Sharp spectral gaps on metric measure spaces

Abstract

In this paper, we extend the sharp lower bounds of spectral gap, due to Andrews-Clutterbuck (Anal PDE 6(5):1013–1024, 2013), Bakry-Qian (Adv Math 155:98–153, 2000), Chen-Wang (Sci Sin A 37:1–14, 1994) and Chen-Wang (Sci Sin A 40:384-394, 1997), from smooth Riemannian manifolds to general metric measure spaces with Riemannian curvature-dimension condition $RCD^*(K,N)$.

1 Introduction

Let (X, d, m) be a compact metric measure space. Given a Lipschitz function $f:X\rightarrow \mathbb R$, its point-wise Lipschitz constant $\mathrm{Lip}f(x)$ is defined as

$$\begin{aligned} \mathrm{Lip}f(x):=\limsup _{y\rightarrow x}\frac{|f(y)-f(x)|}{d(x,y)}. \end{aligned}$$

In this paper, we are concerned with the spectral gap

$$\begin{aligned} \lambda _1(X):=\inf \Big \{\frac{\int _X (\mathrm{Lip}f)^2\mathrm {d}m}{\int _Xf^2\mathrm {d}m}:\ f\in Lip(X)\backslash \{0\} \quad \mathrm{and}\quad \int _Xf\mathrm {d}m=0\Big \}, \end{aligned}$$

(1.1)

where Lip(X) is the space of Lipschitz functions on X.

When M is a compact smooth Riemannian manifold without boundary (or with a convex boundary $\partial M$), the study of the lower bounds of the first eigenvalue $\lambda _1$ of the Laplace-Beltrami operator $\Delta$ has a long history. See for example, Lichnerowicz [27], Cheeger [10], Li-Yau [26], and so on. For an overview the reader is referred to the introduction of [6, 7, 25] and Chapter 3 in book [35], and references therein. In particular the following comparison theorem for $\lambda _1$ has been established by Chen-Wang [12, 13], Bakry-Qian [6] and Andrews-Clutterbuck [5] independently, via three different methods.

Theorem 1.1

(Chen-Wang [12, 13], Bakry-Qian [6], Andrews-Clutterbuck [5]) Let M be an N-dimensional compact Riemannian manifold without boundary (or with a convex boundary). Suppose that the Ricci curvature $Ric(M)\geqslant K$ and that the diameter $\leqslant d$. Let $\lambda _1$ be the first (non-zero) eigenvalue (with Neumann boundary condition if the boundary is not empty). Then

$$\begin{aligned} \lambda _1(M) \geqslant \hat{\lambda }(K,N,d) \end{aligned}$$

where $\hat{\lambda }(K,N,d)$ denotes the first non-zero Neumann eigenvalue of the following one-dimensional model:

$$\begin{aligned} v''(x)-(N-1)T(x)v'(x)=-\lambda v(x) \qquad x\in \left( -\frac{d}{2}, \frac{d}{2}\right) , \qquad v'\left( -\frac{d}{2}\right) =v'\left( \frac{d}{2}\right) =0 \end{aligned}$$

and

$$\begin{aligned} T(x)=\left\{ \begin{array}{ll} \sqrt{\frac{K}{N-1}} \tan \left( \sqrt{\frac{K}{N-1}} x\right) &{} \qquad \mathrm{if }K \geqslant 0,\\ \sqrt{\frac{-K}{N-1}} \tanh \left( \sqrt{\frac{-K}{N-1}} x\right) &{} \qquad \mathrm{if }K < 0. \end{array} \right. \end{aligned}$$

This comparison Theorem 1.1 implies the classical Lichnerowicz estimate [27] for $K=n-1$ and also Zhong-Yang’s estimate [42] for $K=0$. Some lower bounds of the spectral gaps have been extended to singular spaces. In [36], Shioya discussed spectral gaps in Riemannian orbifolds. In [31], Petrunin proved the Linchnerowiz estimate for Alexandrov spaces with curvature $\geqslant 1$ in the sense of Alexandrov. Recently, Theorem 1.1 has been extended to Alexandrov spaces in [32] using a notion of generalized lower Ricci curvature bounds in [41], and by Wang-Xia [40] to Finsler manifolds.

