Spectrum of the Laplacian with weights

Abstract

Given a compact Riemannian manifold (M, g) and two positive functions $\rho$ and $\sigma$, we are interested in the eigenvalues of the Dirichlet energy functional weighted by $\sigma$, with respect to the $L^2$ inner product weighted by $\rho$. Under some regularity conditions on $\rho$ and $\sigma$, these eigenvalues are those of the operator $-\rho ^{-1} \text{ div }(\sigma \nabla u)$ with Neumann conditions on the boundary if $\partial M\ne \emptyset$. We investigate the effect of the weights on eigenvalues and discuss the existence of lower and upper bounds under the condition that the total mass is preserved.

1 Introduction

Let (M, g) be a compact Riemannian manifold of dimension $n\ge 2$, possibly with nonempty boundary. We designate by $\left\{ \lambda _k(M,g)\right\} _{k\ge 0}$ the nondecreasing sequence of eigenvalues of the Laplacian on (M, g) under Neumann conditions on the boundary if $\partial M\ne \emptyset$. The min-max principle tells us that these eigenvalues are variationally defined by

$$\begin{aligned} \lambda _k(M,g)=\inf _{E\in S_{k+1}} \sup _{u\in E{\setminus } \{ 0\} } \frac{\int _M |\nabla u|^2 v_g}{\int _M u^2 v_g}, \end{aligned}$$

where $S_k$ is the set of all k-dimensional vector subspaces of $H^1(M)$ and $v_g$ is the Riemannian volume element associated with g.

The relationships between the eigenvalues $\lambda _k(M,g)$ and the other geometric data of (M, g) constitute a classical topic of research that has been widely investigated in recent decades (the monographs [3, 4, 7, 24, 35] are among basic references on this subject). In the present work, we are interested in eigenvalues of “weighted” energy functionals with respect to “weighted” $L^2$ inner products. Our aim is to investigate the interplay between the geometry of (M, g) and the effect of the weights.

Therefore, let $\rho$ and $\sigma$ be two positive continuous functions on M and consider the Rayleigh quotient

$$\begin{aligned} R_{(g,\rho ,\sigma )}(u)=\frac{\int _M |\nabla u|^2 \sigma \, v_g}{\int _M u^2 \rho \, v_g}. \end{aligned}$$

The corresponding eigenvalues are given by

$$\begin{aligned} \mu _k^g(\rho ,\sigma )=\inf _{E\in S_{k+1}} \sup _{u\in E{\setminus } \{ 0\} } R_{(g,\rho ,\sigma )}(u). \end{aligned}$$

(1)

Under some regularity conditions on $\rho$ and $\sigma$, $\mu _k^g(\rho ,\sigma )$ is the kth eigenvalue of the problem

$$\begin{aligned} -\text{ div }(\sigma \nabla u)=\mu \rho u \quad \mathrm{in }\ M \end{aligned}$$

(2)

with Neumann conditions on the boundary if $\partial M\ne \emptyset$. Here, $\nabla$ and $\text{ div }$ are the gradient and the divergence associated with the Riemannian metric g. When there is no risk of confusion, we will simply write $\mu _k(\rho ,\sigma )$ for $\mu _k^g(\rho ,\sigma )$.

Notice that the numbering of eigenvalues starts from zero. It is clear that the infimum of $R_{(g,\rho ,\sigma )}(u)$ is achieved by constant functions; hence, $\mu _0^g(\rho ,\sigma )=0$ and

$$\begin{aligned} \mu _1^g(\rho ,\sigma )= \inf _{\int _M u\rho v_g=0} R_{(g,\rho ,\sigma )}(u). \end{aligned}$$

(3)

One obviously has $\mu _k^g(1,1)=\lambda _k(M,g)$. When $\sigma =1$, the eigenvalues $\mu _k(\rho ,1)$ correspond to the situation where M has a non-necessarily constant mass density $\rho$ and describe, in dimension 2, the vibrations of a non-homogeneous membrane (see [24, 31] and the references therein). The eigenvalues $\mu _k(1,\sigma )$ are those of the operator $\text{ div }(\sigma \nabla u)$ associated with a conductivity $\sigma$ on M (see [24, Chapter 10] and [2]). In the case where $\rho =\sigma$, the eigenvalues $\mu _k(\rho ,\rho )$ are those of the Witten Laplacian $L_\rho$ (see [12] and the references therein). Finally, when $\sigma$ and $\rho$ are related by $\sigma =\rho ^{\frac{n-2}{n}}$, the corresponding eigenvalues $\mu _k^g(\rho ,\rho ^{\frac{n-2}{n}})$ are exactly those of the Laplacian associated with the conformal metric $\rho ^{\frac{2}{n}}g$, that is, $\mu _k^g(\rho ,\rho ^{\frac{n-2}{n}})=\lambda _k(M,\rho ^{\frac{2}{n}}g)$.

Our goal in this paper is to investigate the behavior of $\mu _k^g(\rho ,\sigma )$, especially in the most significant cases mentioned above, under normalizations that we will specify in the sequel, but which essentially consist in the preservation of the total mass. The last case, corresponding to conformal changes of metrics, has been widely investigated in recent decades (see for instance [9, 22, 23, 26, 28, 29, 33, 34]), and most of the questions we will address in this paper are motivated by results established in the conformal setting. These questions can be listed as follows:

• 1. Can one redistribute the mass density $\rho$ (resp. the conductivity $\sigma$) so that the corresponding eigenvalues become as small as desired?

• 2. Can one redistribute $\rho$ and/or $\sigma$ so that the eigenvalues become as large as desired?

• 3. If Question (1) (resp. (2)) is answered positively, what kind of constraints can one impose in order to get upper or lower bounds for the eigenvalues?

• 4. If Question (1) (resp. (2)) is answered negatively, what are the geometric quantities that bound the eigenvalues?

• 5. If the eigenvalues are bounded, what can one say about their extremal values?

• 6. Is it possible, in some specific situations, to compute or to have sharp estimates for the first positive eigenvalues?

In a preliminary section, we deal with some technical issues concerning the possibility of relaxing the conditions of regularity and positivity of the densities. In the process, we prove a 2-dimensional convergence result (Theorem 2.1) which completes a theorem that Colin de Verdière had established in dimension $n\ge 3$. Question (1) is discussed at the beginning of Sect. 3 where we show that it is possible to fix one of the densities $\rho$ and $\sigma$ and vary the other one, among densities preserving the total mass, in order to produce arbitrarily small eigenvalues (Theorem 3.1). This leads us to get into Question (3) that we tackle by establishing the following Cheeger-type inequality (Theorem 3.2):

$$\begin{aligned} \mu _1(\rho ,\sigma ) \ge \frac{1}{4} h_{\sigma ,\sigma }(M)h_{\rho ,\sigma }(M), \end{aligned}$$

where $h_{\sigma ,\sigma }(M)$ and $h_{\rho ,\sigma }(M)$ are suitably defined isoperimetric constants, in the spirit of what is done in [27].

Whenever a Cheeger-type inequality is proved, a natural question is to investigate a possible reverse inequality under some geometric restrictions (see [6] and the introduction of [32] for a general presentation of this issue). It turns out that in the present situation, such a reverse inequality cannot be obtained without additional assumptions on the densities. Indeed, we prove that on any given Riemannian manifold, there exists families of densities such that the associated Cheeger constants are as small as desired while the corresponding eigenvalues are uniformly bounded from below (Theorem 3.3).

Questions (2) and (4) are addressed in Sect. 4. A. Savo and the authors have proved in [12] that the first positive eigenvalue $\mu _1(\rho ,\rho )$ of the Witten Laplacian is not bounded above as $\rho$ runs over densities of fixed total mass. In Proposition 4.1, we prove that, given a Riemannian metric $g_0$, we can find a metric g, within the set of metrics conformal to $g_0$ and of the same volume as $g_0$, and a density $\rho$, among densities of fixed total mass with respect to $g_0$, so that $\mu _1^g(\rho ,1)$ is as large as desired. The same also holds for $\mu _1^g(1,\sigma )$.

However, if instead of requiring that the total mass of the densities is fixed with respect to $g_0$, we assume that it is fixed with respect to g, and then the situation changes completely. Indeed, Theorem 4.1 gives the following estimate when M is a domain of a complete Riemannian manifold $(\tilde{M},g_0)$ whose Ricci curvature satisfies $\hbox {Ric}_{g_0}\ge -(n-1)$ (including the case $M=\tilde{M}$ if $\tilde{M}$ is compact): For every metric g conformal to $g_0$ and every density $\rho$ on M with $\int _{M}\rho v_g =\vert M\vert _g$, one has

$$\begin{aligned} \mu _k^g(\rho ,1) \le \frac{1}{\vert M\vert _g^{\frac{2}{n}}} \left( A_nk^{\frac{2}{n}} + B_n \vert M\vert _{g_0}^{\frac{2}{n}} \right) , \end{aligned}$$

(4)

where $\vert \ .\ \vert _{g}$ and $\vert \ . \ \vert _{g_0}$ denote the Riemannian volumes with respect to g and $g_0$, respectively, and $A_n$ and $B_n$ are two constants that depend only on the dimension n.

A direct consequence of this theorem is the following inequality satisfied by any density $\rho$ on (M, g) with $\int _{M}\rho v_g =\vert M\vert _g$:

$$\begin{aligned} \mu _k^g(\rho ,1) \le A_n\left( \frac{k}{\vert M\vert _{g}}\right) ^\frac{2}{n}+ B_n \text{ ric }_0, \end{aligned}$$

(5)

where $\text{ ric }_0$ is a positive number such that $\hbox {Ric}_{g}\ge -(n-1)\text{ ric }_0\ g$ (see Corollary 4.1).

Regarding the eigenvalues $\mu _k^g(1,\sigma )$, we are able to prove an estimate of the same type as (5): For every positive density $\sigma$ on (M, g) with $\int _{M}\sigma v_g =\vert M\vert _g$, one has (Theorem 4.2)

$$\begin{aligned} \mu _k^g(1,\sigma ) \le A_n \left( \frac{k}{\vert M\vert _{g}}\right) ^\frac{2}{n} + B_n \text{ ric }_0, \end{aligned}$$

(6)

where $A_n$ and $B_n$ are two constants that depend only on the dimension n. It is worth noting that although the estimates (5) and (6) are similar, their proofs are of different nature. That is why we were not able to decide whether a stronger estimate such as (4) holds for $\mu _k^g(1,\sigma )$.

When M is a bounded domain of a manifold $(\tilde{M},\tilde{g})$ of nonnegative Ricci curvature (e.g., $\mathbb {R}^n$), the inequalities (5) and (6) give the following estimates that can be seen as extensions of Kröger’s inequality [30]: $\mu _k^g(\rho ,1) \le A_n\left( \frac{k}{\vert M\vert _{g}}\right) ^\frac{2}{n}$ and $\mu _k^g(1,\sigma ) \le A_n \left( \frac{k}{\vert M\vert _{g}}\right) ^\frac{2}{n}$, provided that $\int _{M}\rho v_g =\vert M\vert _g$ and $\int _{M}\sigma v_g =\vert M\vert _g$. Notice that if we follow Kröger’s approach, then we get an upper bound of $\mu _k^g(\rho ,1)$ which involves the gradient of $\rho$ and the integral of $\frac{1}{\rho }$ (see [16]).

According to (5) and (6), it is natural to introduce the following extremal eigenvalues on a given Riemannian manifold (M, g):

$$\begin{aligned} \mu _k^*(M,g)=\sup _{\fint _M\rho \, v_g=1}\mu _k^g(\rho ,1) \quad \text{ and } \quad \mu _k^{**}(M,g)=\sup _{\fint _M\sigma v_g=1}\mu _k^g(1,\sigma ). \end{aligned}$$

In Sect. 5, we investigate the qualitative properties of these quantities in the spirit of what we did in [9] for the conformal spectrum, thereby providing some answers to Question (5). For example, when M is of dimension 2, we have the following lower estimate (see [9, Corollary 1]):

$$\begin{aligned} \mu _k^*(M,g)\ge 8\pi \frac{k}{\vert M\vert _g}. \end{aligned}$$

This means that, given any Riemannian surface (M, g), endowed with the constant mass distribution $\rho =1$ (whose eigenvalues can be very close to zero), it is always possible to redistribute the mass density $\rho$ so that the resulting eigenvalue $\mu _k^g(\rho ,1)$ is greater or equal to $8\pi \frac{k}{\vert M\vert _g}$.

It turns out that this phenomenon is specific to the dimension 2. Indeed, we prove (Theorem 5.1) that on any compact manifold M of dimension $n\ge 3$, there exists a one-parameter family of Riemannian metrics $g_\varepsilon$ of volume 1 such that

$$\begin{aligned} \mu _k^*(M,g_\varepsilon )\le C k\varepsilon ^{\frac{n-2}{n}}, \end{aligned}$$

where C is a constant that does not depend on $\varepsilon$. This means that in dimension $n\ge 3$, there exist geometric situations that generate very small eigenvalues, regardless of how the mass density is distributed.

Regarding the extremal eigenvalues $\mu _k^{**}(M,g)$, a similar result is proved (Theorem 5.2) which is, moreover, also valid in dimension 2.

Note, however, that it is possible to construct examples of Riemannian manifolds (M, g) with very small eigenvalues (for the constant densities), for which $\mu _k^*(M,g)$ and $\mu _k^{**}(M,g)$) are sufficiently large (see Proposition 5.2).

The last part of the paper (Sect. 6) is devoted to the study of the first extremal eigenvalues $\mu _1^*$ and $\mu _1^{**}$. We give sharp estimates of these quantities for some standard examples or under strong symmetry assumptions.

2 Preliminary results

This section is dedicated to some preliminary technical results. The reason is that in order to construct examples and counterexamples, it is often more convenient to use densities that are non-smooth or which vanish somewhere in the manifold. The key arguments used in the proof of these results rely on the method developed by Colin de Verdière in [14].

Let (M, g) be a compact Riemannian manifold, possibly with boundary.

Proposition 2.1

Let $\rho \in L^\infty (M)$ and $\sigma \in C^0(M)$ be two positive densities on M. For every $N\in \mathbb {N}^*$, there exist two sequences of smooth positive densities $\rho _p$ and $\sigma _p$ such that $\forall k\le N$,

$$\begin{aligned} \mu _k(\rho _p,\sigma _p)\rightarrow \mu _k(\rho ,\sigma ) \end{aligned}$$

as $p\rightarrow \infty$.

Proof

Using standard density results, let $\rho _p$ and $\sigma _p$ be two sequences of smooth positive densities such that, $\rho _p$ converges to $\rho$ in $L^2(M)$ and $\sigma _p$ converges uniformly toward $\sigma$. Assume furthermore that $\frac{1}{2} \inf \rho \le \rho _p\le 2\sup \rho$ almost everywhere and that (replacing $\sigma _p$ by $\sigma _p+\Vert \sigma _p-\sigma \Vert _\infty$ if necessary) $\sigma \le \sigma _p$ on M. Then, the sequence of quadratic forms $q_p(u)=\int _M \vert \nabla u\vert ^2 \sigma _p v_g$ together with the sequence of norms $\Vert u\Vert _p^2 =\int _M u^2 \rho _p v_g$ satisfies the assumptions of Theorem I.8 of [14] which enables us to conclude. $\square$

Let $M_0$ be a domain in M with $C^1$-boundary, and let $\rho$ be a positive bounded function on $M_0$. In order to state the next result, let us introduce the following quadratic form defined on $H^1(M_0)$:

$$\begin{aligned} Q_0(u)=\int _{M_0}\vert \nabla u\vert ^2 v_g + \int _{M{\setminus } M_0}\vert \nabla H(u)\vert ^2 v_g, \end{aligned}$$

where H(u) is the harmonic extension of u to $M{\setminus } M_0$, with Neumann condition on $\partial M{\setminus }\partial M_0$ if $\partial M{\setminus }\partial M_0\ne \emptyset$ (i.e., H(u) is harmonic on $M{\setminus } M_0$, coincides with u on $\partial M_0{\setminus }\partial M$, and $\frac{\partial H(u)}{\partial \nu }=0$ on $\partial M{\setminus }\partial M_0$. The function H(u) minimizes $\int _{M{\setminus } M_0}\vert \nabla v\vert ^2 v_g$ among all functions v on $M{\setminus } M_0$ which coincide with u on $\partial M_0{\setminus }\partial M$). We denote by $\gamma _k(M_0,\rho )$ the eigenvalues of this quadratic form with respect to the inner product of $L^2(M_0, \rho v_g)$ associated with $\rho$, that is,

$$\begin{aligned} \gamma _k(M_0,\rho )= \inf _{E\in S^0_{k+1}} \sup _{u\in E{\setminus } \{ 0\} }\frac{\int _{M_0}\vert \nabla u\vert ^2 v_g + \int _{M{\setminus } M_0}\vert \nabla H(u)\vert ^2 v_g }{\int _{M_0} u^2 \rho \, v_g}, \end{aligned}$$

where $S^0_k$ is the set of all k-dimensional vector subspaces of $H^1(M_0)$.

Proposition 2.2

Let $M_0\subset M$ be a domain with $C^1$-boundary, and let $\rho \in L^\infty (M_0)$ be a positive density with ${{\mathrm{ess\,inf}}}_{M_0}\rho >0$. Define, for every $\varepsilon >0$, the density $\rho _\varepsilon \in L^\infty (M)$ by

$$\begin{aligned} \rho _\varepsilon (x)= \left\{ \begin{array}{ll} \rho (x)&{} \quad \text {if } x\in M_0\\ \varepsilon &{} \quad \text {otherwise.}\\ \end{array} \right. \end{aligned}$$

Then, for every positive k, $\mu _k(\rho _\varepsilon ,1)$ converges to $\gamma _k(M_0,\rho )$ as $\varepsilon \rightarrow 0$.

