{2}$ must be contained inside the intersection of $\Omega$ with a cone based on the angle $\angle FOX_{1}$. At the same time, this cone is contained inside a small segment of $B_{1}$ based on the line $FK$. However, it is not hard to see that the reference configuration is better. A contradiction. 3. (c)

$X_{1}\in\partial B_{1}$. As in the case (b), we see that $E_{2}$ must be contained in a small segment of $B_{1}$ based on the line $FK$. A contradiction.

Therefore, any segment of $\partial(E_{1}E_{2})$ which is connected with $\partial\Omega$ by an arc of a free boundary must be connected with both $\partial B_{1}$ and $\partial B_{R}$ and it is “perpendicular” to $\partial B_{R}$. Since, in view of step (ii), $\partial E_{i}\cap\partial B_{R}$ has nonempty interior for $i=1,2$, we see that there are at least two segments of $\partial(E_{1}E_{2})$ which are connected with $\partial B_{R}$. Therefore, either $E_{1}$ or $E_{2}$ is contained inside a half-ring. From the uniqueness result of Lemma 6.1 we conclude that $E_{i}$ coincides (up to rotation) with $\mathcal{O}_{r_{0}}$, and therefore the reference configuration is optimal.

6.2. Second Cheeger constant for a disc

Let $\Omega$ be a disc in $\mathbb{R}^{2}$ and let $(E_{1},E_{2})\in\mathcal{E}_{2}$ be a $2$-adjusted Cheeger couple of $\Omega$. Recall that $c_{1}$ and $c_{1}$ are the mean curvatures of the free boundaries $\partial_{f}E_{1}$ and $\partial_{f}E_{2}$, respectively, and $c_{12}$ is the mean curvature of the contact surface $\partial(E_{1}E_{2})$ measured from $E_{1}$.

It was shown in [28] that each $E_{i}$ must be a first Cheeger set of a half-ball of $\Omega$. The last steps of the proof on [28, pp.13-14] rely on [28, Theorem 3.9]. However, as we described in Section 1, this theorem is not correct. Let us show that Proposition 4.7 can be applied to overcome the usage of [28, Theorem 3.9]. From [28, Section 4] we know that $\partial(E_{1}E_{2})$ has nonempty interior and cannot have closed connected components. Let $PQ$ be an arc of $\partial(E_{1}E_{2})$ such that there are arcs $PE\in\partial_{f}E_{1}$ and $PF\in\partial_{f}E_{2}$, where $E,F\in\partial\Omega$, see [28, Figure 3].

If $c_{1}=c_{2}$ and $c_{12}\neq 0$, then, as in [28, p. 13], either $E_{1}$ or $E_{2}$ will be a subset of a circular segment strictly contained in a half-disc. However, the configuration with the Cheeger sets of the two half-disks is better, a contradiction. Therefore, either $c_{1}=c_{2}$ and $c_{12}=0$, or $c_{1}\neq c_{2}$. Let us drop out the second case. Assume, without loss of generality, that $c_{12}\geq 0$. Then Proposition 4.7 implies that $c_{1}\geq c_{2}$. If we suppose that $c_{1}>c_{2}$, then Proposition 4.7 yields $c_{1}=c_{12}$. This means that $E_{1}$ must be a ball of radius $r_{1}=\frac{1}{c_{1}}$. However, it is impossible, since $h_{1}(E_{1})=c_{1}$, but the Cheeger constant of the ball $E_{1}$ equals to $\frac{2}{r_{1}}$, that is, $h_{1}(E_{1})=2c_{1}$. Since $c_{1}>0$, we get a contradiction. Thus, the case where $c_{1}=c_{2}$ and $c_{12}=0$ is the only possible, and this directly leads to the optimal configuration.

Acknowledgments. The article was started during a visit of E.P. at the University of West Bohemia and was continued during a visit of V.B. at Aix-Marseille University. The authors wish to thank the hosting institutions for the invitation and the kind hospitality. V.B. was supported by the project LO1506 of the Czech Ministry of Education, Youth and Sports.