In the last few years, several notions of “the generalized Ricci curvature bounded below” on general metric spaces have been introduced. Sturm [38, 39] and Lott-Villani [28], independently, introduced a so-called curvature-dimension condition, denoted by CD, on metric measure spaces via optimal transportation. A refinement of this notion is given in Ambrosio-Gigli-Savaré [3], which is called Riemannian curvature-dimension condition, denoted by $RCD^*$. Recently, in two remarkable works, Ambrosio-Gigli-Savaré [1] and Erbar-Kuwada-Sturm [16], they proved the equivalence of the Riemannian curvature-dimension condition and of the Bochner formular of Bakry-Émery via an abstract $\Gamma _2$-calculus, denoted by BE. Notice that in the case where M is a (compact) Riemannian manifold. Given two numbers $K\in \mathbb R$ and $N\geqslant 1$, M satisfying the Riemannian curvature-dimension condition $RCD^*(K,N)$ is equivalent to that the Ricci curvature $Ric(M)\geqslant K$ and the dimension $dim\leqslant N$.

We will consider the spectral gap on metric measure spaces under a suitable Riemannian curvature-dimension condition. Lott-Villani [29] and Erbar-Kuwada-Sturm [16] extended Linchnerowicz’s estimate to metric measure spaces with CD(K, N) or $RCD^*(K,N)$ for $K>0$ and $1\leqslant N <\infty$.

In this paper, we will extend Theorem 1.1 to general metric measure spaces. Precisely, we have the following theorem.

Theorem 1.2

Let $K\in \mathbb R$, $1\leqslant N < \infty$ and $d>0$. Let (X, d, m) be a compact metric measure space satisfying the Riemannian curvature-dimension condition $RCD^*(K,N)$ and the diameter $\leqslant d.$ Then the spectral gap $\lambda _1(X)$ has the following lower bound

$$\begin{aligned} \lambda _1(X) \geqslant \hat{\lambda }(K,N,d), \end{aligned}$$

(1.2)

where $\hat{\lambda }(K,N,d)$ is given in Theorem 1.1.

Our proof of Theorem 1.2 relies on the self-improvement of regularity under the Riemannian curvature-dimension condition (Theorem 2.6) and a version of maximum principle, which is similar to the classical maximum principle for $C^2$-functions on manifolds (see Proposition 3.1 and Remark 3.2).

Remark 1.3

(1) When $N>1$ and $K=N-1$, the above Theorem 1.2 implies that

$$\begin{aligned} \lambda _1(X)\geqslant \frac{N}{1-\cos ^N(d/2)}. \end{aligned}$$

In particular, this gives that if $\lambda _1(X)=N$, then $d=\pi$. The combination of this and the maximal diameter theorem in [22] implies an Obata-type rigidity theorem for general metric measure spaces, which is also proved in [23] by Ketterer, independently.

(2) Very recently Cavalletti and Mondino [8, 9] have used a differential method to establish a further generalization of this result. They prove the same sharp spectral gap estimates (and some other sharp isoperimetric and functional inequalities) for non-branching CD-spaces.

2 Preliminaries

In this section, we recall some basic notions and the calculus on metric measure spaces. For our purpose in this paper, we will focus only on the case of compact spaces. Let (X, d) be a compact metric space, and let m be a Radon measure with $\mathrm{supp}(m)=X$.

2.1 Riemannian curvature-dimension condition $RCD^*(K,N)$

Let (X, d, m) be a compact metric measure space. The Cheeger energy is given in [2] from the relaxation in $L^2(X,m)$ of the point-wise Lipschitz constant of Lipschitz functions. That is, given a function $f\in L^2(X,m)$, the Cheeger energy of f is defined [2] by

$$\begin{aligned} \mathrm{Ch}(f):=\inf \Big \{\liminf _{j\rightarrow \infty }\frac{1}{2}\int _X(\mathrm{Lip} f_j)^2dm\Big \}, \end{aligned}$$

where the infimum is taken over all sequences of Lipschitz functions $\{f_j\}$ converging to f in $L^2(X,m)$. If $\mathrm{Ch}(f)<\infty$, then there is a (unique) so-called minimal relaxed gradient $|Df|_w$ such that

$$\begin{aligned} \mathrm{Ch}(f)=\frac{1}{2}\int _X|Df|_w^2\mathrm {d}m. \end{aligned}$$

The domain of $\mathrm{Ch}$ in $L^2(X,m)$, $\mathrm{D(Ch)}$, is a Banach space with norm $\sqrt{\Vert f\Vert _{L^2}^2+\Vert |Df|_w\Vert _{L^2}^2}$.