Proof

The eigenvalues $\mu _k(\rho _\varepsilon ,1)$ are those of the quadratic form $q(u)=\int _M \vert \nabla u\vert ^2 v_g$, $u\in H^{1}(M)$, with respect to the inner product $\Vert u\Vert _\varepsilon ^2 = \int _M u^2 \rho _\varepsilon v_g$. Set $M_\infty =M{\setminus } M_0$ and $\Gamma =\partial M_0\cap \partial M_\infty =\partial M_0{\setminus }\partial M$. We identify $H^1(M)$ with the space through the map . We endow $\mathcal H_\varepsilon$ with the inner product given by $\Vert (v_0,v_\infty )\Vert _\rho ^2 = \int _{M_0} v_0^2 \rho v_g+\int _{M_\infty } v_\infty ^2 v_g$ and consider the quadratic form $q_\varepsilon (v_0,v_\infty )=\int _{M_0} \vert \nabla v_0\vert ^2 v_g+\frac{1}{\varepsilon }\int _{M_\infty } \vert \nabla v_\infty \vert ^2 v_g$, so that, for every $u\in H^1(M)$

$$\begin{aligned} \Vert \Psi _\varepsilon (u)\Vert _\rho = \Vert u\Vert _\varepsilon \quad \text{ and } \quad q_\varepsilon (\Psi _\varepsilon (u))=q(u). \end{aligned}$$

Therefore, the eigenvalues of the quadratic form $q:H^1(M)\rightarrow \mathbb {R}$ with respect to $\Vert \ \Vert _\varepsilon$ (i.e., $\mu _k^g(\rho _\varepsilon ,1)$ ) coincide with those of $q_\varepsilon :{\mathcal {H}}_\varepsilon \rightarrow \mathbb {R}$ with respect to $\Vert \ \Vert _\rho$.

The space ${\mathcal {H}}_\varepsilon$ decomposes into the direct sum ${\mathcal {H}}_\varepsilon =\mathcal K_0^\varepsilon \oplus \mathcal K_\infty ^\varepsilon$ with $\mathcal K_0^\varepsilon =\{(v_0,v_\infty )\in {\mathcal {H}}_\varepsilon \ :\ v_\infty \text{ is } \text{ harmonic, } \text{ and } \frac{\partial v_\infty }{\partial \nu }=0 \text{ on } \partial M{\setminus }\partial M_0 \text{ if } \partial M{\setminus }\partial M_0\ne \emptyset \}$, and $\mathcal K_\infty ^\varepsilon =\{(v_0,v_\infty )\in \mathcal H_\varepsilon \ :\ v_0=0 \}$ (Indeed, $v=(v_0,v_\infty )=(v_0, \sqrt{\varepsilon }H(v_0))+ (0, v_\infty -\sqrt{\varepsilon }H(v_0))$). These two subspaces are $q_\varepsilon$-orthogonal and, denoting by $\lambda _1(M_\infty )$ the first eigenvalue of $M_\infty$ under Dirichlet boundary conditions on $\Gamma$ and Neumann boundary conditions on $\partial M_\infty {\setminus }\Gamma$, we have, for every $v=(0,v_\infty )\in \mathcal K_\infty$,

$$\begin{aligned} q_\varepsilon (v)= \frac{1}{\varepsilon }\int _{M_\infty } \vert \nabla v_\infty \vert ^2 v_g \ge \frac{1}{\varepsilon }\lambda _1(M_\infty ) \int _{M_\infty } v_\infty ^2 v_g = \frac{1}{\varepsilon }\lambda _1(M_\infty ) \Vert v\Vert _\rho ^2. \end{aligned}$$

Theorem I.7 of [14] then implies that, given any integer $N>0$, theNfirst eigenvalues$\mu _k(\rho _\varepsilon ,1)$of$q_\varepsilon$on${\mathcal {H}}_\varepsilon$are, for sufficiently small$\varepsilon$, as close as desired to the eigenvalues of the restriction of$q_\varepsilon$on$\mathcal K_0^\varepsilon$.

We still have to compare the eigenvalues of $q_\varepsilon$ on $\mathcal K_0^\varepsilon$, that we denote $\gamma _k(\varepsilon )$, with the eigenvalues $\gamma _k(M_0,\rho )$ of $Q_0$ on $L^2(M_0,\rho v_g)$. For this, we make use of Theorem I.8 of [14]. Indeed, $\mathcal K_0^\varepsilon$ can be identified to $H^1(M_0)$ through $\Psi _\varepsilon ^0: u\in H^1(M_0) \mapsto (u, \sqrt{\varepsilon }H(u))\in \mathcal K_0^\varepsilon$, which satisfies $\Vert \Psi _\varepsilon ^0(u)\Vert _\varepsilon ^2= \int _{M_0} u^2\rho v_g + \varepsilon \int _{M_\infty } H(u)^2 v_g$ and $q_\varepsilon (\Psi _\varepsilon ^0(u)) = Q_0(u) = \int _{M_0} \vert \nabla u\vert ^2 v_g+\int _{M_\infty } \vert \nabla H(u)\vert ^2 v_g$. Hence, we are led to compare, on $L^2(M_0)$, the eigenvalues of the quadratic form $Q_0$ with respect to the following two scalar products: $\Vert u\Vert _\rho ^2 = \int _{M_0} u^2\rho v_g$ and $\Vert u\Vert _\varepsilon ^2=\int _{M_0} u^2\rho v_g + \varepsilon \int _{M_\infty } H(u)^2 v_g$.

Now, since H(u) is a harmonic extension of to $M_\infty$, there exists a constant C, which does not depend on $\varepsilon$, such that $\int _{M_\infty } H(u)^2 v_g \le C \int _{\Gamma } u^2 v_{\bar{g}}$, where $\bar{g}$ is the metric induced on $\Gamma$ by g. Indeed, let $\eta$ be the solution in $M_\infty$ of $\Delta \eta =-1$ with and $\frac{\partial \eta }{\partial \nu } =0$ on $\partial M_\infty {\setminus }\Gamma$. Observe that we have $\eta \ge 0$ (maximum principle and Hopf lemma) and, since $\int _{M_\infty } g(\nabla (\eta H(u)),\nabla H(u)) v_g =0$, $\int _{M_\infty } g(\nabla \eta ,\nabla H(u)^2) v_g = -2\int _{M_\infty } \eta \vert \nabla H(u)\vert ^2 v_g\le 0$. Thus,

$$\begin{aligned} \int _{M_\infty } H(u)^2 v_g =- \int _{M_\infty } H(u)^2 \Delta \eta \, v_g=\int _{M_\infty } g(\nabla \eta ,\nabla H(u)^2) v_g + \int _{\Gamma } u^2 \frac{\partial \eta }{\partial \nu } v_{\bar{g}} \le c \int _{\Gamma } u^2 v_{\bar{g}}, \end{aligned}$$

where c is an upper bound of $\frac{\partial \eta }{\partial \nu }$ on $\Gamma$. On the other hand, $\int _{\Gamma } u^2 v_{\bar{g}}$ is controlled by $\Vert u\Vert ^2_{H^{\frac{1}{2}}(\Gamma )}$ which in turn is controlled (using boundary trace inequalities in $M_0$) by $\Vert u\Vert ^2_{H^1(M_0)}$. Finally, there exists a constant C (which depends on ${{\mathrm{ess\,inf}}}_{M_0} \rho$ but not on $\varepsilon$) such that $\int _{M_\infty } H(u)^2 v_g\le C (\int _{M_0} u^2\rho v_g + \int _{M_0} \vert \nabla u\vert ^2 v_g)$ and, then

$$\begin{aligned} \Vert u\Vert _\varepsilon ^2\le C(\Vert u\Vert _\rho ^2 +Q_0(u)). \end{aligned}$$

Since $\Vert u\Vert _\varepsilon ^2$ converges to $\Vert u\Vert _\rho ^2$ as $\varepsilon \rightarrow 0$, this implies, according to [14, Theorem I.8] (see also [25, Remark 2.14]), that, for sufficiently small $\varepsilon$, the N first eigenvalues $\gamma _k(\varepsilon )$ of $Q_0$ with respect to $\Vert \ \Vert _\varepsilon$ are as close as desired to those, $\gamma _k(M_0,\rho )$, of $Q_0$, with respect to $\Vert \ \Vert _\rho$. $\square$

Recall that in dimension 2, one has

$$\begin{aligned} \mu _k^g(\rho ,1)=\lambda _k(M,\rho g). \end{aligned}$$

(7)

An immediate consequence of Proposition 2.2 is the following result which completes Theorem III.1 of Colin de Verdière [14].

Theorem 2.1

Let (M, g) be a compact Riemannian manifold of dimension $n\ge 2$ and let $M_0\subset M$ be a domain with boundary of class $C^1$. Let $g_{\varepsilon }$ be a family of Riemannian metrics on M, with $g_{\varepsilon }=g$ on $M_0$ and $g_{\varepsilon }=\varepsilon g$ outside $M_0$. Let $k\ge 1$.

• 1. (Theorem III.1 of [14]) If $n\ge 3$, then $\lambda _k(M,g_{\varepsilon })$ converges to $\lambda _k(M_0,g)$ as $\varepsilon \rightarrow 0$

• 2. If $n=2$, then $\lambda _k(M,g_{\varepsilon })$ converges to $\gamma _k(M_0, 1)$ as $\varepsilon \rightarrow 0$.

From Propositions 2.1 and 2.2, we can deduce the following two corollaries:

Corollary 2.1

Let $\rho \in L^\infty (M_0)$ be a positive density on a domain $M_0\subset M$ with boundary of class $C^1$. There exists a family of smooth positive densities $\rho _\varepsilon$ on M such that $\int _M \rho _\varepsilon v_g$ tends to $\int _{M_0} \rho v_g$ and, for every $k\in \mathbb {N}^*$, $\mu _k(\rho _\varepsilon ,1)$ converges to $\gamma _k(M_0,\rho )$ as $\varepsilon \rightarrow 0$.

Corollary 2.2

Let (M, g) be a compact manifold possibly with boundary, and let $M_0\subset M$ be a domain with boundary of class $C^1$. For every integer $k>0$ and every $\varepsilon >0$, there exists a positive smooth density $\rho _\varepsilon$ on M such that $\int _M \rho _\varepsilon v_g=\vert M\vert _g$ and

$$\begin{aligned} \mu _k(\rho _\varepsilon ,1)\ge \frac{\vert M_0\vert _g}{\vert M\vert _g}\lambda _k(M_0,g) -\varepsilon . \end{aligned}$$

Proof

Let $\rho$ be the density on $M_0$ defined by $\rho = \frac{\vert M\vert _g}{\vert M_0\vert _g}$. We apply Corollary 2.1 taking into account that $\gamma _k(M_0, \rho )= \frac{\vert M_0\vert _g}{\vert M\vert _g}\gamma _k(M_0, 1) \ge \frac{\vert M_0\vert _g}{\vert M\vert _g}\lambda _k(M_0,g)$. $\square$

Remark 2.1

In dimension 2, it is clear from (7) that the problem of minimizing or maximizing $\mu _k^{g}(\rho ,1)$ w.r.t. $\rho$ is equivalent to the problem of minimizing or maximizing $\lambda _k(M,g)$ w.r.t. conformal deformations of the metric g. In dimension $n\ge 3$, the two problems are completely different. To emphasize this difference, observe that, given a positive constant c, one has

$$\begin{aligned} \inf _{\rho \le c} \mu _k^{g}(\rho ,1)\ge \frac{1}{c} \mu _k^{g}(1,1)=\frac{1}{c} \lambda _k(M,g)>0 \end{aligned}$$

while

$$\begin{aligned} \inf _{\rho \le c} \lambda _k(M,\rho g)=0.\ \end{aligned}$$

Indeed, let $B_j$, $j\le k+1$ be a family of mutually disjoint balls in M and consider the density $\rho _\varepsilon$ which is equal to c on each $B_j$ and equal to $\varepsilon$ elsewhere. According to [14, Theorem III.1], $\lambda _k(M,\rho _\varepsilon g)$ converges as $\varepsilon \rightarrow 0$ to the $(k+1)$th Neumann eigenvalue of the union of balls which is zero.

3 Bounding the eigenvalues from below

3.1 Nonexistence of “density-free” lower bounds

Let (M, g) be a compact Riemannian manifold of dimension $n\ge 2$, possibly with boundary, and denote by [g] the set of all Riemannian metrics $g'$ on M which are conformal to g with $\vert M\vert _{g'}=\vert M \vert _{g}$. It is well known that $\lambda _k(M,g')$ can be as small as desired when $g'$ varies within [g], i.e., $\inf _{g'\in [g]}\lambda _k(M,g) =0$ (Cheeger dumbbells). Since $\mu _k^{g}(\rho ,\rho ^{\frac{n-2}{n}})= \lambda _k(M, \rho ^{\frac{2}{n}}g)$, this property is equivalent to

$$\begin{aligned} \inf _{\int _M \rho v_{g}= \vert M\vert _g} \mu _k^{g}(\rho ,\rho ^{\frac{n-2}{n}}) =0. \end{aligned}$$

(8)

Let us denote by $\mathcal R_0$ the set of positive smooth functions $\phi$ on M satisfying $\fint _M \phi v_{g} =1$, where

$$\begin{aligned} \fint _M \phi v_{g} =\frac{1}{\vert M\vert _{g}} \int _M \phi v_{g}. \end{aligned}$$

The following theorem shows that $\mu _k(\rho ,\sigma )$ is not bounded below when one of the densities $\rho ,\sigma$ is fixed and the second one is varying within $\mathcal R_0$. We also deal with the case $\sigma =\rho ^p$, $p\ge 0$, which includes (8) and the case of the Witten Laplacian.

Theorem 3.1

For every positive integer k, one has, $\forall p>0$

$$\begin{aligned}&\mathrm{i.} \quad \inf _{ \rho \in \mathcal R_0 } \mu _k(\rho ,1) = 0 \\&\mathrm{ii.} \quad \inf _{\sigma \in \mathcal R_0} \mu _k(1,\sigma ) = 0 \\&\mathrm{iii.} \quad \inf _{\rho \in \mathcal R_0}\mu _k(\rho ,\rho ^p) =0. \end{aligned}$$

Proof of Theorem 3.1

(i) In dimension 2, one has $\mu _k(\rho ,1) = \lambda _k(M,\rho g)$ and the problem is equivalent to that of deforming conformally the metric g into a metric $\rho g$ whose kth eigenvalue is as small as desired. The existence of such a deformation is well known.

Assume now that the dimension of M is at least 3. Let us choose a point $x_0$ in M. The Riemannian volume of a geodesic ball B(x, r) of radius r in M is asymptotically equivalent, as $r\rightarrow 0$, to $\omega _{n} r^{n}$, where $\omega _{n}$ is the volume of the unit ball in the n-dimensional Euclidean space. Therefore, there exists $\varepsilon _0 \in (0,1)$ sufficiently small and $N\in \mathbb {N}$ so that, for every $r < \frac{\varepsilon _0}{N}$ and every $x\in B(x_0,\varepsilon _0)$,

$$\begin{aligned} \frac{1}{2} \omega _{n} r^{n}\le |B(x,r)| \le 2\omega _{n} r^{n}. \end{aligned}$$

(9)

Fix a positive integer k and let $\delta =\frac{n-2}{4}$ so that $\delta <\frac{n}{2} -1$. One can choose $N\in \mathbb {N}$ sufficiently large so that, for every $\varepsilon < \frac{\varepsilon _0}{N}$, the ball $B(x_0,\varepsilon )$ contains k mutually disjoint balls of radius $2\varepsilon ^{\frac{n}{2} -\delta }$ (indeed, since $\frac{n}{2} -\delta >1$, $2\varepsilon ^{\frac{n}{2} -\delta }$ is very small compared to $\varepsilon$ as the latter tends to zero). We consider a smooth positive density $\rho _\varepsilon$ such that $\rho _\varepsilon = \frac{1}{\varepsilon ^n}$ inside $B(x_0,\varepsilon )$, $\rho _\varepsilon =\varepsilon$ in $M{\setminus } B(x_0,2\varepsilon )$, and $\rho _\varepsilon \le \frac{1}{\varepsilon ^n}$ elsewhere. Thanks to (9), one has

$$\begin{aligned} \int _M \rho _\varepsilon v_{g} \le \frac{1}{\varepsilon ^n}\vert B(x_0,2\varepsilon )\vert _g + \varepsilon \vert M\vert _g\le 2^{n+1}\omega _{n} + \varepsilon \vert M\vert _g . \end{aligned}$$

For simplicity, we set $\alpha = {\frac{n}{2} -\delta }= \frac{n+2}{4}$ and denote by $x_1,\dots , x_k$ the centers of k mutually disjoint balls of radius $2\varepsilon ^{\alpha }$ contained in $B(x_0,\varepsilon )$.

For each $i\le k$, we denote $f_i$ the function which vanishes outside $B(x_i,2\varepsilon ^{\alpha })$, equals 1 in $B(x_i,\varepsilon ^{\alpha })$, and $f_i(x)=2-\frac{1}{\varepsilon ^{\alpha }} d_{g}(x,x_i)$ for every x in the annulus $B(x_i,2\varepsilon ^{\alpha }) {\setminus } B(x_i,\varepsilon ^{\alpha })$. The norm of the gradient of $f_i$ vanishes everywhere except inside the annulus where we have $|\nabla f_i|= \frac{1}{\varepsilon ^{\alpha }}$. Thus, using (9),

$$\begin{aligned} \int _M f_i^2 \rho _\varepsilon v_g\ge \frac{1}{\varepsilon ^{n}} \int _{B(x_i,\varepsilon ^{\alpha })} f_i^2 v_g = \frac{|B(x_i,\varepsilon ^{\alpha })|}{\varepsilon ^{n}}\ge \frac{1}{2} \omega _n \varepsilon ^{n(\alpha -1)} \end{aligned}$$

and

$$\begin{aligned} \int _M |\nabla f_i|^2 v_g\le \frac{|B(x_i,2\varepsilon ^{\alpha })|}{\varepsilon ^{2\alpha }}\le 2^{n+1} \omega _n \varepsilon ^{\alpha (n-2)} . \end{aligned}$$

Thus,

$$\begin{aligned} R_{(g,\rho _\varepsilon ,1)}(f_i)\le {2^{n+2}}\varepsilon ^{n-2\alpha }={2^{n+2}}\varepsilon ^{\frac{n-2}{2}}. \end{aligned}$$

In conclusion, we have

$$\begin{aligned} \mu _k(\rho _\varepsilon ,1)\le {2^{n+2}}\varepsilon ^{\frac{n-2}{2}} \end{aligned}$$

and

$$\begin{aligned} \mu _k\left( \frac{\rho _\varepsilon }{\fint _M \rho _\varepsilon v_g},1\right) =\mu _k(\rho _\varepsilon ,1)\fint _M \rho _\varepsilon v_g\le {2^{n+2}}\left( \frac{2^{n+1}\omega _{n}}{\vert M\vert _g}\varepsilon ^{\frac{n-2}{2}} + \varepsilon ^{\frac{n}{2}} \right) . \end{aligned}$$

Letting $\varepsilon$ tend to zero we get the result.

(ii) The proof is similar to the previous one. For $\varepsilon$ sufficiently small, we may assume that there exist $k+1$ mutually disjoint balls $B(x_i,\varepsilon ^2)$ inside a ball $B(x_0,\varepsilon )$ and consider any function $\sigma _\varepsilon \in \mathcal R_0$ such that $\sigma _\varepsilon =\varepsilon ^5$ inside $B(x_0,\varepsilon )$. For each $i\le k+1$, let $f_i$ be the function which vanishes outside $B(x_i,2\varepsilon ^{2})$, equals 1 in $B(x_i,\varepsilon ^{2})$, and $f_i(x)=2-\frac{1}{\varepsilon ^{2}} d_{g}(x,x_i)$ in $B(x_i,2\varepsilon ^{2}) {\setminus } B(x_i,\varepsilon ^2)$. As before,

$$\begin{aligned} \int _M f_i^2 v_g\ge \int _{B(x_i,\varepsilon ^{2})} f_i^2 \hbox {d}x \ge |B(x_i,\varepsilon ^{2})| \ge \frac{1}{2}\omega _n \varepsilon ^{2n} \end{aligned}$$

and

$$\begin{aligned} \int _M |\nabla f_i|^2\sigma _\varepsilon v_g\le \frac{1}{\varepsilon ^4 } \int _{B(x_i,2\varepsilon ^{2})}\sigma _\varepsilon v_g \le \varepsilon |B(x_i,2\varepsilon ^{2})|\le 2^{n+1} \omega _n \varepsilon ^{2n+1}. \end{aligned}$$

Thus,

$$\begin{aligned} \mu _k(1,\sigma _\varepsilon )\le \max _{i\le k+1} \frac{\int _M |\nabla f_i|^2\sigma _\varepsilon v_g}{\int _M f_i^2 v_g}\le 2^{n+2} \varepsilon . \end{aligned}$$

(iii) For sufficiently small $\varepsilon$, let $B(x_i,4\varepsilon )$, $i\le k+1$, be $k+1$ mutually disjoint balls of radius $4\varepsilon$ in M. As before, we can assume that, $\forall r\le 4\varepsilon$, $\frac{1}{2} \omega _{n} r^{n}\le |B(x_i,r)| \le 2\omega _{n} r^{n}$. We define $\rho _\varepsilon$ to be equal to $\frac{1}{\varepsilon ^n}$ on each of the balls $B(x_i,\varepsilon )$ and equal to $\varepsilon ^n$ in the complement of $\cup _{i\le k} B(x_i,2\varepsilon )$. For every $i\le k+1$, the function $f_i$ defined to be equal to 1 on $B(x_i,2\varepsilon )$ and $f_i(x)=2-\frac{1}{2\varepsilon } d_{g}(x,x_i)$ in the annulus $B(x_i,4\varepsilon ){\setminus } B(x_i,2\varepsilon )$ and zero in the complement of $B(x_i,4\varepsilon )$ satisfies

$$\begin{aligned} \int _M f_i^2 \rho _\varepsilon v_g\ge \int _{B(x_i,\varepsilon )} f_i^2 \rho _\varepsilon \hbox {d}x = \frac{1}{\varepsilon ^n} |B(x_i,\varepsilon )| \ge \frac{1}{2}{ \omega _{n}}. \end{aligned}$$

On the other hand, $\forall p>0$,

$$\begin{aligned} \int _M |\nabla f_i|^2\rho _\varepsilon ^p v_g =\varepsilon ^{pn} \int _{B(x_i,4\varepsilon ){\setminus } B(x_j,2\varepsilon )} |\nabla f_i|^2 v_g = \varepsilon ^{pn} \frac{1}{4\varepsilon ^{2}}|B(x_i,4\varepsilon )| \le 2^{2n-1}\omega _n\varepsilon ^{(p+1)n-2}. \end{aligned}$$

Thus,

$$\begin{aligned} \mu _k(\rho _\varepsilon ,\rho _\varepsilon ^p )\le \max _{i\le k+1} \frac{\int _M |\nabla f_i|^2\sigma _\varepsilon v_g}{\int _M f_i^2 v_g}\le 2^{2n}\varepsilon ^{(p+1)n-2}. \end{aligned}$$

Regarding $\fint _M \rho _\varepsilon v_g$, it is clear that it is bounded both from above and from below by positive constants that are independent of $\varepsilon$, which enables us to conclude. $\square$

3.2 Cheeger-type inequality

Theorem 3.1 tells us that it is necessary to involve other quantities than the total mass in order to get lower bounds for the eigenvalues. Our next theorem gives a lower estimate which is modeled on Cheeger’s inequality, with suitably defined isoperimetric constants, as was done by Jammes for Steklov eigenvalues [27].

Let (M, g) be a compact Riemannian manifold, possibly with boundary. The classical Cheeger constant is defined by

$$\begin{aligned} h(M)=\inf _{{\vert D\vert _g} \le \frac{1}{2}\vert M\vert _g }\frac{\vert \partial D{\setminus }\partial M\vert _g}{\vert D\vert _g}=\inf _{D\subset M} \frac{\vert \partial D{\setminus }\partial M \vert _g}{\min \{\vert D\vert _g, \vert M\vert _g - \vert D\vert _g\}}. \end{aligned}$$

Given two positive densities $\rho$ and $\sigma$ on M, we introduce the following Cheeger-type constant:

$$\begin{aligned} h_{\rho , \sigma }(M)=\inf _{\vert D\vert _\sigma \le \frac{1}{2} \vert M\vert _\sigma }\frac{\vert \partial D{\setminus }\partial M\vert _\sigma }{\vert D\vert _\rho } \end{aligned}$$

with $\vert D\vert _\sigma$ (resp. $\vert \partial D{\setminus }\partial M\vert _\sigma$) is the n-volume of D (resp. the $(n-1)$-volume of $\partial D{\setminus }\partial M$) with respect to the measure induced by $\sigma v_g$.

Theorem 3.2

One has

$$\begin{aligned} \mu _1(\rho ,\sigma )\ge \frac{1}{4} h_{\sigma ,\sigma }(M) h_{\rho , \sigma }(M). \end{aligned}$$

Proof

The proof follows the same general outline as the original proof by Cheeger (see [5, 8]). We give here a complete proof in the case where M is a closed manifold. The proof in the case $\partial M\ne \emptyset$ can be done analogously. Let f be a Morse function such that the $\sigma$-volume of its positive nodal domain $\Omega _+(f)=\{f>0\}$ is less or equal to half the $\sigma$-volume of M. For every $t\in (0,\sup f)$ excepting a finite number of values, the set $f^{-1}(t)$ is a regular hypersurface of M. We denote by $v^t_g$ the measure induced on $f^{-1}(t)$ by $v_g$ and set $P_\sigma (t)= \int _{f^{-1}(t)}\sigma v^t_g$. The level sets of f are denoted $\Omega (t)=\{f>t\}$ and we set $V_\sigma (t)=\int _{\Omega (t)} \sigma v_g$ and $V_\rho (t)=\int _{\Omega (t)} \rho \, v_g$ . Using the co-area formula, one gets

$$\begin{aligned} \int _{\Omega _+(f)} |\nabla f| \sigma v_g= \int _0^{+\,\infty } P_\sigma (t)\hbox {d}t . \end{aligned}$$

On the other hand, the same co-area formula gives

$$\begin{aligned} V_\rho (t)= \int _t^{+\,\infty } ds\int _{f^{-1}(s)} \frac{\rho }{|\nabla f| } v^s_g . \end{aligned}$$

Thus,

$$\begin{aligned} V_\rho '(t) = -\int _{f^{-1}(t)} \frac{\rho }{|\nabla f| } v^t_g. \end{aligned}$$

Now,

$$\begin{aligned} \int _{\Omega _+(f)} f \rho \, v_g = \int _0^{+\,\infty }\hbox {d}t \int _{f^{-1}(t)} \frac{ f\rho }{|\nabla f| } v^t_g = \int _0^{+\,\infty } tdt\int _{f^{-1}(t)} \frac{\rho }{|\nabla f| } v^t_g = -\int _0^{+\,\infty } tV_\rho '(t) \hbox {d}t \end{aligned}$$

which gives after integration by parts

$$\begin{aligned} \int _{\Omega _+(f)} f\rho \, v_g= \int _0^{+\,\infty } V_\rho (t) \hbox {d}t . \end{aligned}$$

Similarly, one has

$$\begin{aligned} \int _{\Omega _+(f)} f \sigma v_g = \int _0^{+\,\infty } V_\sigma (t) \hbox {d}t . \end{aligned}$$

Since $P_\sigma (t)\ge h_{\sigma ,\sigma }(M) V_\sigma (t)$ and $P_\sigma (t)\ge h_{\rho , \sigma }(M) V_\rho (t)$, we deduce

$$\begin{aligned} \int _{\Omega _+(f)} |\nabla f| \sigma v_g\ge \max \left\{ h_{\sigma ,\sigma }(M) \int _{\Omega _+(f)} f \sigma v_g \ , \ h_{\rho , \sigma }(M) \int _{\Omega _+(f)} f \rho \, v_g \right\} . \end{aligned}$$

Using the Cauchy–Schwarz inequality, we get

$$\begin{aligned} \int _{\Omega _+(f)} |\nabla f|^2 \sigma v_g\ge & {} \frac{1}{4}\frac{\left( \int _{\Omega _+(f)} |\nabla f^2| \sigma v_g\right) ^2 }{\int _{\Omega _+(f)} f^2 \sigma v_g}\nonumber \\\ge & {} \frac{1}{4}\frac{ h_{\sigma ,\sigma }(M)h_{\rho , \sigma }(M) \int _{\Omega _+(f)} f^2\sigma v_g \int _{\Omega _+(f)} f^2 \rho \, v_g }{\int _{\Omega _+(f)} f^2\sigma v_g}\nonumber \\= & {} \frac{1}{4} h_{\sigma ,\sigma }(M)h_{\rho , \sigma }(M) \int _{\Omega _+(f)} f^2 \rho \, v_g . \end{aligned}$$

(10)

Now, let $m\in \mathbb {R}$ be such that $\vert \{f>m\}\vert _\sigma = \vert \{f<m\}\vert _\sigma =\frac{1}{2} \vert M\vert _\sigma$ (such an m is called a median of f for $\sigma$). Applying (10) to $f-m$ and $m-f$, we get

$$\begin{aligned} \int _{\{f>m\}} |\nabla f|^2 \sigma v_g \ge \frac{1}{4} h_{\sigma ,\sigma }(M)h_{\rho , \sigma }(M) \int _{\{f>m\}} (f-m)^2 \rho \, v_g \end{aligned}$$

and

$$\begin{aligned} \int _{\{f<m\}} |\nabla f|^2 \sigma v_g \ge \frac{1}{4} h_{\sigma ,\sigma }(M)h_{\rho , \sigma }(M) \int _{\{f<m\}} (f-m)^2 \rho \, v_g . \end{aligned}$$

Summing up we obtain

$$\begin{aligned} \int _{M} |\nabla f|^2 \sigma v_g \ge \frac{1}{4} h_{\sigma ,\sigma }(M)h_{\rho , \sigma }(M) \int _{M} (f-m)^2 \rho \, v_g. \end{aligned}$$

Since $\int _{M} (f-m)^2 \rho \, v_g= \int _{M} f^2 \rho \, v_g +m^2\vert M\vert _\rho -2m\int _{M} f \rho \, v_g$, we deduce that, for every f such that $\int _{M} f \rho \, v_g =0$,

$$\begin{aligned} \int _{M} |\nabla f|^2 \sigma v_g \ge \frac{1}{4} h_{\sigma ,\sigma }(M)h_{\rho , \sigma }(M) \int _{M} f^2 \rho \, v_g \end{aligned}$$

which, thanks to (3), implies the desired inequality. $\square$

Remark 3.1

In dimension 2, Theorem 3.2 can be restated as follows: If (M, g) is a compact Riemannian surface, then

$$\begin{aligned} \lambda _1(M,g)\ge \frac{1}{4} \sup _{g'\in [g]}h_{g',g'}(M) h_{g,g'}(M), \end{aligned}$$

(11)

where $h_{g,g'}(M)=\inf _{\vert D\vert _{g'}\le \frac{1}{2} \vert M\vert _{g'} }\frac{\vert \partial D\vert _{g'}}{\vert D\vert _{g}}$. Indeed, for any $g'\in [g]$ there exists a positive $\rho \in C^\infty (M)$ such that $g=\rho g'$. Thus, $\lambda _1(M,g)=\mu _1^{g'}(\rho ,1)$ and (11) follows from Theorem 3.2. This inequality can be seen as an improvement in Cheeger’s inequality since the right-hand side is obviously bounded below by $h_{g,g}(M)^2$. Notice that in [6], Buser gives an example of a family of metrics on the 2-torus such that the Cheeger constant goes to zero, while the first eigenvalue is bounded below. The advantage of (11) is that its right-hand side does not go to zero for Buser’s example.

A natural question is to investigate a possible reverse inequality of Buser’s type (see [6, 32]). The following theorem provides a negative answer to this question.

Theorem 3.3

Let (M, g) be a compact Riemannian manifold, possibly with boundary.

• i. There exists a family of positive densities $\sigma _\varepsilon$, $\varepsilon >0$, on M with $\fint _M \sigma _\varepsilon v_{g}=1$ and such that $h_{1,\sigma _\varepsilon }(M) h_{\sigma _\varepsilon ,\sigma _\varepsilon }(M)$ goes to zero with $\varepsilon$, while $\mu _1(1,\sigma _\varepsilon )$ stays bounded below by a constant C which does not depend on $\varepsilon$.

• ii. There exists a family of positive densities $\rho _\varepsilon$, $\varepsilon >0$, on M with $\fint _M \rho _\varepsilon v_{g}=1$ and such that $h_{\rho _\varepsilon ,1}(M)$ goes to zero with $\varepsilon$, while $\mu _1(\rho _\varepsilon ,1)$ stays bounded below by a constant C which does not depend on $\varepsilon$.

Proof

We start by proving the result for the unit ball $B^n\subset \mathbb {R}^n$ and then explain how to deduce it for any compact Riemannian manifold. For every $r \in (0, 1)$, we denote by B(r) the ball of radius r centered at the origin and by $A_r$ the annulus $B^n{\setminus } B( r )$. In the sequel, whenever we integrate over a Euclidean set, the integration is implicitly made with respect to the standard Lebesgue’s measure.

Proof of (i): For every $\varepsilon \in (0,\frac{1}{2})$, we define a smooth nonincreasing radial density $\sigma _\varepsilon$ on $B^n$ such that $\sigma _\varepsilon = \frac{1}{\varepsilon ^{1+a}}$, with $a\in (0,1)$ (e.g., $a=\frac{1}{2}$) inside $B^n(\varepsilon )$ and $\sigma _\varepsilon = b_\varepsilon$ in $B^n{\setminus } B(2\varepsilon )$, where $b_\varepsilon$ is chosen so that $\int _{B^n}\sigma _\varepsilon = \omega _n,$ the volume of $B^n$. We then have

$$\begin{aligned} \int _{B(\varepsilon )} \sigma _\varepsilon = \omega _n \varepsilon ^{n-1-a}\quad \text{ and } \quad \int _{A_{2\varepsilon }}\sigma _\varepsilon =\omega _n(1-2^n\varepsilon ^n)b_\varepsilon . \end{aligned}$$

Since $\int _{B^n} \sigma _\varepsilon = \omega _n$ and $b_\varepsilon \le \sigma _\varepsilon \le \varepsilon ^{-1-a}$ on $B(2\varepsilon ){\setminus } B(\varepsilon )$, we have

$$\begin{aligned} \omega _n \varepsilon ^{n-1-a}+b_\varepsilon \omega _n(1-\varepsilon ^n) \le \omega _n\le \omega _n 2^n \varepsilon ^{n-1-a}+b_\varepsilon \omega _n(1-2^n\varepsilon ^n), \end{aligned}$$

that is,

$$\begin{aligned} \frac{1-2^n\varepsilon ^{n-1-a}}{1-2^n\varepsilon ^n}\le b_\varepsilon \le \frac{1-\varepsilon ^{n-1-a}}{1-\varepsilon ^n}. \end{aligned}$$

(12)

Now, the Cheeger constant $h_{\sigma _\varepsilon ,\sigma _\varepsilon }(B^n)$ satisfies

$$\begin{aligned} h_{\sigma _\varepsilon ,\sigma _\varepsilon }(B^n)\le \frac{\vert \partial B(2\varepsilon )\vert _{\sigma _\varepsilon }}{\vert B(2\varepsilon )\vert _{\sigma _\varepsilon } }\le \frac{\vert \partial B(2\varepsilon )\vert _{\sigma _\varepsilon }}{\vert B(\varepsilon )\vert _{\sigma _\varepsilon } } = \frac{nb_\varepsilon \omega _n(2\varepsilon )^{n-1}}{\omega _n\varepsilon ^{n-1-a}}\le n2^{n-1}\varepsilon ^{a}. \end{aligned}$$

On the other hand, for $r_0 = \left( \frac{1}{4}\right) ^{\frac{1}{n}}$ we have $\vert B( r_0 )\vert _{\sigma _\varepsilon }<\omega _n(\varepsilon ^{n-1-a}+ \frac{1}{4}b_\varepsilon ) <\frac{1}{2} \omega _n$ when $\varepsilon$ is sufficiently small, so that

$$\begin{aligned} h_{1,\sigma _\varepsilon }(B^n)\le \frac{\vert \partial B( r_0 )\vert _{\sigma _\varepsilon } }{\vert B( r_0 )\vert } = \frac{n\omega _n r_0^{n-1}b_\varepsilon }{ \omega _n r_0^n}\le 4^{\frac{1}{n}}n. \end{aligned}$$

Hence, the product $h_{1,\sigma _\varepsilon } (B^n)h_{\sigma _\varepsilon ,\sigma _\varepsilon }(B^n)$ tends to zero as $\varepsilon \rightarrow 0$. Regarding the first positive eigenvalue $\mu _1( 1,\sigma _\varepsilon )$, if f is a corresponding eigenfunction, then $\int _{B^n} f =0$ and

$$\begin{aligned} \mu _1( 1,\sigma _\varepsilon ) = \frac{\int _{B^n} \vert \nabla f\vert ^2\sigma _\varepsilon }{\int _{B^n} f^2 } \ge b_\varepsilon \frac{\int _{B^n} \vert \nabla f\vert ^2 }{\int _{B^n} f^2 }\ge b_\varepsilon \lambda _1(B^n, g_E) \end{aligned}$$

with $b_\varepsilon \ge \frac{1}{2}$ for sufficiently small $\varepsilon$ according to (12).