Definition 2.1

([2]) A metric measure space (X, d, m) is called infinitesimally Hilbertian if the associated Cheeger energy $\mathrm{Ch}$ is a quadratic form.

Let (X, d, m) be an infinitesimally Hilbertian space. It is proved in [3] that the scalar product

$$\begin{aligned} \Gamma (f,g):=\lim _{\epsilon \rightarrow 0^+}\frac{|D(f+\epsilon g)|^2_w-|Df|^2_w}{2\epsilon }\qquad f,g\in \mathrm{D(Ch)} \end{aligned}$$

exists in $L^1(X,m).$ In the following we denote by $\mathbb V$ the Hilbert space $\mathrm{D(Ch)}$ with the scalar product

$$\begin{aligned} (f,g)_{\mathbb V}:=\int _X\big (fg+\Gamma (f,g)\big )\mathrm {d}m. \end{aligned}$$

The quadratic form $\mathrm{Ch}$ canonically induces a symmetric, regular, strongly local Dirichlet form $(\mathrm{Ch},\mathbb V)$. The regular property of $(\mathrm{Ch},\mathbb V)$ comes from that X is always assumed to be compact. Moreover, for any $f,g\in \mathbb V$, $\Gamma (f,g)$ provides an explicit expression of the Carré du champ of the Dirichlet form $(\mathrm{Ch},\mathbb V)$. The associated energy measure of f is absolutely continuous with respect to m with density $\Gamma (f)=|Df|^2_w$.

Denote by $(H_t)_{t>0}$ and $\Delta$ the associated Markov semigroup in $L^2(X,m)$ and its generator respectively. Since X is compact, according to [33], the $RCD^*(K,N)$ condition implies that (X, d, m) supports a global Poincaré inequality. Moreover, the operator $(-\Delta )^{-1}$ is a compact operator. Then the spectral theorem gives that the $\lambda _1(X)$ in (1.1) is the first non-zero eigenvalue of $-\Delta .$ (See, for example, [15].)

We adopt the notations given in [4]:

$$\begin{aligned} D_{\mathbb V}(\Delta ):= \big \{f\in \mathbb V :\Delta f \in \mathbb V \big \} \end{aligned}$$

and, for every $p\in [1,\infty ]$,

$$\begin{aligned} D_{L^p}(\Delta ):= \big \{f\in \mathbb V\cap L^p(X,m) :\Delta f \in L^2\cap L^p(X,m) \big \}. \end{aligned}$$

Definition 2.2

([4, 16]) Let $K\in \mathbb R$ and $N\geqslant 1$. An infinitesimally Hilbertian space (X, d, m) is said to satisfy the condition BE(K, N) if the associated Dirichlet form $(\mathrm{Ch},\mathbb V)$ satisfies

$$\begin{aligned} \int _X\Big (\frac{1}{2} \Gamma (f)\Delta \phi -\Gamma (f,\Delta f)\phi \Big )\mathrm {d}m\geqslant K\int _X\Gamma (f)\phi \mathrm {d}m +\frac{1}{N}\int _X(\Delta f)^2\phi \mathrm {d}m \end{aligned}$$

for all $f\in D_{\mathbb V}(\Delta )$ and all non-negative $\phi \in D_{L^\infty }(\Delta ).$

According to [4, 16], the Riemannian curvature-dimension condition $RCD^*(K,N)$ is equivalent to the corresponding Bakry-Émery condition BE(K, N) with a slight regularity. We shall use the following definition for $RCD^*(K,N)$ (Notice that X is always assumed to be compact in the paper).