Now, given a Riemannian manifold (M, g), we fix a point $x_0$ and choose $\delta >0$ so that the geodesic ball $B(x_0,\delta )$ is 2-quasi-isometric to the Euclidean ball of radius $\delta$. In the Riemannian manifold $(M,\frac{1}{\delta ^2}g)$, the ball $B(x_0,1)$ is 2-quasi-isometric to the Euclidean ball $B^n$. We define $\sigma _\varepsilon$ in $B(x_0,1)$ as the pullback of the function $\sigma _\varepsilon$ constructed above, and extend it by $b_\varepsilon$ in $M{\setminus } B(x_0,1)$. Because of (12), we easily see that $\fint _M \sigma _\varepsilon v_g$ stays bounded independently from $\varepsilon$. We can also check that $h_{1,\sigma _\varepsilon }(M)$ and $h_{\sigma _\varepsilon ,\sigma _\varepsilon }(M)$ have the same behavior as before and that (since $\sigma _\varepsilon \ge b_\varepsilon \ge \frac{1}{2}$) the eigenvalue $\mu _1^{\delta ^{-2}g}(1,\sigma _{\varepsilon })$ is bounded from below by $\frac{1}{2}\lambda _1(M,\delta ^{-2}g)$ which is a positive constant C independent of $\varepsilon$. Thus, $\mu _1^{g}(1,\sigma _{\varepsilon })= \delta ^2\mu _1^{\delta ^{-2}g}(1,\sigma _{\varepsilon })\ge C\delta ^2.$

Proof of (ii) As before we define the density $\rho _\varepsilon \in L^\infty (B^n)$, $\varepsilon \in (0,\frac{1}{2})$, by

$$\begin{aligned} \rho _\varepsilon = \left\{ \begin{array}{lll} \frac{1}{\varepsilon ^{1+a}} &{} \ &{} \text {if} \ x\in B(\varepsilon )\\ b_\varepsilon =\frac{1-\varepsilon ^{n-1-a}}{1-\varepsilon ^n} &{} \ &{} \text {if} \ x\in B^n{\setminus } B(\varepsilon ) \end{array} \right. \end{aligned}$$

(13)

so that $\int _{B^n}\rho _\varepsilon \hbox {d}x = \omega _n$ and $b_\varepsilon <1$. The corresponding Cheeger constant satisfies

$$\begin{aligned} h_{\rho _\varepsilon , 1} \le \frac{\vert \partial B(\varepsilon ) \vert }{\vert B(\varepsilon )\vert _{\rho _\varepsilon }} = \frac{n\omega _n \varepsilon ^{n-1}}{\omega _n \varepsilon ^{n-1-a}}= n\varepsilon ^{a}, \end{aligned}$$

which goes to zero as $\varepsilon \rightarrow 0$.

To prove that the first positive Neumann eigenvalue $\mu _1(\rho _\varepsilon ,1)$ is uniformly bounded below we will first prove that the first Dirichlet eigenvalue $\lambda _1(\rho _\varepsilon )$ satisfies

$$\begin{aligned} \lambda _1(\rho _\varepsilon ) \ge \frac{1}{4} \lambda ^*, \end{aligned}$$

(14)

where $\lambda ^*$ is the first Dirichlet eigenvalue of the Laplacian on $B^n$. Indeed, let f be a positive eigenfunction associated to $\lambda _1(\rho _\varepsilon )$. Such a function is necessarily a nonincreasing radial function and it satisfies (with $b_\varepsilon \le 1$)

$$\begin{aligned} \lambda _1(\rho _\varepsilon )= \frac{\int _{B(\varepsilon )}\vert \nabla f\vert ^2 + \int _{A_\varepsilon }\vert \nabla f\vert ^2}{\int _{B(\varepsilon )}f^2\rho _\varepsilon + \int _{A_\varepsilon }f^2\rho _\varepsilon }\ge \frac{\int _{B(\varepsilon )}\vert \nabla f\vert ^2 + \int _{A_\varepsilon }\vert \nabla f\vert ^2}{\varepsilon ^{-1-a}\int _{B(\varepsilon )}f^2 + \int _{A_\varepsilon }f^2}. \end{aligned}$$

(15)

For convenience, we assume that $f(\varepsilon )=1$.

If we denote by $\nu (A_\varepsilon )$ the first eigenvalue of the mixed eigenvalue problem on the annulus $A_\varepsilon$, with Dirichlet conditions on the outer boundary and Neumann conditions on the inner boundary, then it is well known that $\nu (A_\varepsilon )$ converges to $\lambda ^*$ as $\varepsilon \rightarrow 0$ (see[1]). Thus, using the min-max, we will have for sufficiently small $\varepsilon$,

$$\begin{aligned} \int _{A_\varepsilon }\vert \nabla f\vert ^2\ge \nu (A_\varepsilon ) { \int _{A_\varepsilon }f^2}\ge \frac{1}{2} \lambda ^*{ \int _{A_\varepsilon }f^2}. \end{aligned}$$

(16)

On the other hand, since $f-1$ vanishes along $\partial B(\varepsilon )$, its Rayleigh quotient is bounded below by $\frac{1}{\varepsilon ^2}\lambda ^*$, the first Dirichlet eigenvalue of $B(\varepsilon )$. Thus,

$$\begin{aligned} \int _{B(\varepsilon )} \vert \nabla f\vert ^2\ge \frac{1}{\varepsilon ^2} \lambda ^* \int _{B(\varepsilon )}(f-1)^2 \ge \frac{1}{\varepsilon ^2} \lambda ^*\left( \int _{B(\varepsilon )} f^2 -2\int _{B(\varepsilon )}f\right) \end{aligned}$$

(17)

with

$$\begin{aligned} \int _{B(\varepsilon )}f\le \left( \omega _n\varepsilon ^n \int _{B(\varepsilon )} f^2\right) ^{\frac{1}{2}} . \end{aligned}$$

Thus, if $\omega _n\varepsilon ^n \le \frac{1}{16} \int _{B(\varepsilon )} f^2$, then (17) yields

$$\begin{aligned} \int _{B(\varepsilon )} \vert \nabla f\vert ^2\ge \frac{1}{2\varepsilon ^2} \lambda ^* \int _{B(\varepsilon )}f^2 > \frac{1}{2} \lambda ^* \varepsilon ^{-1-a}\int _{B(\varepsilon )}f^2 \end{aligned}$$

which, combined with (16) and (15), implies (14).

Assume now that $\omega _n\varepsilon ^n \ge \frac{1}{16} \int _{B(\varepsilon )} f^2$ and let us prove the following:

$$\begin{aligned} \int _{A_\varepsilon }\vert \nabla f\vert ^2\ge \left\{ \begin{array}{lll} \frac{n(n-2)}{16\varepsilon ^{1-a}} \ \varepsilon ^{-1-a}\int _{B(\varepsilon )}f^2 &{} \ &{} \text {if} \ n\ge 3\\ \frac{1}{8 \varepsilon ^{1-a}\ln (1/ \varepsilon )} \ \varepsilon ^{-1-a}\int _{B(\varepsilon )}f^2 &{} \ &{} \text {if} \ n=2 \end{array}\right. \end{aligned}$$

(18)

which implies for sufficiently small $\varepsilon$,

$$\begin{aligned} \int _{A_\varepsilon }\vert \nabla f\vert ^2\ge \frac{1}{2} \lambda ^*\varepsilon ^{-1-a}\int _{B(\varepsilon )}f^2 \end{aligned}$$

(19)

enabling us to deduce (14) from (15) and (16). Indeed, since $f(\varepsilon ) =1$ and $f(1)=0$, one has $\int _{\varepsilon }^{1} f' =-1.$ Therefore, applying the Cauchy–Schwarz inequality to the product $f' = \left( f'r^{(n-1)/2}\right) r^{-(n-1)/2}$, we get

$$\begin{aligned} \frac{1}{n\omega _n}\int _{A_\varepsilon }\vert \nabla f\vert ^2 =\int _{\varepsilon }^{1} f'^2 r^{n-1} \ge \left( \int _{\varepsilon }^{1} f' \right) ^2 \left( \int _{\varepsilon }^{1}\frac{1}{r^{n-1}}\right) ^{-1} \ge \frac{1}{ \int _{\varepsilon }^{1}\frac{1}{r^{n-1}}} \end{aligned}$$

with

$$\begin{aligned} \int _{\varepsilon }^{1}\frac{1}{r^{n-1}} = \left\{ \begin{array}{lll} \frac{1}{n-2}\left( \frac{1}{\varepsilon ^{n-2}}-1 \right) < \frac{1}{n-2} \frac{1}{\varepsilon ^{n-2}} &{} \ &{} \text {if} \ n\ge 3\\ \ln ({1} /\varepsilon ) &{} \ &{} \text {if} \ n=2 \end{array}\right. . \end{aligned}$$

(20)

Therefore,

$$\begin{aligned} \int _{A_\varepsilon }\vert \nabla f\vert ^2\ge \left\{ \begin{array}{lll} {n(n-2)\omega _n}\varepsilon ^{n-2} &{} \ &{} \text {if} \ n\ge 3\\ \frac{2\pi }{\ln (1/ \varepsilon )} &{} \ &{} \text {if} \ n=2 \end{array}\right. \end{aligned}$$

(21)

which gives (18) since $\omega _n\varepsilon ^n \ge \frac{1}{16} \int _{B(\varepsilon )} f^2$.

Let us check now that the first positive Neumann eigenvalue is also uniformly bounded from below. Indeed, let f be a Neumann eigenfunction with $\Delta f=-\mu _1(\rho _\varepsilon ,1) \rho _\varepsilon f$. If f is radial, then $\mu _1(\rho _\varepsilon ,1)\ge \lambda _1(\rho _\varepsilon )\ge \frac{1}{4}\lambda ^*$ (there exists $r_0<1$ with $f(r_0)=0$ so that f is a Dirichlet eigenfunction on the ball $B(r_0)$). If f is not radial, then, up to averaging (or assuming that f is orthogonal to radial functions), one can assume w.l.o.g. that $\int _{\mathbb {S}^{n-1}( r )} f \hbox {d}\theta =0$ for every $r<1$. Thus, $\int _{\mathbb {S}^{n-1}( r )} \vert \nabla ^{0} f\vert ^2 \hbox {d}\theta \ge \frac{n-1}{r^2}\int _{\mathbb {S}^{n-1}( r )} f^2 \hbox {d}\theta$, where $\nabla ^0f$ is the tangential part of $\nabla f$. Hence,

$$\begin{aligned} \int _{B^n} \vert \nabla f\vert ^2= & {} \int _0^1 r^{n-1} dr\int _{\mathbb {S}^{n-1}( r )} \vert \nabla f\vert ^2 \hbox {d}\theta \ge (n-1) \int _0^1 r^{n-1} dr\int _{\mathbb {S}^{n-1}( r )} \left( \frac{f}{r}\right) ^2 \hbox {d}\theta \\= & {} (n-1)\int _{B^n} \left( \frac{f}{r}\right) ^2 \ge (n-1) \int _{B^n} f^2 \rho _\varepsilon \end{aligned}$$

since $\rho _\varepsilon ( r )\le \frac{1}{r^2}$ everywhere. Thus, in this case, $\mu _1(\rho _\varepsilon ,1)\ge n-1$. Finally,

$$\begin{aligned} \mu _1(\rho _\varepsilon ,1)\ge \min (n-1,\frac{1}{4}\lambda ^*). \end{aligned}$$

As before, this construction can be implemented in any Riemannian manifold (M, g), using a quasi-isometry argument, Proposition 2.2 and Corollary 2.1. $\square$

A relevant problem is to know if a Buser’s type inequality can be obtained in this context under assumptions on the volume of balls with respect to $\sigma$ and $\rho$.

4 Bounding the eigenvalues from above

4.1 Unboundedness of eigenvalues if only one parameter among $g, \rho ,\sigma$ is fixed

Let $(M,g_0)$ be a compact Riemannian manifold, possibly with boundary. Our first observation in this section is that the eigenvalues $\mu _k^g( \rho ,\sigma )$ are not bounded from above when one quantity among $g\in [g_0], \rho \in \mathcal R_0,\sigma \in \mathcal R_0$ is fixed and the two others are varying (here $\mathcal R_0=\{\phi \in C^\infty (M)\ : \phi >0 \text{ and } \fint _M\phi \, v_{g_0}=1\}$).

Let us first recall that the authors and Savo have proved in [12] that on any compact Riemannian manifold $(M,g_0)$ there exists a sequence of densities $\rho _j\in \mathcal R_0$ such that $\mu _1^{g_0}(\rho _j,\rho _j)$ tends to $+\,\infty$ with j. In particular,

$$\begin{aligned} \sup _{\fint _M \rho v_{g_0}= 1, \fint _M \sigma v_{g_0}=1} \mu _1^{g_0}(\rho ,\sigma ) \ge \sup _{\fint _M \rho v_{g_0}= 1} \mu _1^{g_0}(\rho ,\rho ) =+\,\infty . \end{aligned}$$

(22)

A natural subsequent question is: Can one construct examples of $g\in [g_0]$ and $\rho \in \mathcal R_0$ (resp. $\sigma \in \mathcal R_0$) so that $\mu _1^g(\rho ,1)$ (resp. $\mu _1^g(1,\sigma )$) is as large as desired ?

Proposition 4.1

Let $(M,g_0)$ be a compact Riemannian manifold, possibly with boundary. Then,

$$\begin{aligned} \sup _{g\in [g_0], \, \rho \in \mathcal R_0} \mu _1^g(\rho ,1) =+\,\infty \end{aligned}$$

(23)

and

$$\begin{aligned} \sup _{g\in [g_0], \, \sigma \in \mathcal R_0} \mu _1^g(1,\sigma ) =+\,\infty . \end{aligned}$$

(24)

Proof

To prove (23), the idea is to deform both the metric and the density so that $\rho _\varepsilon v_{g_\varepsilon }$ becomes everywhere small. Indeed, let V be an open set of M with $\vert V\vert _{g_0} \ge \frac{1}{10}\vert M\vert _{g_0}$. For every $\varepsilon \in (0,1)$, we consider a continuous density $\rho _\varepsilon$ such that $\rho _\varepsilon =\varepsilon$ on V, $\varepsilon \le \rho _\varepsilon \le 2$ everywhere on M, and $\fint _M \rho _\varepsilon v_{g_0}= 1$. Define $g_\varepsilon = \phi _\varepsilon ^2 g_0$ with

$$\begin{aligned} \phi _\varepsilon ^n=\frac{\vert M\vert _{g_0} }{\int _M \rho _\varepsilon ^{-1} v_{g_0}} \ \frac{1}{\rho }_\varepsilon \end{aligned}$$

so that $\vert M\vert _{g_\varepsilon } =\int _M \phi _\varepsilon ^n v_{g_0}= \vert M\vert _{g_0}$ (here n denotes the dimension of M). Now, we observe that

$$\begin{aligned} \frac{1}{\varepsilon } \vert M\vert _{g_0}\ge {\int _M \rho _\varepsilon ^{-1} v_{g_0}}\ge {\int _V \rho _\varepsilon ^{-1} v_{g_0}} =\frac{1}{\varepsilon } \vert V\vert _{g_0}\ge \frac{1}{10 \varepsilon } \vert M\vert _{g_0}. \end{aligned}$$

Thus,

$$\begin{aligned} \phi _\varepsilon ^n\le \frac{10 \varepsilon }{\rho _\varepsilon } \end{aligned}$$

and, since $\rho _\varepsilon \le 2$,

$$\begin{aligned} \phi _\varepsilon ^n\ge \frac{\varepsilon }{\rho _\varepsilon } \ge \frac{\varepsilon }{2}. \end{aligned}$$

Now, for any smooth function u on M one has (with $\frac{\varepsilon }{2}\le \phi _\varepsilon ^n\le \frac{10 \varepsilon }{\rho _\varepsilon }$)

$$\begin{aligned} \frac{\int _M \vert \nabla u\vert ^2 v_{g_\varepsilon }}{\int _M u^2 \rho _\varepsilon v_{g_\varepsilon }} = \frac{\int _M \vert \nabla u\vert ^2 \phi _\varepsilon ^{n-2} v_{g_0}}{\int _M u^2 \rho _\varepsilon \phi _\varepsilon ^n v_{g_0}} \ge \frac{1}{2^{\frac{n-2}{n}}10\varepsilon ^{\frac{2}{n}}} \frac{\int _M \vert \nabla u\vert ^2 v_{g_0}}{\int _M u^2 v_{g_0}}. \end{aligned}$$

Therefore,

$$\begin{aligned} \mu _1^{g_\varepsilon }(\rho _\varepsilon ,1)\ge \frac{1}{2^{\frac{n-2}{n}}10\varepsilon ^{\frac{2}{n}}} \mu _1^{g_0}(1 , 1) \end{aligned}$$

which tends to infinity as $\varepsilon$ goes to zero.

To prove (24) we first observe that, for any positive density $\sigma$, one has, $\forall u\in C^2(M)$,

$$\begin{aligned} R_{(\sigma g_0, 1, \sigma )}(u) =R_{(g_0, \sigma ^{\frac{n}{2}}, \sigma ^{\frac{n}{2}})}(u). \end{aligned}$$

Thus,

$$\begin{aligned} \mu _k^{\sigma g_0}( 1, \sigma )= \mu _k^{ g_0}( \sigma ^{\frac{n}{2}}, \sigma ^{\frac{n}{2}}). \end{aligned}$$

According to [12], there exists on M a sequence $\sigma _j$ of positive densities such that $\int _M \sigma _j^{\frac{n}{2}}v_{g_0}=\vert M\vert _{g_0}$ and $\mu _k^{ g_0}(\sigma _j^{\frac{n}{2}}, \sigma _j^{\frac{n}{2}})$ tends to infinity with j. We set $g_j=\sigma _j g_0 \in [g_0]$. Hölder’s inequality implies that

$$\begin{aligned} \int _M \sigma _j v_{g_0}\le \left( \int _M \sigma _j^{\frac{n}{2}}v_{g_0}\right) ^{\frac{2}{n}} \vert M\vert _{g_0}^{1-\frac{2}{n}} =\vert M\vert _{g_0}. \end{aligned}$$

Setting $\sigma '_j=\frac{\sigma _j}{\fint _M \sigma _j v_{g_0}} \in \mathcal R_0$, we get

$$\begin{aligned} \mu _k^{ g_j}(1, \sigma '_j)=\frac{1}{\fint _M \sigma _j v_{g_0}} \mu _k^{ \sigma _jg_0}( 1, \sigma _j)\ge \mu _k^{ \sigma _jg_0}( 1, \sigma _j)=\mu _k^{g_0}( \sigma _j^{\frac{n}{2}}, \sigma _j^{\frac{n}{2}}) \end{aligned}$$

which proves that $\mu _k^{g_j} (1, \sigma '_j)$ tends to infinity with j. $\square$

4.2 Upper bounds for $\mu _k(\rho ,1)$ and $\mu _k(1,\sigma )$

Let (M, g) be a compact Riemannian manifold of dimension $n\ge 2$, possibly with boundary. According to the result by Hassannezhad [23], one has, when M is a closed manifold,

$$\begin{aligned} \lambda _k(M,g)\le \frac{1}{\vert M\vert _g^{\frac{2}{n}}} \left( A_nk^{\frac{2}{n}}+B_n V([g])^{\frac{2}{n}}\right) , \end{aligned}$$

(25)

where $A_n$ and $B_n$ are two constants which only depend on n, and V([g]) is a conformally invariant geometric quantity defined as follows:

$$\begin{aligned} V([g])=\inf \{\vert M\vert _{g_0} \ : \ g_0 \text{ is } \text{ conformal } \text{ to } g \text{ and } Ric_{g_0}\ge -(n-1)g_0\}, \end{aligned}$$

where $\hbox {Ric}_{g_0}$ is the Ricci curvature of $g_0$. Now, for every positive $\rho$ such that $\fint _M \rho v_g =1$, we have $V([\rho ^{\frac{2}{n}}g])=V([g])$, $\vert M\vert _{\rho ^{\frac{2}{n}}g}=\vert M\vert _g$ and $\lambda _k(M,\rho ^{\frac{2}{n}}g)=\mu _k^g(\rho , \rho ^{\frac{n-2}{n}} )$. Hence, inequality (25) implies that for every positive $\rho$ such that $\fint _M \rho v_g =1$,

$$\begin{aligned} \mu _k^g(\rho , \rho ^{\frac{n-2}{n}} )\le \frac{1}{\vert M\vert _g^{\frac{2}{n}}} \left( A_nk^{\frac{2}{n}}+B_n V([g])^{\frac{2}{n}}\right) . \end{aligned}$$

(26)

This estimate is in contrast to what happens for the Witten Laplacian where we have $\sup _{\fint _M \rho v_{g}= 1} \mu _1^g(\rho ,\rho ) =+\,\infty$ (see [12]).

Our aim in this section is to discuss the boundedness of $\mu _k^g(\rho , \sigma )$ in the two remaining important cases: $\mu _k^g(\rho ,1)$ and $\mu _k^g(1,\sigma )$. In [12, Theorem 2.1], it has been shown that the use of the GNY (Grigor’yan–Netrusov–Yau) method [22] leads to the following estimate

$$\begin{aligned} \mu _k^g(\rho ,1) \fint _M \rho v_{g}\le C( [g])\left( \frac{k}{\vert M\vert _g}\right) ^{\frac{2}{n}}, \end{aligned}$$

(27)

where C( [g]) is a constant which only depends on the conformal class of the metric g.

This approach fails in the dual situation where $\sigma$ is varying, while $\rho$ is fixed. Indeed, the GNY method leads to an upper bound of $\mu _k^g(1,\sigma )$ in terms of the $L^{\frac{n-2}{n}}$-norm of $\sigma$ (instead of the $L^1$-norm). However, using the techniques developed by Colbois and Maerten in [13], it is possible to obtain an inequality of the form

$$\begin{aligned} \mu _k^g(1,\sigma ) \le C( M,g)\left( \frac{k}{\vert M\vert _g}\right) ^{\frac{2}{n}} \fint _M \sigma v_{g}, \end{aligned}$$

(28)

where C(M, g) is a geometric constant which does not depend on $\sigma$ (unlike (27), this method of proof does not allow us to obtain a conformally invariant constant instead of C(M, g)).

In what follows, we will establish inequalities of the type (26) for $\mu _k(\rho ,1)$ and $\mu _k(1,\sigma )$.

Theorem 4.1

Let M be a bounded open domain possibly with boundary of class $C^1$ of a complete Riemannian manifold $(\tilde{M}, \tilde{g}_0)$ of dimension $n\ge 2$ (with $\tilde{M}=M$ if $\partial M=\emptyset$). Assume that $\hbox {Ric}_{\tilde{g}_0}\ge -(n-1)\tilde{g}_0$ and let $g_0=\tilde{g}_0 \vert _M$. For every metric g conformal to $g_0$ and every density $\rho$ with $\fint _{M}\rho v_g =1$, one has

$$\begin{aligned} \mu _k^g(\rho ,1) \le \frac{1}{\vert M\vert _g^{\frac{2}{n}}}\left( A_n k^{\frac{2}{n}}+ B_n\vert M\vert _{g_0}^{\frac{2}{n}}\right) , \end{aligned}$$

(29)

where $A_n$ and $B_n$ are two constants which depend only on the dimension n.

In the particular case where (M, g) is a compact manifold without boundary, we can apply Theorem 4.1 with $M =\tilde{M}$ and get immediately the following estimate which extends (25):

$$\begin{aligned} \mu _k^g(\rho ,1) \le \frac{1}{\vert M\vert _g^{\frac{2}{n}}}\left( A_n k^{\frac{2}{n}}+ B_nV([g])^{\frac{2}{n}}\right) . \end{aligned}$$

(30)

On the other hand, if $\tilde{g}$ is a metric on $\tilde{M}$ and if $\text{ ric }_0$ is a positive number such that $\hbox {Ric}_{\tilde{g}}\ge -(n-1)\text{ ric }_0\ \tilde{g}$, then the metric $\tilde{g}_0=\text{ ric }_0 \tilde{g}$ satisfies $\hbox {Ric}_{\tilde{g}_0}\ge -(n-1)\tilde{g}_0$ and $\vert M\vert _{g_0}= \text{ ric }_0^{n/2} \vert M\vert _g$, where $g=\tilde{g} \vert _M$ and $g_0=\tilde{g}_0 \vert _M$. Thus, we get

Corollary 4.1

Let M be a bounded open domain possibly with boundary of class $C^1$ of a complete Riemannian manifold $(\tilde{M}, \tilde{g})$ of dimension $n\ge 2$ (with $\tilde{M}=M$ if $\partial M=\emptyset$) and let $g=\tilde{g} \vert _M$. For every density $\rho$ with $\fint _{M}\rho v_g =1$, one has

$$\begin{aligned} \mu _k^g(\rho ,1) \le A_n\left( \frac{k}{\vert M\vert _{g}}\right) ^\frac{2}{n}+B_n \text{ ric }_0, \end{aligned}$$

(31)

where $\text{ ric }_0>0$ is such that $\hbox {Ric}_{\tilde{g}}\ge -(n-1)\text{ ric }_0\ \tilde{g}$. In particular, $\forall k\ge \vert M\vert _{g} \text{ ric }_0^{\frac{n}{2}}$,

$$\begin{aligned} \mu _k^g(\rho ,1) \le C_n\left( \frac{k}{\vert M\vert _{g}}\right) ^\frac{2}{n} \end{aligned}$$

(32)

with $C_n=A_n+B_n$.

Inequalities (30) and (31) are conceptually much stronger than (27), especially since they lead to a Kröger-type inequality (32) for every k exceeding an explicit geometric threshold, independent of $\rho$ (it is well known that if the Ricci curvature is not nonnegative, then an inequality like (32) cannot hold for every k, see [13, Remark 1.2(iii)]).

Theorem 4.2

Let M be a bounded open domain possibly with boundary of class $C^1$ of a complete Riemannian manifold $(\tilde{M}, \tilde{g})$ of dimension $n\ge 2$ (with $\tilde{M}=M$ if $\partial M=\emptyset$) and let $g=\tilde{g} \vert _M$. For every positive density $\sigma$ on M with $\fint _{M}\sigma v_g =1$, one has

$$\begin{aligned} \mu _k^g(1,\sigma ) \le A_n \left( \frac{k}{\vert M\vert _{g}}\right) ^\frac{2}{n} +B_n \text{ ric }_0, \end{aligned}$$

(33)

where $\text{ ric }_0>0$ is such that $\hbox {Ric}_{\tilde{g}}\ge -(n-1)\text{ ric }_0\ \tilde{g}$ and where $A_n$ and $B_n$ are two constants which depend only on n. In particular, $\forall k\ge \vert M\vert _{g} \text{ ric }_0^{\frac{n}{2}}$,

$$\begin{aligned} \mu _k^g(1,\sigma ) \le C_n\left( \frac{k}{\vert M\vert _{g}}\right) ^\frac{2}{n} \end{aligned}$$

(34)

with $C_n=A_n+B_n$.

Proof of Theorem 4.1

We consider the metric measured space $(M,d_0,\nu )$ where $d_0$ is the restriction to M of the Riemannian distance on $(\tilde{M}, \tilde{g}_0)$, and $\nu = \rho v_{g}$. Since $\hbox {Ric}_{g_0}\ge -(n-1)g_0$, the space $(M,d_0,\nu )$ satisfies a (2, N; 1)-covering property for some fixed N (see [23]). Therefore, we can apply Theorem 2.1 of [23] and find a family of $3(k+1)$ pairs of sets $(F_j,G_j)$ of M with $F_j\subset G_j$, such that the $G_j$’s are mutually disjoint and $\nu (F_j)\ge \frac{\nu (M)}{c^2{(k+1)}}$, where $c=c(n)$ is a constant that depends only on n. Moreover, each pair $(F_j,G_j)$ satisfies one of the following properties:

• $F_j$ is an annulus A of the form $A=\{r<d_0(x,a)<R\}$, and $G_j=2A=\{\frac{r}{2}<d_0(x,a)<2R\}$, with outer radius 2R less than 1,

• $F_j$ is an open set $V\subset M$ and $G_j=V^{r_0}=\{x\in M\ ; \ d_0(x,V)<r_0\}$, with $r_0= \frac{1}{1600}$.

Let us start with the case where $F_j$ is an annulus $A=A(a,r,R)=\{r<d_0(x,a)<R\}$ and $G_j=2A$. To such an annulus, we associate the function $u_A$ supported in $2A=\{\frac{r}{2}<d_0(x,a)<2R\}$ and such that

$$\begin{aligned} u_A( x )=\left\{ \begin{array}{ll} \frac{2}{r}d_0(x,a)-1 &{} \quad \text {if } \frac{r}{2}\le d_0(x,a)\le r\\ 1&{} \quad \text {if } x\in A\\ 2-\frac{1}{R}d_0(x,a) &{} \quad \text {if } R\le d_0(x,a)\le 2R\\ \end{array}\right. . \end{aligned}$$

(35)

Since $u_A$ is supported in 2A we get, using Hölder’s inequality and the conformal invariance of $\vert \nabla ^{g} u_A\vert ^n v_{g}$,

$$\begin{aligned} \int _{M} \vert \nabla ^{g}u_A\vert ^2 v_{g}= & {} \int _{2A } \vert \nabla ^{g}u_A\vert ^2 v_{g} \le \left( \int _{2A } \vert \nabla ^{g} u_A\vert ^n v_{g}\right) ^{\frac{2}{n}}{\left( \int _{2A} v_{g}\right) ^{1-\frac{2}{n}}}\\= & {} \left( \int _{2A} \vert \nabla ^{g_0} u_A\vert ^n v_{g_0}\right) ^{\frac{2}{n}} \vert 2A\vert _{g}^{1-\frac{2}{n}}. \end{aligned}$$

Since

$$\begin{aligned} \vert \nabla ^{g_0} u_A\vert \overset{a.e.}{=} \left\{ \begin{array}{ll} \frac{2}{r} &{} \quad \text {if } \frac{r}{2}\le d_0(x,a)\le r\\ 0&{} \quad \text {if } r\le d_0(x,a)\le R\\ \frac{1}{R}&{} \quad \text {if } R\le d_0(x,a)\le 2R\\ \end{array}\right. , \end{aligned}$$

we get

$$\begin{aligned} \int _{2A} \vert \nabla ^{g_0} u_A\vert ^n v_{g_0} \le \left( \frac{2}{r} \right) ^n \vert B(a,r)\vert _{g_0} + \left( \frac{1}{R} \right) ^n \vert B(a,2R)\vert _{g_0} \le 2^{n+1}\Gamma (g_0), \end{aligned}$$

where

$$\begin{aligned} \Gamma (g_0)=\sup _{x\in M, t\in (0,1)}\frac{ \vert B(x,t)\vert _{g_0}}{t^n} \end{aligned}$$

(here B(x, t) stands for the ball of radius t centered at x in $(M,d_0)$). Notice that since $\hbox {Ric}_{\tilde{g}_0}\ge -(n-1)\tilde{g}_0$, the constant $\Gamma (g_0)$ is bounded above by a constant that depends only on n (Bishop–Gromov inequality). Hence,

$$\begin{aligned} \int _{M} \vert \nabla ^{g}u_A\vert ^2 v_{g}\le C(n) \vert 2A \vert _{g}^{1-\frac{2}{n}}, \end{aligned}$$

where $C(n)\ge 2^{n+1}\Gamma (g_0)$. On the other hand, we have

$$\begin{aligned} \int _{M} u_A^2\rho \, v_{g} \ge \int _{A } \rho \, v_{g} =\nu (A)\ge \frac{\nu (M)}{c^2{(k+1)}}. \end{aligned}$$

Thus,

$$\begin{aligned} R_{(g,\rho ,1)}(u_A)=\frac{\int _{M} \vert \nabla ^{g}u_A\vert ^2 v_{g}}{\int _{M} u_A^2\rho \, v_{g}}\le A_n \frac{ \vert 2A \vert _{g}^{1-\frac{2}{n}} }{ \nu (M) } (k+1) \end{aligned}$$

for some constant $A_n$.

Now, in the second situation, where $F_j$ is an open set V and $G_j=V^{r_0}$, we introduce the function $u_V$ defined to be equal to 1 inside V, 0 outside $V^{r_0}$ and proportional to the $d_0$-distance to the outer boundary in $V^{r_0}{\setminus } V$. We have, since $u_V=1$ in V and $\vert \nabla ^{g_0} u_V\vert$ is equal to $\frac{1}{r_0}$ almost everywhere in $V^{r_0}{\setminus } V$ and vanishes in V and in $M{\setminus } V^{r_0}$,

$$\begin{aligned} \int _{M} u_V^2\rho \, v_{g}\ge \int _{V} \rho \, v_{g} =\nu (V)\ge \frac{\nu (M)}{c^2{(k+1)}} \end{aligned}$$

and

$$\begin{aligned} {\int _{M} \vert \nabla ^{g} u_V\vert ^2v_{g}}\le & {} {\left( \int _{V^{r_0}} \vert \nabla ^{g} u_V\vert ^nv_{g}\right) ^{\frac{2}{n}}\vert V^{r_0}\vert _{g}^{1-\frac{2}{n}}}= {\left( \int _{V^{r_0}} \vert \nabla ^{g_0} u_V\vert ^nv_{g_0}\right) ^{\frac{2}{n}}\vert V^{r_0}\vert _{g}^{1-\frac{2}{n}}}\\\le & {} \frac{ \vert V^{r_0} \vert _{g_0}^{\frac{2}{n}}\vert V^{r_0} \vert _{g}^{1-\frac{2}{n}}}{{r_0}^2 }. \end{aligned}$$

Thus,

$$\begin{aligned} R_{(g,\rho ,1)}(u_V) \le B_n\frac{ { \vert V^{r_0} \vert _{g_0}^{\frac{2}{n}}} \vert V^{r_0} \vert _{g}^{1-\frac{2}{n}}}{\nu (M)} (k+1), \end{aligned}$$

where $B_n= \frac{c^2}{r_0^2}$ is a constant which depends only on n.