Definition 2.3

([4, 16]), Let $K\in \mathbb R$ and $N\geqslant 1$. A compact, infinitesimally Hilbertian geodesic space (X, d, m) is said to satisfy the $RCD^*(K,N)$-condition (or metric BE(K, N) condition) if it satisfies BE(K, N) and that every $f\in \mathbb V$ with $\Vert \Gamma (f)\Vert _{L^\infty }\leqslant 1$ has a 1-Lipschitz representative.

Recall that a (locally) compact metric (X, d) is a geodesic space if the distance between any two points in X can be realized as the length of some curve connecting them. Notice that if (X, d, m) satisfies $RCD^*(K,N)$ condition then $d=d_\mathrm{Ch}$, where $d_\mathrm{Ch}$ is the induced metric by the Dirichlet form $(\mathrm{Ch},\mathbb V)$. For any $f\in \mathbb V$ with $\Gamma (f)\in L^\infty (X,m)$, we always identify f with its Lipschitz representative. Moreover, $H_tf$, $H_t(|\nabla f|^2_w)$ and $\Delta H_tf$ have continuous representatives (see Proposition 4.4 of [16]).

2.2 The self-improvement of regularity on $RCD^*(K,N)$-spaces

Let $K\in \mathbb R$ and $1\leqslant N<\infty$, and let (X, d, m) be a compact metric measure space satisfying the $RCD^*(K,N)$ condition.

Let us recall an extension of the generator $\Delta$ of $(\mathrm{Ch},\mathbb V)$, which is introduced in [4, 34]. Denote by $\mathbb V'$ the set of continuous linear functionals $\ell :\mathbb {V}\rightarrow \mathbb R,$ and $\mathbb {V}'_+$ denotes the set of positive linear functionals $\ell \in \mathbb {V}'$ such that $\ell ( \varphi ) \geqslant 0$ for all $\varphi \in \mathbb {V}$ with $\varphi \geqslant 0$ m-a.e. in X . An important characterization of functionals in $\mathbb V'_+$ is that, for each $\ell \in \mathbb V'_+$ there exists a unique corresponding Radon measure $\mu _\ell$ on X such that

$$\begin{aligned} \ell (\varphi )=\int _X\tilde{\varphi }\mathrm {d}\mu _\ell \quad \forall \varphi \in \mathbb V, \end{aligned}$$

where $\tilde{\varphi }$ is a quasi continuous representative of $\varphi$. Denote by

$$\begin{aligned} \mathbb {M}_{\infty }:=\Big \{f\in \mathbb {V}\cap L^{\infty }(X,m): \exists \ \mu \ \ \mathrm{such\ that}\ \ -\mathcal {E}(f,\varphi )=\int _X \tilde{\varphi } \mathrm {d}\mu \quad \forall \varphi \in \mathbb {V}\Big \}, \end{aligned}$$

where $\mu =\mu _+-\mu _-$ with $\mu _{+},\mu _- \in \mathbb {V}'_+$. When a function $f\in \mathbb {M}_{\infty }$, we set $\Delta ^*f:=\mu ,$ and denote its Lebesgue decomposition w.r.t m as $\Delta ^* f=\Delta ^{ab} f\cdot m+\Delta ^sf$. It is clear that if $f\in D(\Delta )\cap L^{\infty }(X,m)$ then $f\in \mathbb {M}_{\infty }$ and $\Delta ^*f=\Delta f\cdot m.$

Lemma 2.4

Let $K\in \mathbb R$ and $N\geqslant 1$, and let (X, d, m) be a compact metric measure space satisfying $RCD^*(K,N)$ condition.

• (i) (Chain rule, [34, Lemma3.2]) If $g\in D(\Delta )\cap Lip(X)$ and $\phi \in C^{2}(\mathbb R)$ with $\phi (0)=0$, then we have

  $$\begin{aligned} \phi \circ g\in D(\Delta )\cap Lip(X)\qquad \mathrm{and}\qquad \Delta (\phi \circ g)=\phi '\circ g\cdot \Delta g +\phi ''\circ g\cdot \Gamma (g); \end{aligned}$$

• (ii) (Leibniz rule, [34, Corollary2.7]) If $g_1\in \mathbb {M}_{\infty }$ and $g_2\in D(\Delta )\cap Lip(X)$, then we have

  $$\begin{aligned} g_1\cdot g_2\in \mathbb {M}_{\infty }\qquad \mathrm{and}\qquad \Delta ^*(g_1\cdot g_2)=g_2\cdot \Delta ^*g_1+g_1\cdot \Delta g_2\cdot m+2\Gamma (g_1,g_2)\cdot m. \end{aligned}$$

Remark 2.5

We can take $\phi \in C^{2}(\mathbb R)$ without the restriction $\phi (0)=0$ in the Chain rule. This comes from the fact that $1\in D(\Delta )$ and $\Delta 1=0$, because X is assumed to be compact.

The following self-improvement of regularity is given in Lemma 3.2 of [34]. (See also Theorem 2.7 of [17]).

Theorem 2.6

([17, 34]) Let $K\in \mathbb R$ and $1\leqslant N<\infty$, and let (X, d, m) be a compact metric measure space satisfying $RCD^*(K,N)$ condition. If $f \in D_{\mathbb {V}}(\Delta )\cap Lip(X)$, then we have $\Gamma (f) \in \mathbb {M}_{\infty }$ and

$$\begin{aligned} \frac{1}{2} \Delta ^*\Gamma (f)-\Gamma (f,\Delta f)\cdot m \geqslant K\Gamma (f)\cdot m +\frac{1}{N} (\Delta f)^2\cdot m. \end{aligned}$$

(2.1)

A crucial fact, which is implied by the above inequality, is that the singular part of $\Delta ^*\Gamma (f)$ has a correct sign: $\Delta ^s \Gamma (f)$ is non-negative.

Using the same trick as in the proof of Bakry-Qian [6, Thm 6] and [34, Thm 3.4], one can prove the following Corollary of Theorem 2.6 (see [21, Lemma 2.3] for a detailed proof):

Corollary 2.7

Let $K\in \mathbb R$ and $1\leqslant N<\infty$, and let (X, d, m) be a compact metric measure space satisfying $RCD^*(K,N)$ condition. If $f \in D_{\mathbb {V}}(\Delta )\cap Lip(X)$, then $\Delta ^s f\geqslant 0$ and the following holds m-a.e. on $\{x\in X:\Gamma (f)(x)\ne 0\}$,

$$\begin{aligned} \left( \frac{1}{2}\Delta ^{ab}\Gamma (f)-\Gamma (f,\Delta f)-K\Gamma (f)-\frac{1}{N}(\Delta f)^2\right) \geqslant \frac{N}{N-1}\left( \frac{\Delta f}{N} -\frac{\Gamma (f,\Gamma (f))}{2\Gamma (f)}\right) ^2. \end{aligned}$$

(2.2)

For $\kappa \in \mathbb {R}$ and $\theta \geqslant 0$ we denote the function

$$\begin{aligned} \mathfrak {s}_\kappa (\theta )=\left\{ \begin{array}{ll} \frac{1}{\sqrt{\kappa }} \sin (\sqrt{\kappa }\theta ),&{}\qquad \kappa >0,\\ \theta ,&{}\qquad \kappa =0,\\ \frac{1}{\sqrt{ -\kappa }} \sinh (\sqrt{-\kappa } \theta ),&{}\qquad \kappa <0. \end{array} \right. \end{aligned}$$

Proposition 2.8

(Bishop-Gromov inequality, [18, 39]) For each $x_0 \in X$ and $0<r<R \leqslant \pi \sqrt{(N-1)/(K\vee 0)}$, we have

$$\begin{aligned} \frac{m(B_r(x_0))}{m(B_R(x_0))} \geqslant \frac{\int _0^r \mathfrak {s}_{\frac{K}{N-1}}(t)^{N-1} \mathrm {d}t}{\int _0^R \mathfrak {s}_{\frac{K}{N-1}}(t)^{N-1} \mathrm {d}t}. \end{aligned}$$

(2.3)

Proof

By Corollary of 1.5 in [18], (X, d, m) satisfies MCP(K, N) condition. The desired Bishop-Gromov inequality (2.3) holds on MCP(K, N)-spaces by Remark 5.3 of [39]. $\square$

We need also the following mean value inequality in [30]. See also Lemma 2.1 of [14].