In conclusion, to each pair $(F_j,G_j)$ we associate a test function $u_j$ supported in $G_j$ and satisfying either $R_{(g,\rho ,1)}(u_j) \le A_n \frac{ \vert G_j \vert _{g}^{1-\frac{2}{n}} }{ \nu (M) } (k+1)$ or $R_{(g,\rho ,1)}(u_j) \le B_n\frac{ { \vert G_j \vert _{g_0}^{\frac{2}{n}}} \vert G_j \vert _{g}^{1-\frac{2}{n}}}{\nu (M)} (k+1)$, that is,

$$\begin{aligned} R_{(g,\rho ,1)}(u_j) \le A_n \frac{ \vert G_j \vert _{g}^{1-\frac{2}{n}} }{ \nu (M) } (k+1) + B_n\frac{ { \vert G_j \vert _{g_0}^{\frac{2}{n}}} \vert G_j \vert _{g}^{1-\frac{2}{n}}}{\nu (M)} (k+1). \end{aligned}$$

Now, observe that since $\sum _{j\le 3(k+1)}\vert G_j \vert _{g_0}\le \vert M\vert _{g_0}$ and $\sum _{j\le 3(k+1)}\vert G_j\vert _{g}\le \vert M\vert _{g}$, there exist at least $k+1$ sets among $G_1,\dots , G_{3(k+1)}$ satisfying both $\vert G_j\vert _{g_0}\le \frac{\vert M\vert _{g_0}}{k+1}$ and $\vert G_j\vert _{g}\le \frac{\vert M\vert _{g}}{k+1}$. This leads to a subspace of $k+1$ disjointly supported functions $u_j$ whose Rayleigh quotients are such that

$$\begin{aligned} R_{(g,\rho ,1)}(u_j)\le & {} A_n \frac{ \vert G_j \vert _{g}^{1-\frac{2}{n}} }{ \nu (M) } (k+1) + B_n\frac{ { \vert G_j \vert _{g_0}^{\frac{2}{n}}} \vert G_j \vert _{g}^{1-\frac{2}{n}}}{\nu (M)} (k+1)\\\le & {} A_n \frac{ {\vert M\vert _{g}}^{1-\frac{2}{n}} }{ \nu (M) } (k+1)^{\frac{2}{n}} + B_n \frac{ \vert M\vert _{g_0}^{\frac{2}{n}}}{ \nu (M) }\vert M\vert _{g}^{1-\frac{2}{n}} \end{aligned}$$

with $\nu (M)=\int _M \rho v_{g}=\vert M\vert _g$. The desired inequality then immediately follows thanks to (1). $\square$

Proof of Theorem 4.2

First, observe that it suffices to prove the theorem when $\text{ ric }_0 =1$ (i.e., $\hbox {Ric}_{\tilde{g}}\ge -(n-1)\tilde{g}$). Indeed, the Riemannian metric $\tilde{g}_0= {\text{ ric }_0} \tilde{g}$ satisfies $\hbox {Ric}_{\tilde{g}_0}\ge -(n-1)\tilde{g}_0$ and $\vert M\vert _{g_0}= (\text{ ric }_0)^{n/2} \vert M\vert _g$, with $g_0=\tilde{g}_0\vert M$. Hence, the inequality

$$\begin{aligned} \mu _k^{g_0}(1,\sigma ) \le A_n \left( \frac{k}{\vert M \vert _{g_0}}\right) ^\frac{2}{n} +B_n \end{aligned}$$

implies

$$\begin{aligned} \mu _k^g(1,\sigma ) = \text{ ric }_0 \mu _k^{g_0}(1,\sigma )\le \text{ ric }_0\left( A_n \left( \frac{k}{\vert M\vert _{g_0}}\right) ^\frac{2}{n}+B_n \right) = A_n \left( \frac{k}{\vert M\vert _{g}}\right) ^\frac{2}{n} + B_n \text{ ric }_0 . \end{aligned}$$

Therefore, assume that $\text{ ric }_0 =1$ and consider the metric measured space $(M,d,v_{g})$ where d is the restriction to M of the Riemannian distance of $(\tilde{M}, \tilde{g})$. The proof relies on the method developed by Colbois and Maerten [13] as presented in Lemma 2.1 of [11]. Applying the Bishop–Gromov theorem, we deduce that there exist two constants, $C_n$ and $N_n$, depending only on n, such that, $\forall x\in M$ and $\forall r\le 1$,

• $\vert B(x,r)\vert _{g}\le C_n r^n$

• B(x, 4r) can be covered by $N_n$ balls of radius r

where B(x, r) stands for the ball in M of radius r with respect to the distance d.

Let $k_0$ be the smallest integer such that $2(k_0+1)> \frac{ \vert M \vert _{g}}{4 C_n N_n^2}$. For every $k\ge k_0$, we define $r_k$ by

$$\begin{aligned} r_k^n = \frac{\vert M \vert _{g}}{8C_nN_n^2 (k+1)}\le 1 \end{aligned}$$

which means that, $\forall x\in M$,

$$\begin{aligned} \vert B(x,r_k) \vert _{g} \le C_n r_k^n\le \frac{\vert M\vert _{g}}{8N_n^2 (k+1)}. \end{aligned}$$

Thus, we can apply Lemma 2.1 of [11] and deduce the existence of $2(k+1)$ measurable subsets $A_1,\dots ,A_{2(k+1)}$ of M such that, $\forall i\le 2(k+1)$, $\vert A_i \vert _{g}\ge \frac{\vert M\vert _{g}}{4N_n(k+1)}$ and, for $i\not =j$, $d(A_i,A_j) \ge 3r_k$. To each set $A_j$, we associate the function $f_j$ supported in $A_j^{r_k}=\{x\in M\, :\, d(x,A_j)<r_k\}$ and defined to be equal to 1 inside $A_j$ and proportional to the distance to the outer boundary in $A_j^{r_k}{\setminus } A_j$. The length of the gradient $\vert \nabla ^{g} f_j\vert$ is then equal to $\frac{1}{r_k}$ almost everywhere in $A_j^{r_k}{\setminus } A_j$ and vanishes elsewhere, so that we get

$$\begin{aligned} R_{(g,1,\sigma )}(f_j) = \frac{\int _{A_j^{r_k}} \vert \nabla ^{g} f_j\vert ^2\sigma v_{g}}{\int _{A_j^{r_k}} f_j^2 v_{g}}\le \frac{\frac{1}{{r_k}^2 } \int _{A_j^{r_k}} \sigma v_{g}}{ \vert A_j\vert _{g}}\le \frac{4N_n}{{r_k}^2 } \frac{ \int _{A_j^{r_k}} \sigma v_{g} }{ \vert M\vert _{g}} (k+1) \end{aligned}$$

which gives, after replacing $r_k$ by its explicit value,

$$\begin{aligned} R_{(g,1,\sigma )}(f_j) \le A_n \frac{ \int _{A_j^{r_k}} \sigma v_{g} }{ \vert M \vert _{g}^{1+\frac{2}{n}}} (k+1)^{1+\frac{2}{n}}. \end{aligned}$$

for some constant $A_n$. Now, since $\sum _{j\le 2(k+1)} \int _{A_j^{r_k}} \sigma v_{g}\le \int _{M} \sigma v_{g}$, there exist at least $k+1$ sets among the $A_j$’s such that $\int _{A_j^{r_k}} \sigma v_{g}\le \frac{\int _{M} \sigma v_{g}}{k+1}$. This leads to a subspace of $k+1$ disjointly supported functions $f_j$ whose Rayleigh quotients are such that

$$\begin{aligned} R_{(g,1,\sigma )}(f_j) \le A_n \frac{ \int _{M} \sigma v_{g} }{ \vert M\vert _{g}^{1+\frac{2}{n}}} (k+1)^{\frac{2}{n}}. \end{aligned}$$

Consequently, we have thanks to (1), for all $k\ge k_0$,

$$\begin{aligned} \mu _k^g(1,\sigma ) \le A_n \frac{ \int _{M} \sigma v_{g} }{ \vert M\vert _{g}^{1+\frac{2}{n}}} (k+1)^{\frac{2}{n}} = A_n\left( \frac{ k+1 }{ \vert M\vert _{g}}\right) ^{\frac{2}{n}} \end{aligned}$$

since we have assumed that $\int _{M} \sigma v_{g} =\vert M\vert _{g}$. On the other hand, for every $k\le k_0$, one obviously has (since $k_0+1\le \frac{ \vert M\vert _{g}}{4 C_n N_n^2}$)

$$\begin{aligned} \mu _k^g(1,\sigma ) \le \mu _{k_0}^g(1,\sigma ) \le A_n\left( \frac{ k_0+1 }{ \vert M\vert _{g}}\right) ^{\frac{2}{n}} \le A_n \left( \frac{ 1}{4 C_n N_n^2}\right) ^{\frac{2}{n}}. \end{aligned}$$

Denoting by $B_n$ the latter constant we obtain, for every $k\ge 0$,

$$\begin{aligned} \mu _k^g(1,\sigma )\le A_n\left( \frac{ k}{ \vert M\vert _{g}}\right) ^{\frac{2}{n}}+ B_n . \end{aligned}$$

$\square$

5 Extremal eigenvalues

Let (M, g) be a compact Riemannian manifold of dimension $n\ge 2$, possibly with boundary. In [9], we introduced the following conformally invariant quantities that we named “conformal eigenvalues”: For every $k\in \mathbb {N}$, $\lambda _k^c(M,[g])$ is defined as the supremum of $\lambda _k(M,g')$ when $g'$ runs over all metrics of unit volume which are conformal to g (or, equivalently, $\lambda _k^c(M,[g]) =\sup \lambda _k(M,g')\vert M\vert _{g'}^{\frac{2}{n}}$ when $g'$ runs over all metrics conformal to g). Thus, we can write

$$\begin{aligned} \lambda _k^c(M,[g])=\sup _{\int _M\rho \, v_g=1 }\lambda _k(M,{\rho ^{\frac{2}{n}} g})=\sup _{\int _M\rho \, v_g=1}\mu _k^g(\rho ,\rho ^{\frac{n-2}{n}}). \end{aligned}$$

We investigated in [9] some of the properties of the conformal eigenvalues such as the existence of a universal lower bound, and proved that

$$\begin{aligned} \lambda _k^c(M,[g])\ge \lambda _k^c(\mathbb {S}^n,[g_s])\ge n\alpha _n^{\frac{2}{n}}k^{\frac{2}{n}}, \end{aligned}$$

(36)

where $\alpha _n=(n+1)\omega _{n+1}$ is the volume of the standard sphere. Moreover, we proved that the gap between two consecutive conformal eigenvalues satisfies the following estimate:

$$\begin{aligned} \lambda _{k+1}^c(M,[g])^{\frac{n}{2}}-\lambda _k^c(M,[g])^{\frac{n}{2}}\ge n^{\frac{n}{2}}\alpha _n. \end{aligned}$$

(37)

Actually, these properties were established in the context of closed manifolds. However, they remain valid in the context of bounded domains, under Neumann boundary conditions, without the need to change anything to the proofs. In this regard, we can point out the following curious phenomenon that all bounded Euclidean domains have the same conformal spectrum.

Proposition 5.1

For every bounded domain $\Omega \subset \mathbb {R}^n$ with $C^1$-boundary, one has

$$\begin{aligned} \lambda _k^c(\Omega ,[g_E])=\lambda _k^c(B^n,[g_{E}]), \end{aligned}$$

where $g_E$ is the Euclidean metric.

For $k=1$, we have $\lambda _1^c(\Omega ,[g_E])=n\alpha _n^{\frac{2}{n}}$ (see Corollary 6.1).

Proof

Let us first observe that if $\Omega$ is a proper subset of $\Omega '$, then $\lambda _k^c(\Omega ,[g_E])\le \lambda _k^c (\Omega ',[g_E])$. Indeed, given a metric $g=fg_E$ conformal to $g_E$ on $\Omega$, we extend it to a metric $g'$ on $\Omega '$, conformal to $g_E$. For every $\varepsilon >0$, we multiply $g'$ by the function $f_\varepsilon$ which is equal to 1 on $\Omega$ and equal to $\varepsilon$ on $\Omega '{\setminus }\Omega$ and apply Theorem 2.1. In dimension $n\ge 3$, this theorem tells us that $\lambda _k(\Omega ',f_\varepsilon g')$ converges to $\lambda _k(\Omega ,g)$. Since the volume of $(\Omega ',f_\varepsilon g')$ converges to the volume of $(\Omega ,g)$, we deduce that $\lambda _k(\Omega ,g)\vert \Omega \vert _g^{2/n}\le \lambda _k^c (\Omega ',[g_E])$. In dimension 2, we obtain that $\lambda _k(\Omega ',f_\varepsilon g')$ converges to the kth eigenvalue of the quadratic form $\int _\Omega \vert \nabla u\vert ^2v_g+\int _{\Omega '{\setminus }\Omega }\vert \nabla H(u)\vert ^2v_g$. This quadratic form is clearly larger than the Dirichlet energy $\int _\Omega \vert \nabla u\vert ^2v_g$ on $\Omega$ so that its k-th eigenvalue is bounded below by $\lambda _k(\Omega ,g)$. Again, this implies that $\lambda _k(\Omega ,g)\le \lambda _k^c(\Omega ',[g_E]).$

Now, since $\Omega$ is open and bounded, there exist two positive radii $r_1$ and $r_2$ so that

$$\begin{aligned} B^n(r_1)\subset \Omega \subset B^n(r_2), \end{aligned}$$

where $B^n(r_1)$ and $B^n(r_2)$ are two concentric Euclidean balls. Using the observation above we get

$$\begin{aligned} \lambda _{k}^c(B^n(r_1),[g_E]) \le \lambda _{k}^c(\Omega ,[g_E]) \le \lambda _{k}^c(B^n(r_2),[g_E]) . \end{aligned}$$

Since the balls $B^n(r_1)$ and $B^n(r_2)$ are homothetic to the unit ball $B^n$, one necessarily has $\lambda _{k}^c(B^n(r_1),[g_E])= \lambda _{k}^c(B^n(r_2),[g_E])=\lambda _{k}^c(B^n,[g_E])$ which enables us to conclude. $\square$

As a consequence of the upper bounds given in the previous section, it is natural to introduce the following extremal eigenvalues:

$$\begin{aligned} \mu _k^*(M,g)= & {} \sup _{\fint _M\rho \, v_g=1}\mu _k^g(\rho ,1) = \sup _{\rho }\mu _k^g(\rho ,1)\fint _M\rho \, v_g\\ \mu _k^{**}(M,g)= & {} \sup _{\fint _M\sigma v_g=1}\mu _k^g(1,\sigma )=\sup _{\sigma }\frac{\mu _k^g(1,\sigma )}{\fint _M\sigma v_g}. \end{aligned}$$

A natural question is whether properties such as (36) and (37) may occur for $\mu _k^*(M,g)$ and $\mu _k^{**}(M,g)$. Observe that these quantities are not invariant under metric scaling since

$$\begin{aligned} \mu _k^*(M,r^2g)=r^{-2}\mu _k^*(M,g)\quad and \quad \mu _k^{**}(M,r^2g)=r^{-2}\mu _k^{**}(M,g). \end{aligned}$$

Hence, we will assume that the volume of the manifold is fixed.

In the particular case of manifolds (M, g) of dimension 2, one has for every $\rho$, $\mu _k^g(\rho , 1)=\lambda _k(M,\rho g)$. Thus,

$$\begin{aligned} \mu _k^*(M,g) = \frac{\lambda _k^c(M,g)}{\vert M\vert _g} \end{aligned}$$

(38)

and we deduce from (36) and (37) that any 2-dimensional Riemannian manifold (M, g) satisfies

$$\begin{aligned} \mu _k^*(M,g) \ge \frac{8\pi k}{\vert M\vert _g} \end{aligned}$$

and

$$\begin{aligned} \mu _{k+1}^*(M,g)-\mu _k^*(M,g) \ge \frac{8\pi }{\vert M\vert _g}. \end{aligned}$$

The following theorem shows that the 2-dimensional case is in fact exceptional. Indeed, it turns out that any compact manifold of dimension $n\ge 3$ can be deformed in such a way that $\mu _k^*(M,g)$ becomes as small as desired.

Theorem 5.1

Let M be a compact manifold of dimension $n\ge 3$. There exists on M a one-parameter family of metrics $g_\varepsilon$, $\varepsilon >0$, of volume 1 such that

$$\begin{aligned} \mu _k^*(M,g_\varepsilon )\le C k\, \varepsilon ^{\frac{n-2}{n}}, \end{aligned}$$

where C is a constant which does not depend on $\varepsilon$ or k.

Similarly, we have the following result for the supremum with respect to $\sigma$.

Theorem 5.2

Let M be a compact manifold of dimension $n\ge 2$. There exists on M a one-parameter family of metrics $g_\varepsilon$, $\varepsilon >0$, of volume 1 such that

$$\begin{aligned} \mu _k^{**}(M,g_\varepsilon )\le C k^2 \varepsilon ^{2\frac{n-1}{n}}, \end{aligned}$$

where C is a constant which depends only on n.