Lemma 2.9

([30, Lemma 3.4]) Let $f \in D(\Delta )$ be a non-negative, continuous function with $\Delta f \leqslant c_0$ m-a.e.. Then there exists a constant $C(K,N,\text {diam}X)$ such that the following holds:

$$\begin{aligned} \int \!\!\!\!\!\!-_{B_r(x)} f \mathrm {d}m \leqslant C (f(x)+c_0 r^2). \end{aligned}$$

(2.4)

At last, we need the following Sobolev inequality, whose proof is similar to that of Theorem 13.1 of [19]. For the reader’s convenience, we include a proof here.

Lemma 2.10

Let $E \subset X$ be an m-measurable subset with $m(E)>0$. Then there exist constants $\nu > 2$ and $\widetilde{C_S}$ which depend only on K, N, X and E, such that for any $f \in \mathbb {V}$ with $f=0$ m-a.e. in E, the following Sobolev inequality holds:

$$\begin{aligned} \Vert f\Vert _{L^{\nu }(X)}\leqslant \widetilde{C_S} \left( \int _X \Gamma (f)\mathrm {d}m\right) ^{\frac{1}{2}}. \end{aligned}$$

(2.5)

Proof

The above Bishop-Gromov inequality (2.3) implies the doubling property; and by Theorem 2.1 of [33], a Poincaré inequality holds. These two ingredients imply the following Sobolev inequality by Theorem 9.7 of [19]: there exist constants $\nu >2$ and $C_S>0$, depending on K, N and $\text {diam}X$, such that for all $f\in \mathbb {V}$,

$$\begin{aligned} \left( \int \!\!\!\!\!\!-_X|f-\int \!\!\!\!\!\!-_Xf|^{\nu }\right) ^{\frac{1}{\nu }} \leqslant C_S\left( \int \!\!\!\!\!\!-_X \Gamma (f)\mathrm {d}m\right) ^{\frac{1}{2}}, \end{aligned}$$

(2.6)

where $\int \!\!\!\!\!\!-_X \Gamma (f):=\frac{1}{m(X)}\int _X \Gamma (f) \mathrm {d}m$.

Note that $\int \!\!\!\!\!\!-_X f$ is a constant and that $f=0$ on E, thus we have $\Vert f-\int \!\!\!\!\!\!-_X f\Vert _{L^{\nu }(E)}=(m(E))^{\frac{1}{\nu }}\cdot |\int \!\!\!\!\!\!-_X f|$ and

$$\begin{aligned} \Vert \int \!\!\!\!\!\!-_X f\Vert _{L^{\nu }(X)}&={m(X)}^{\frac{1}{\nu }}|\int \!\!\!\!\!\!-_X f|\\&=\left( \frac{m(X)}{m(E)}\right) ^{\frac{1}{\nu }}\Vert f-\int \!\!\!\!\!\!-_X f\Vert _{L^{\nu }(E)}\\&\leqslant \left( \frac{m(X)}{m(E)}\right) ^{\frac{1}{\nu }}\Vert f-\int \!\!\!\!\!\!-_X f\Vert _{L^{\nu }(X)}. \end{aligned}$$

Then, by Minkowski inequality, we have

$$\begin{aligned} \Vert f\Vert _{L^{\nu }(X)}&\leqslant \Vert f-\int \!\!\!\!\!\!-_X f\Vert _{L^{\nu }(X)}+\Vert \int \!\!\!\!\!\!-_X f\Vert _{L^{\nu }(X)}\\&\leqslant \left[ 1+(\frac{m(X)}{m(E)})^{\frac{1}{\nu }}\right] \Vert f-\int \!\!\!\!\!\!-_X f\Vert _{L^{\nu }(X)}\\&\mathop {\leqslant }\limits ^{(2.6)} \left[ 1+(\frac{m(X)}{m(E)})^{\frac{1}{\nu }}\right] \cdot C_S {m(X)}^{\frac{1}{\nu }-\frac{1}{2}} \left( \int _X \Gamma (f)\mathrm {d}m\right) ^{\frac{1}{2}}. \end{aligned}$$

Let $\widetilde{C_S}= C_S {m(X)}^{\frac{1}{\nu }-\frac{1}{2}} \left[ 1+\frac{m(X)}{m(E)})^{\frac{1}{\nu }}\right]$, thus we have completed the proof. $\square$

3 Eigenvalue estimate for $RCD^*(K,N)$-spaces

Let $K\in \mathbb R$ and $1\leqslant N<\infty$, and let (X, d, m) be a compact $RCD^*(K,N)$-space. We need a version of maximum principle on X as follows.