The proofs of these theorems rely on the construction below. It is worth noticing that the one-parameter family of metrics $g_\varepsilon$ we will exhibit can be chosen within a fixed conformal class. Actually, we start with a Riemannian metric $g_0$ on M that we conformally deform in the neighborhood of a point.

The construction. We start with a metric $g_0$ on M and choose a sufficiently small open set $V\subset M$ so that $g_0$ is 2-quasi-isometric to a flat metric in V. Since the eigenvalues corresponding to two quasi-isometric metrics are “comparable,” we can assume w.l.o.g. that the metric $g_0$ is flat inside V. Therefore, there exists a positive $\delta$ so that V contains a flat (Euclidean) ball of radius $\delta$. After a possible dilation, we can assume that $\delta =1$. We deform this unit Euclidean ball into a long capped cylinder (i.e., an Euclidean cylinder of radius $\delta$ closed by a spherical cap). This construction is standard and is explained, for example, in [20, pp. 3856–3857]. We can even do it through a conformal deformation of $g_0$, as explained in [10, pp. 718–719]. Therefore, we obtain a family of Riemannian manifolds $(M,g_\varepsilon )$ so that M is the union of three parts

$$\begin{aligned} M=M_0 \cup C \cup S_0^n \end{aligned}$$

with

• $M_0$ is an open subset of M and $g_\varepsilon$ does not vary with $\varepsilon$ on $M_0$,

• $(C, g_\varepsilon )$ is isometric to the cylinder $[0,\frac{1}{\varepsilon }]\times \mathbb {S}^{n-1}$ of length $\frac{1}{\varepsilon }$ (with $0 < \varepsilon \le 1$),

• $S_0^n$ is a round hemisphere of radius 1 which closes the end of the cylinder C and $g_\varepsilon \vert _{ S_0^n}$ is the round metric (and is independent of $\varepsilon$).

The only varying parameter in this construction is the length $\frac{1}{\varepsilon }$ of the cylinder $(C,g_{\varepsilon })$. Notice that the volume of $(M,g_{\varepsilon })$ is not equal to 1, but we will make a suitable scaling at the end of the proof.

In order to bound the eigenvalues $\mu _k^{g_\varepsilon }(\rho , 1)$ from above, we will use the GNY method [22]. To this end, we need a uniform control (w.r.t. $\varepsilon$) of the packing constant (see [22, Definition 3.3 and Theorem 3.5]) and of the volume growth of balls in $(M,g_{\varepsilon })$. This will be done in the following lemmas. For this purpose, we introduce the connected open subset $\tilde{M}_0\subset M$ obtained as the union of $M_0$ and the part of the cylinder which corresponds to $(0,3d_0)\times \mathbb {S}^{n-1}\subset [0,\frac{1}{\varepsilon }]\times \mathbb {S}^{n-1}$, where $d_0$ is the diameter of $M_0$.

Lemma 5.1

(volume growth of balls) There exist two positive constants $C_1$ and $C_2$, independent of $\varepsilon$, such that, for every ball $B_\varepsilon (x,r)$ in $(M,g_{\varepsilon })$ we have

$$\begin{aligned} \vert B_\varepsilon (x,r)\vert _{g_\varepsilon } \le \left\{ \begin{array}{lll} C_1 r^n &{} \ &{} \text {if} \ r \le 2 d_0\\ C_2r &{} \ &{} \text {if} \ r \ge 2 d_0 \end{array}\right. \end{aligned}$$

(39)

Proof

If $B_\varepsilon (x,r)\cap M_0= \emptyset$, then $B_\varepsilon (x,r)$ is isometric to a geodesic ball of radius r of the capped cylinder and an obvious calculation shows that (39) holds true with two constants $C_1$ and $C_2$ independent of $\varepsilon$ (in fact, we can compare the volume of $B_\varepsilon (x,r)$ with the volume of $(-r,r)\times \mathbb {S}^{n-1}$ to get $\vert B_\varepsilon (x,r)\vert _{g_\varepsilon }\le A r$ for some positive A). If $B_\varepsilon (x,r)\cap M_0\ne \emptyset$ and $r< 2d_0$, then $B_\varepsilon (x,r)$ is contained in $\tilde{M}_0$. Hence, there exists a constant C, depending only on $\tilde{M}_0$, such that $\vert B_\varepsilon (x,r)\vert _{g_\varepsilon }\le C r^n$. If $B_\varepsilon (x,r)\cap M_0\ne \emptyset$ and $r\ge 2d_0$, then $B_\varepsilon (x,r)$ is contained in the union of a ball $B(x_0,2d_0)\subset \tilde{M}_0$ centered at a point $x_0\in M_0$ and a ball of radius $r'\le r$ contained in the cylindrical part. Thus, $\vert B_\varepsilon (x,r)\vert _{g_\varepsilon }\le C 2^n d_0^n + A r\le C_2r$ for some positive $C_2$ which does not depend on $\varepsilon$. $\square$

Lemma 5.2

There exists a constant N, independent of $\varepsilon$, such that any ball of radius $r>0$ in $(M,g_{\varepsilon })$ can be covered by N balls of radius $\frac{r}{2}$.

Proof

Let $B_\varepsilon (x,r)$ be a ball of radius r in $(M,g_{\varepsilon })$. If $B_\varepsilon (x,r)\cap M_0= \emptyset$, then, since $(M{\setminus } M_0, g_\varepsilon )$ is isometric to the capped cylinder whose Ricci curvature is everywhere nonnegative, $B_\varepsilon (x,r)$ can be covered by $N_E$ balls of radius $\frac{r}{2}$, where $N_E$ is the packing constant of the Euclidean space $\mathbb {R}^n$ (Bishop–Gromov theorem).

Assume that $B_\varepsilon (x,r) \cap M_0\ne \emptyset$. If $r< 2d_0$, then $B_\varepsilon (x,r)$ is contained in $\tilde{M}_0$. Thus, $B_\varepsilon (x,r)$ can be covered by $N(\tilde{M}_0)$ balls of radius $\frac{r}{2}$, where $N(\tilde{M}_0)$ is the packing constant of $\tilde{M}_0$. If $r\ge 2d_0$, then $B_\varepsilon (x,r)$ is contained in the union of a ball $B_\varepsilon (x_0,2d_0)\subset \tilde{M}_0$ centered at a point $x_0\in M_0$ and a ball of radius $r'\le r$ contained in the capped cylinder. Again, $B_\varepsilon (x,r)$ can be covered by $N_E + N(\tilde{M}_0)$ balls of radius $\frac{r}{2}$. $\square$

Proof of Theorem 5.1

Let $\rho$ be a positive density on M with $\fint _M\rho v_{g_\varepsilon } =1$. Applying [22, Theorem 3.5] to the metric measured space $(M,d_\varepsilon , \rho v_{g_\varepsilon } )$, where $d_\varepsilon$ is the Riemannian distance associated to $g_\varepsilon$, we deduce the existence of $k+1$ annuli $A_1,\dots , A_{k+1}$ such that $\int _{A_j} \rho v_{g_\varepsilon }\ge \frac{\vert M\vert _{g_\varepsilon }}{Ck}$ and $2A_1,\dots 2A_{k+1}$ are mutually disjoint. Here, C should depend on the packing constant of $(M,g_\varepsilon )$, but since the latter is dominated independently of $\varepsilon$, thanks to Lemma 5.2, we can assume that C is independent of $\varepsilon$.

To each annulus of the form $A=B_\varepsilon (x,R) {\setminus } B_\varepsilon (x,r)$, we associate a function $u_A$ defined as in (35). We obtain

$$\begin{aligned} R_{(g_\varepsilon , \rho , 1)}(u_A) =\frac{\int _{2A} \vert \nabla ^\varepsilon u_A\vert _{g_\varepsilon }^2 v_{g_\varepsilon }}{\int _{2A} u_A^2 v_{g_\varepsilon }}\le \frac{\frac{4}{r^2}{\vert B_\varepsilon (x,r)\vert _{g_\varepsilon }} +\frac{1}{R^2}{\vert B_\varepsilon (x,2R)\vert _{g_\varepsilon }} }{\int _{A} \rho v_{g_\varepsilon }}. \end{aligned}$$

Using Lemma 5.1, we get for every $r>0$,

$$\begin{aligned} \frac{1}{r^2}\vert B_\varepsilon (x,r)\vert _{g_\varepsilon } \le \left\{ \begin{array}{lll} C_1 r^{n-2}\le C_1 d_0^{n-2} &{} \ &{} \text {if} \ r \le 2 d_0\\ \frac{C_2}{r} \le \frac{C_2}{2d_0} &{} \ &{} \text {if} \ r \ge 2 d_0 \end{array}\right. . \end{aligned}$$

(40)

Therefore, there exists a constant $C'$ which depends on $C_1$, $C_2$ and $d_0$ (but independent of $\varepsilon$), such that

$$\begin{aligned} R_{(g_\varepsilon , \rho , 1)}(u_A) \le \frac{ C'}{\int _{A} \rho v_{g_\varepsilon }}. \end{aligned}$$

Consequently, the $k+1$ annuli $A_1,\dots , A_{k+1}$ provide $k+1$ disjointly supported functions satisfying $R_{(g_\varepsilon , \rho , 1)}(u_{A_j})\le \frac{ C'}{\int _{A_j} \rho v_{g_\varepsilon }}\le \frac{CC' k}{\vert M\vert _{g_\varepsilon }}$. Thus,

$$\begin{aligned} \mu _k^{g_\varepsilon }( \rho , 1)\le C'' \frac{k}{\vert M\vert _{g_\varepsilon }}. \end{aligned}$$

In order to obtain a family of metrics of volume 1, we set $g'_{\varepsilon } = \frac{1}{\vert M\vert _{g_\varepsilon }^{2/n}} g_\varepsilon$. Hence, for any $\rho$ such that $\fint _M \rho \, v_{g'_\varepsilon }=\fint _M \rho \, v_{g_\varepsilon } =1$, we have

$$\begin{aligned} \mu _k^{ g'_\varepsilon }( \rho , 1)= {\vert M\vert _{g_\varepsilon }^{2/n}} \mu _k^{g_\varepsilon }( \rho , 1)\le C'' \frac{k}{\vert M\vert _{g_\varepsilon }^{1-\frac{2}{n}}}. \end{aligned}$$

But $\vert M\vert _{g_\varepsilon }\ge \vert C\vert _{g_\varepsilon }\ge \frac{n\omega _n}{\varepsilon }$. Thus,

$$\begin{aligned} \mu _k^*(M,g'_\varepsilon )\le C {k}\varepsilon ^{1-\frac{2}{n} }. \end{aligned}$$

$\square$

Proof of Theorem 5.2

Let $(M,g_\varepsilon )$ be as in the construction above and let $\sigma$ be such that $\int _M \sigma v_{g_\varepsilon }=\vert M\vert _{g_\varepsilon }$. The cylindrical part $(C,g_{\varepsilon })$ of $(M,g_\varepsilon )$ can be decomposed into $2(k+1)$ small cylinders $C_j\approx [\frac{j}{2(k+1)\varepsilon },\frac{j+1}{2(k+1)\varepsilon }] \times \mathbb {S}^{n-1}$, $j=0,...,2k+1$, of length $\frac{1}{2(k+1)\varepsilon }$. At least $(k+1)$ cylinders among $C_0, \dots , C_{2k+1}$ have a measure with respect to $\sigma$ which is less or equal to $\frac{\vert M\vert _{g_\varepsilon }}{k+1}$. To each such $C_j$, we associate a function f with support in $C_j$ and which is defined in $C_j$, through the obvious identification between $C_j$ and $[0,\frac{1}{2(k+1)\varepsilon }]\times \mathbb {S}^{n-1}$, as follows: $\forall (t,z)\in [0,\frac{1}{2(k+1)\varepsilon }]\times \mathbb {S}^{n-1}\approx C_j$,

$$\begin{aligned} f(t,z)= \left\{ \begin{array}{lll} 6(k+1)\varepsilon t &{} \ &{} \text {if} \ \ 0 \le t\le \frac{1}{6(k+1)\varepsilon }\\ 1 &{} \ &{} \text {if} \ \ \frac{1}{6(k+1)\varepsilon } \le t\le \frac{2}{6(k+1)\varepsilon }\\ -6(k+1)\varepsilon t+3 &{} \ &{} \text {if} \ \ \frac{2}{6(k+1)\varepsilon }\le t\le \frac{3}{6(k+1)\varepsilon }. \end{array} \right. \end{aligned}$$

(41)

We have

$$\begin{aligned} \int _Mf^2 v_{g_\varepsilon }\ge \int _{ [\frac{1}{6(k+1)\varepsilon },\frac{2}{6(k+1)\varepsilon }]\times \mathbb {S}^{n-1}}f^2 \ v_E = \frac{n\omega _n}{6(k+1)\varepsilon }, \end{aligned}$$

where $v_E$ is the standard product measure. On the other hand, the norm of the gradient of f is supported in $C_j$ and is dominated by $6(k+1)\varepsilon$. Thus,

$$\begin{aligned} \int _M \vert \nabla ^\varepsilon f\vert _{g_\varepsilon }^2\sigma v_{g_\varepsilon } \le (6(k+1)\varepsilon )^2 \int _{C_j} \sigma v_{g_\varepsilon } \le (6(k+1)\varepsilon )^2 \frac{\vert M\vert _{g_\varepsilon }}{k+1}= 36(k+1)\varepsilon ^2\vert M\vert _{g_\varepsilon } \end{aligned}$$

and the Rayleigh quotient of f satisfies

$$\begin{aligned} R_{(g_\varepsilon , 1,\sigma )}(f) \le \frac{216(k+1)^2\varepsilon ^3 \vert M\vert _{g_\varepsilon }}{n\omega _n}. \end{aligned}$$

Consequently, the $k+1$ chosen cylinders provide $k+1$ disjointly supported functions satisfying the last inequality, which yields

$$\begin{aligned} \mu _k^{g_\varepsilon }(1,\sigma ) \le C \vert M\vert _{g_\varepsilon }(k+1)^2 \varepsilon ^3 \end{aligned}$$

with $C= \frac{216}{n\omega _n}.$ Setting $g'_{\varepsilon }= \frac{1}{\vert M\vert _{g_\varepsilon }^{\frac{2}{n}}}g_{\varepsilon }$, we get

$$\begin{aligned} \mu _k^{g'_\varepsilon }(1,\sigma )=\vert M\vert _{g_\varepsilon }^{\frac{2}{n}} \mu _k^{g_\varepsilon }(1,\sigma ) \le C \varepsilon ^3 \vert M\vert _{g_\varepsilon }^{1+\frac{2}{n}}(k+1)^2 \end{aligned}$$

with $\vert M\vert _{g_\varepsilon }= \vert \tilde{M}_0\vert _{g} +\vert C \vert _{g_\varepsilon } +\frac{1}{2} n\omega _n\le \frac{A}{\varepsilon }$ for some constant A. Thus,

$$\begin{aligned} \mu _k^{**}(M,g'_\varepsilon )\le C'\varepsilon ^{2-\frac{2}{n}} (k+1)^2. \end{aligned}$$

$\square$

Remark 5.1

The same type of construction used in the proof of Theorems 5.1 and 5.2 allows us to prove the existence of a family of bounded domains $\Omega _\varepsilon \subset \mathbb {R}^n$ of volume 1 such that $\mu _k^*(\Omega _\varepsilon ,g_E)$ (resp. $\mu _k^{**}(\Omega _\varepsilon ,g_E)$) goes to zero with $\varepsilon$. This is to be compared with the result of Proposition 5.1.

We end this section with the following proposition in which we show how to produce examples of manifolds $(M,g_\varepsilon )$ of fixed volume for which the ratio $\frac{\mu _1^*(M,g_\varepsilon )}{\lambda _1(M,g_\varepsilon )}$ (resp. $\frac{\mu _1^{**}(M,g_\varepsilon )}{\lambda _1(M,g_\varepsilon )}$) tends to infinity as $\varepsilon \rightarrow 0$.

Proposition 5.2

Let M be a compact manifold and let A be a positive constant.

• i. There exists a family of metrics $g_\varepsilon$of volume 1 on M and a constant $A>0$ such that $\forall \varepsilon \in (0,1)$, $\lambda _1(M,g_\varepsilon )\le \varepsilon$ while $\mu _1^*(M,g_\varepsilon )\ge A$.

• ii. There exists a family of metrics $g_\varepsilon$of volume 1 on M and a constant $A>0$ such that, $\forall \varepsilon \in (0,1)$, $\lambda _1(M,g_\varepsilon )\rightarrow 0$ while $\mu _1^{**}(M,g_\varepsilon )\ge A$.

Proof

• (i) Let us start with a Riemannian metric g of volume one on M such that an open set V of M is isometric to the Euclidean ball of volume $\frac{1}{2}$. By a standard argument (Cheeger Dumbbell construction), one can deform the metric g outside V into a metric $g_\varepsilon$ of volume 1 such that $\lambda _1(M,g_\varepsilon )\le \varepsilon$. Applying Corollary 2.2 with $M_0=V$, we get $\mu _1^*(M, g_\varepsilon )\ge \vert V\vert _{g_\varepsilon } \lambda _1(V,g_\varepsilon )=\frac{1}{2} \lambda _1(V,g)$. Since $\lambda _1(V,g) = (2\omega _n)^{\frac{2}{n}} \lambda _1(B^n,g_E)$, where $B^n$ is the unit Euclidean ball, we get the desired inequality with $A= \frac{1}{2}(2\omega _n)^{\frac{2}{n}} \lambda _1(B^n,g_E)$.