Proposition 3.1

Let $u\in \mathbb M_\infty$ and let $\varepsilon _0>0$. If the measure $\Delta ^*u$ satisfies that the singular part $\Delta ^su\geqslant 0$ on X and that the absolutely continuous part

$$\begin{aligned} \Delta ^{ab}u\geqslant C_1\cdot u-C_2\cdot \sqrt{\Gamma (u)}\quad m\mathrm{-a.e. \ on }\quad \{x: u(x)\geqslant \varepsilon _0\} \end{aligned}$$

(3.1)

holds for some positive constants $C_1$ and $C_2$ (they may depend on $\varepsilon _0)$. Then $u\leqslant \varepsilon _0$ m-a.e. on X.

Remark 3.2

If X is a smooth Riemannian manifold, and if u is a $C^2$-function, then the Proposition 3.1 is a corollary of the classical maximum principle. In fact, if the assertion is false in this case, we assume that u achieves its maximum at point p, where $u(p)>\epsilon _0$. By using the maximum principle on $C^2$-functions, we have

$$\begin{aligned} \Delta u(p)\leqslant 0\quad \mathrm{and} \quad \Gamma (u)(p)=0. \end{aligned}$$

Hence, by (3.1), we have $u(p)\leqslant 0.$ This contradicts to $u(p)> \varepsilon _0.$

In the setting of metric measure spaces, we need a new argument.

Proof of Proposition 3.1

Since $u\in L^\infty (X,m)$, we have $\sup _{X} u<\infty$, where $\sup _Xu=\inf \{l: (u-l)_+=0, m\mathrm{-}a.e. \ \mathrm{in}\ X\}.$

Let us argue by contradiction. Suppose that $\varepsilon _0<\sup _Xu.$

Take any constant $k\in [\varepsilon _0,\sup _X u)$ and set $\phi _k=(u-k)^+$. Then $\phi _k \in \mathbb {V}$. Since the singular part $\Delta ^s u \geqslant 0$, we have

$$\begin{aligned} -\int _X \Gamma (u,\phi _k)\,\mathrm {d}m&=\int _X \widetilde{\phi _k} d\Delta ^*u\\&\geqslant \int _X \widetilde{\phi _k} \Delta ^{ab}u\,\mathrm {d}m \\&= \int _{\{x: u(x)\geqslant k\}} \phi _k \Delta ^{ab}u\,\mathrm {d}m \\&\overset{(3.1)}{\geqslant } -C_2\int _{X_k} \phi _k \sqrt{\Gamma (u)}\mathrm {d}m\\&\geqslant -C_2 \left( \int _{X_k} {\phi _k}^2\right) ^{\frac{1}{2}}\left( \int _{X_k} \Gamma (u)\right) ^{\frac{1}{2}}, \end{aligned}$$

where $X_k:=\{x: \Gamma (u)\not =0\}\cap \{x: u(x)> k\}.$

By the truncation property in [37] and $\Gamma (u,\phi _k) =\Gamma (u)=\Gamma (\phi _k)$ m-a.e. in $X_k$, we have

$$\begin{aligned} \int _X \Gamma (u,\phi _k)=\int _{X_k} \Gamma (u,\phi _k)=\int _{X_k} \Gamma (u)=\int _{X_k} \Gamma (\phi _k). \end{aligned}$$

The combination of the above two equations implies that

$$\begin{aligned} \int _{X_k}\Gamma (\phi _k) \leqslant C_2^2 \int _{X_k} {\phi _k}^2. \end{aligned}$$

(3.2)

Now we claim that there exists a constant $k_0\in [\varepsilon _0,\sup _X u)$ such that

$$\begin{aligned} m(\{x: u(x)<k_0\})>0. \end{aligned}$$

(3.3)

Suppose that (3.3) fails for any $k\in [\varepsilon _0,\sup _Xu)$. That is, $m(\{x: u(x)<k\})=0$ for any $k\in [\varepsilon _0,\sup _Xu).$ Letting k tend to $\sup _Xu$, we get $m(\{x: u(x)<\sup _Xu\})=0$. Thus $u=\sup _Xu$ m-a.e. in X. Now, we have $\Delta ^*u=0$ and $\Gamma (u)=0$ m-a.e. in X. This contradicts (3.1) and proves the claim.