• (ii) Let g be a Riemannian metric on M such that an open subset V of M is isometric to the capped cylinder $C=(-2,2)\times \mathbb {S}^{n-1}$ closed by a spherical cap. We will deform the metric g inside V so that $(M,g_{\varepsilon })$ looks like a Cheeger dumbbell (thus $\lambda _1(M,g_{\varepsilon })\rightarrow 0$ as $\varepsilon \rightarrow 0$) and associate with $g_{\varepsilon }$ a family of densities such that $\mu _1^{g_{\varepsilon }}(1,\sigma _\varepsilon ) \ge A>0$. Indeed, the metric on the cylinder $C=(-2,2)\times \mathbb {S}^{n-1}$ is given in coordinates $(t,x)\in (-2 , 2)\times \mathbb {S}^{n-1}$ by $g_\varepsilon (t,x)=\hbox {d}t^2+ \gamma _{\varepsilon }^2(t) g_{\mathbb {S}^{n-1}}$ with $\gamma _{\varepsilon } (-t)=\gamma _{\varepsilon } (t)$ and

  $$\begin{aligned} \gamma _\varepsilon (t)= \left\{ \begin{array}{lll} \varepsilon &{} \ &{} \text {if} \ t\in [0 , \frac{1}{2}]\\ \in (\varepsilon , 1) &{} \ &{} \text {if} \ t\in [\frac{1}{2},1 ]\\ 1 &{} \ &{} \text {if} \ t\in [1 , 2) \end{array}\right. . \end{aligned}$$

  (42)

  We do not change the metric g outside V. We endow $(M,g_{\varepsilon })$ with the density $\sigma _\varepsilon$ given by $\sigma _\varepsilon (t,x)=\frac{1}{\gamma _{\varepsilon } (t)^{n-1}}$ on the cylinder C and extended by 1 outside C.

It is well known that $\lambda _1(M,g_{\varepsilon }) \rightarrow 0$ as $\varepsilon \rightarrow 0$. Let us study $\mu _1^{g_{\varepsilon }}(1,\sigma _{\varepsilon })$. One has for every $f\in C^\infty (M)$

$$\begin{aligned} \int _M \vert \nabla ^\varepsilon f\vert _{g_\varepsilon }^2 \sigma _\varepsilon v_{g_\varepsilon }= & {} \int _{M{\setminus } C} \vert \nabla f\vert _{g}^{2} v_{g}+ \int _{-2}^{2}\hbox {d}t \int _{\mathbb {S}^{n-1}} \vert \nabla ^\varepsilon f\vert _{g_\varepsilon }^2 \sigma _\varepsilon (t)\gamma _\varepsilon (t)^{n-1} v_{\mathbb {S}^{n-1}}\\= & {} \int _{M{\setminus } C} \vert \nabla f\vert _{g}^{2} v_{g}+ \int _{-2}^{2}\hbox {d}t \int _{\mathbb {S}^{n-1}} \vert \nabla ^\varepsilon f\vert _{g_\varepsilon }^2 v_{\mathbb {S}^{n-1}}, \end{aligned}$$

where $v_{\mathbb {S}^{n-1}}$ denotes the volume form on the sphere $\mathbb {S}^{n-1}$. Now, observe that $\vert \nabla ^\varepsilon f\vert _{g_\varepsilon }^2$ can be estimated as follows:

$$\begin{aligned} \vert \nabla ^\varepsilon f\vert _{g_\varepsilon }^2= \left( \frac{\partial f}{\partial t}\right) ^2+\vert \nabla _{0} f\vert ^2 \gamma _{\varepsilon }(t)^{-2} \ge \left( \frac{\partial f}{\partial t}\right) ^2+\vert \nabla _{0} f\vert ^2 = \vert \nabla f\vert _{g}^2, \end{aligned}$$

where $\nabla _0f$ is the tangential part of the gradient of f w.r.t. $\mathbb {S}^{n-1}$. Therefore,

$$\begin{aligned} \int _M \vert \nabla ^\varepsilon f\vert _{g_\varepsilon }^2 \sigma _\varepsilon v_{g_\varepsilon } \ge \int _{M{\setminus } C} \vert \nabla f\vert _{g}^{2} v_{g}+\int _{-2}^{2}\hbox {d}t \int _{\mathbb {S}^{n-1}} \vert \nabla f\vert _{g}^2 v_{\mathbb {S}^{n-1}} = \int _M \vert \nabla f\vert _{g}^2 v_{g}. \end{aligned}$$

On the other hand (since $\gamma _\varepsilon (t)^{2}\le 1$)

$$\begin{aligned} \int _M f^2 v_{g_\varepsilon } \le \int _{M }f^2 v_{g}. \end{aligned}$$

In conclusion, for every $f\in C^\infty (M)$, one has

$$\begin{aligned} R_{(g_{\varepsilon },1,\sigma _{\varepsilon })}(f)\ge R_{(g,1,1)}(f). \end{aligned}$$

It follows, thanks to the min-max principle, that

$$\begin{aligned} \mu _1^{g_{\varepsilon }}(1,\sigma _{\varepsilon })\ge \lambda _1(M,g). \end{aligned}$$

The last point is to suitably rescale $g_{\varepsilon }$ and $\sigma _\varepsilon$. For this purpose, just observe that $\int _M \sigma _{\varepsilon }v_{g_{\varepsilon }} =\vert M\vert _g$ and $\frac{1}{2} \vert M\vert _g\le \vert M\vert _{g_\varepsilon }\le \vert M\vert _g$. $\square$

6 Examples

In this section, we describe situations in which we can compute or give explicit estimates for the first extremal eigenvalues. Let (M, g) be a compact Riemannian manifold of dimension $n\ge 2$, possibly with a nonempty boundary.

Proposition 6.1

Assume that there exists a conformal map $\phi$ from (M, g) to the standard n-dimensional sphere $\mathbb {S}^n$. Then,

$$\begin{aligned} \lambda _1^c(M,g)=n {\alpha _n}^{\frac{2}{n}} \end{aligned}$$

(43)

and

$$\begin{aligned} \mu _1^*(M,g)\le n \left( \frac{\alpha _n}{|M|_g}\right) ^{\frac{2}{n}}, \end{aligned}$$

(44)

where $\alpha _n$ is the volume of the unit Euclidean n-sphere. Moreover, if $n=2$, then the equality holds in (44).

Notice that when (M, g) is the standard sphere $\mathbb {S}^n$, then equality holds in (44) (see Corollary 6.3).

Proof of Propositon 6.1

Let us first prove (44). Let $\rho$ be a density on M with $\int _M\rho v_g=1$. Given any nonconstant map $\phi =(\phi _1,\cdots ,\phi _{n+1}):(M,g)\rightarrow \mathbb {S}^n$, a standard argument tells us that there exists a conformal diffeomorphism ${\gamma \in Conf(\mathbb {S}^n)}$ such that $\psi =\gamma \circ \phi$ satisfies $\int _M \psi _j \rho \, v_g =0$, $j=1\dots , n+1$ (see for instance [21, Proposition 4.1.5]). Thus, $\forall j\le n+1$,

$$\begin{aligned} \mu _1( \rho ,1)\int _M\psi _j^2 \rho \, v_g \le \int _M \vert \nabla \psi _j \vert ^2 v_g \end{aligned}$$

(see (3)) and, summing up w.r.t. j,

$$\begin{aligned} \mu _1( \rho ,1)\int _M\rho \, v_g \le \int _M \vert d\psi \vert ^2 v_g \le \left( \int _M \vert d\psi \vert ^n v_g\right) ^{\frac{2}{n}} \vert M\vert _g^{1-\frac{2}{n}}. \end{aligned}$$

Since $\psi =\gamma \circ \phi$ is a conformal map, $\int _M \vert d\psi \vert ^n v_g$ is nothing but $n^{\frac{n}{2}}$ times the volume of $\psi (M)\subset \mathbb {S}^n$ with respect to the standard metric $g_s$ of $\mathbb {S}^n$ (indeed, $\psi ^*g_s=\frac{1}{n} \vert d\psi \vert ^2g$). Therefore,

$$\begin{aligned} \mu _1( \rho ,1)\fint _M\rho v_g\le n\vert \psi (M)\vert _{g_s}^{\frac{2}{n}} \vert M\vert _g^{-\frac{2}{n}}\le n \left( \frac{\alpha _n}{|M|_g}\right) ^{\frac{2}{n}} \end{aligned}$$

which proves (44).

Using the same arguments, we can prove the inequality $\lambda _1^c(M,g)\le n {\alpha _n}^{\frac{2}{n}}$. The reverse inequality follows from [9, Theorem A]. $\square$

It is well known that the Euclidean space $\mathbb {R}^n$ and the hyperbolic space $\mathbb H^n$ are conformally equivalent to open parts of the sphere $\mathbb {S}^n$. This leads to the following corollary.

Corollary 6.1

Let $\Omega$ be a bounded domain of the Euclidean space $\mathbb {R}^n$, the hyperbolic space $\mathbb H^n$ or the sphere $\mathbb {S}^n$, endowed with the induced metric $g_s$. One has

$$\begin{aligned} \lambda _1^c(\Omega ,g_{s})=n {\alpha _n}^{\frac{2}{n}} \end{aligned}$$

and

$$\begin{aligned} \mu _1^*(\Omega ,g_{s})\le n \left( \frac{\alpha _n}{|\Omega |}\right) ^{\frac{2}{n}}. \end{aligned}$$

Moreover, the following equality holds in dimension 2: $\mu _1^*(\Omega ,g_{s})=\lambda _1^c(\Omega ,g_{s}){|\Omega |}^{-1} =\frac{8\pi }{|\Omega |}.$

Remark 6.1

Let D be the unit disk in $\mathbb {R}^2$ and let $\rho _t=\frac{4t}{(t^2\vert z\vert ^2+1)^2}$. Then,

$$\begin{aligned} \mu _1^*(D,g_{E}) =\lim _{t\rightarrow \infty }\mu _1^{g_{E}} (\frac{\rho _t}{\fint _D\rho _t \hbox {d}x},1) = 8. \end{aligned}$$

Indeed, the map $\phi _t(z)= \frac{1}{t^2\vert z\vert ^2+1}(2 tz, t^2\vert z\vert ^2-1)$ identifies $(D, \frac{4t}{(t^2\vert z\vert ^2+1)^2} g_E)$ with a spherical cap $C_t$ in $\mathbb {S}^2$ whose radius goes to $\pi$ as $t\rightarrow \infty$. Hence, $\mu _1^{g_{E}}(\rho _t,1)\int _D\rho _t \hbox {d}x = \mu _1(C_t)\vert C_t\vert$ which converges to $8\pi$ as $t\rightarrow \infty$.

Proposition 6.2

Assume that there exists a map $\phi :(M,g)\rightarrow \mathbb {S}^p$ from (M, g) to the standard p-dimensional sphere $\mathbb {S}^p$ satisfying both $\int _M\phi v_g=0$ and $\vert d\phi \vert ^2\le \Lambda$ for some positive constant $\Lambda$. Then,

$$\begin{aligned} \mu _1^{**}(M,g)\le \Lambda . \end{aligned}$$

(45)

Proof

One has, for every $j\le p+1$,

$$\begin{aligned} \mu _1( 1,\sigma )\int _M\phi _j^2 \, v_g \le \int _M \vert \nabla \phi _j \vert ^2\sigma v_g \end{aligned}$$

and, summing up w.r.t. j,

$$\begin{aligned} \mu _1( 1,\sigma )\vert M\vert _g \le \int _M \vert d\phi \vert ^2 \sigma v_g \le \Lambda \int _M \sigma v_g \end{aligned}$$

which implies (45). $\square$

If (M, g) is a compact homogeneous Riemannian manifold, and if $\phi _1,\dots ,\phi _p$ is an $L^2$-orthonormal basis of the first eigenspace of the Laplacian, then both $\sum _{i\le p} \phi _i^2$ and $\vert d\phi \vert ^2=\sum _{i\le p} \vert d\phi _i\vert ^2$ are constant on M. This enables us to apply Proposition 6.2 and get the following

Corollary 6.2

Let (M, g) be a compact homogeneous Riemannian manifold. Then,

$$\begin{aligned} \mu _1^{**}(M,g)=\mu _1(M,g). \end{aligned}$$

In other words, on a compact homogeneous Riemannian manifold, $\mu _1(1,\sigma )$ is maximized when $\sigma$ is constant.

Example 6.1

In [19], it is proved that if $\Gamma ={\mathbb Z}e_1+ {\mathbb Z}e_2 \subset {\mathbb R}^2$ is a lattice such that $|e_1 | =|e_2|$, then the corresponding flat metric $g_{_\Gamma }$ on the torus ${\mathbb T}^2$ satisfies $\mu _1 ^c ({\mathbb T}^2, g_{_\Gamma })=\lambda _1 ({\mathbb T}^2,g_{_\Gamma })\vert {\mathbb T}^2 \vert _{g_{_\Gamma }}$. A higher-dimensional version of this result was also established in [18]. Since a flat torus is a 2-dimensional homogeneous Riemannian manifold, we have the following equalities

$$\begin{aligned} \lambda _1^c ({\mathbb T}^2, g_{_\Gamma })\vert {\mathbb T}^2 \vert _{g_{_\Gamma }}^{-1} =\mu _1^{*}({\mathbb T}^2, g_{_\Gamma })=\mu _1^{**}({\mathbb T}^2, g_{_\Gamma })=\lambda _1 ({\mathbb T}^2, g_{_\Gamma }). \end{aligned}$$

Nevertheless, whereas we always have $\mu _1^{**}({\mathbb T}^2, g_{_\Gamma })=\mu _1 ({\mathbb T}^2, g_{_\Gamma })$, it follows from [9, Theorem A] that when the length ratio $|e_2| / |e_1 |$ of the vectors $e_1$ and $e_2$ is sufficiently far from 1, then $\mu _1^{*}({\mathbb T}^2, g_{_\Gamma })= \lambda _1 ^c ({\mathbb T}^2, g_{_\Gamma })\vert {\mathbb T}^2 \vert _{g_{_\Gamma }}^{-1}>\lambda _1 ({\mathbb T}^2,g_{_\Gamma })$.

Recall that a map $\phi =(\phi _1,\cdots ,\phi _{p+1}):(M,g)\rightarrow \mathbb {S}^p$ is harmonic if and only if its components $\phi _1,\cdots ,\phi _{p+1}$ satisfy

$$\begin{aligned} \Delta _g\phi _j = -\vert d\phi \vert ^2 \phi _j, \quad j=1\cdots , p+1. \end{aligned}$$

The stress-energy tensor of a map $\phi$ is a symmetric covariant 2-tensor defined for every tangent vector field X on M by: $S_\phi (X,X)=\frac{1}{2} \vert d\phi \vert ^2 \vert X\vert ^2_g-\vert d\phi (X)\vert ^2$. In [15, Theorem 3.1], it is proved that if the stress-energy tensor of a harmonic map $\phi$ is nonnegative, then for every conformal diffeomorphism $\gamma$ of the sphere $\mathbb {S}^p$ one has

$$\begin{aligned} \int _M \vert d(\gamma \circ \phi )\vert ^2 v_g \le \int _M \vert d\phi \vert ^2 v_g. \end{aligned}$$

Moreover, the strict inequality holds if $\gamma$ is not an isometry and if $S_\phi$ is positive definite at some point. Observe that if $\phi :(M,g)\rightarrow \mathbb {S}^p$ is a conformal map or a horizontally conformal map, then $S_\phi$ is nonnegative (see [15]).

Proposition 6.3

Assume that there exists a harmonic map $\phi :(M,g)\rightarrow \mathbb {S}^p$ with nonnegative stress-energy tensor. Then,

$$\begin{aligned} \mu _1^*(M,g)\le \fint _M \vert d\phi \vert ^2 v_g. \end{aligned}$$

(46)

Proof

Let $\rho$ be a positive density on M. As before, we know that there exists ${\gamma \in Conf(\mathbb {S}^n)}$ such that $\psi =\gamma \circ \phi$ satisfies $\int _M \psi _j \rho \, v_g =0$, $j=1\dots , n+1$. Thus,

$$\begin{aligned} \mu _1( \rho ,1)\int _M\psi _j^2 \rho \, v_g \le \int _M \vert \nabla \psi _j \vert ^2 v_g \end{aligned}$$

and, summing up w.r.t. j,

$$\begin{aligned} \mu _1( \rho ,1)\int _M\rho \, v_g \le \int _M \vert d(\gamma \circ \phi )\vert ^2 v_g \le \int _M \vert d\phi \vert ^2 v_g \end{aligned}$$

which implies (46). $\square$

A particular case of Proposition 6.3 is when there exists a harmonic map $\phi :(M,g)\rightarrow \mathbb {S}^p$ which is homothetic. In this case, $S_\phi =\frac{n-2}{n} \vert d\phi \vert ^2 g$ and $\vert d\phi \vert ^2$ is constant and coincides with an eigenvalue $\lambda _k(M,g)$ for some $k\ge 1$. For example, if (M, g) is a compact isotropy irreducible homogeneous space (e.g., a compact rank-one symmetric space) and if $\phi _1,\dots ,\phi _p$ is an $L^2$-orthonormal basis of the first eigenspace of the Laplacian, then $\phi =\left( \frac{\vert M\vert _g}{p}\right) ^{\frac{1}{2}}(\phi _1,\dots ,\phi _p)$ is a harmonic map from (M, g) to $\mathbb {S}^p$ which is homothetic and satisfies $\vert d\phi \vert ^2=\lambda _1(M,g)$. Proposition 6.3 then implies that $\mu _1^{*}(M,g)=\lambda _1(M,g)$. On the other hand, the second author and Ilias [17] proved that in this situation we also have $\lambda _1^c (M, g)= \lambda _1 (M,g)\vert M\vert _g^{\frac{2}{n}}$. Consequently, we have the following

Corollary 6.3

Let (M, g) be a compact isotropy irreducible homogeneous space. Then,

$$\begin{aligned} \lambda _1^{c}(M,g)\vert M\vert _g^{-\frac{2}{n}}=\mu _1^{*}(M,g)=\mu _1^{**}(M,g)=\lambda _1(M,g). \end{aligned}$$