Fix such a constant $k_0\in [\varepsilon _0,\sup _Xu)$ such that (3.3) holds. Denote $E=\{x: u(x)<k_0\}$. For all $k\in (k_0,\sup _Xu)$, we have $\phi _k=0$ m-a.e. in E. By applying Lemma 2.10, we conclude that

$$\begin{aligned} \Vert \phi _k\Vert _{L^{\nu }(X)}\leqslant \widetilde{C_S} \left( \int _X \Gamma (\phi _k)\mathrm {d}m\right) ^{\frac{1}{2}},\quad \ \forall k\in (k_0,\mathrm{sup}_Xu). \end{aligned}$$

(3.4)

We shall show that $m(X_k)>0$ for all $k\in (k_0,\mathrm{sup}_Xu)$. Fix any $k\in (k_0,\mathrm{sup}_Xu)$, the set $\{x: u(x)>k\}$ has positive measure, because $k<\sup _Xu$. Hence, $\Vert \phi _k\Vert _{L^{\nu }(X)}>0.$ By using (3.4), we get $m(\{x: \Gamma (\phi _k)\not =0\})>0.$ Note that

$$\begin{aligned}\Gamma (\phi _k)= {\left\{ \begin{array}{ll} \Gamma (u) &{} m\mathrm{-}a.e. \ \mathrm{in}\ \{x: u(x)>k\}\\ 0 &{} m\mathrm{-}a.e.\ \mathrm{in}\ \{x: u(x)\leqslant k\}, \end{array}\right. } \end{aligned}$$

we have $\{x: \Gamma (\phi _k)\not =0\}\subset X_k$ up to a zero measure set. Thus, we get $m(X_k)\geqslant m(\{x: \Gamma (\phi _k)\not =0\}) >0$.

On the other hand, we have

$$\begin{aligned} \Vert \phi _k\Vert _{L^2(X_k)}&\leqslant \Vert \phi _k\Vert _{L^{\nu }(X_k)}\cdot (m(X_k))^{1/2-1/{\nu }} \leqslant \Vert \phi _k\Vert _{L^{\nu }(X )}\cdot (m(X_k))^{1/2-1/{\nu }}\\&\leqslant \widetilde{C_S} \cdot \Big (\int _{X_k} \Gamma (\phi _k)\mathrm {d}m\Big )^{1/2} (m(X_k))^{1/2-1/{\nu }}\\&\overset{(3.2)}{\leqslant } \widetilde{C_S} \cdot C_2\cdot \Vert \phi _k\Vert _{L^2(X_k)} (m(X_k))^{1/2-1/{\nu }}, \end{aligned}$$

where we have used that $\{x: \Gamma (\phi _k)\not =0\}\subset X_k$ up to a zero measure set again. Note that $m(X_k)>0$, hence $\Vert \phi _k\Vert _{L^2(X_k)}\not =0$, for all $k\in (k_0,\mathrm{sup}_Xu)$, there is a constant $C>0$, such that $m(X_k)>C$ for all $k_0 \leqslant k< \sup _X u.$ Recall that $X_k=\{x: \Gamma (u)\not =0\}\cap \{x: u(x)> k\}$, by letting $k\rightarrow \sup _X u$, we have

$$\begin{aligned} m(\{x: \Gamma (u)\not =0\}\cap \{u=\sup _X u \}) \geqslant C. \end{aligned}$$

This contradicts the fact that $\Gamma (u)=0$ a.e. in $\{u=\sup _X u \}$ (see Proposition 2.22 of [11]), and proves the proposition. $\square$

Let us recall the one-dimensional model operators $L_{R,l}$ in [6]. Given $R\in \mathbb {R}$ and $l>1$, the one-dimensional models $L_{R,l}$ are defined as follows: let $L=R/(l-1)$,

• (1) If $R>0$, $L_{R,l}$ defined on $(-\pi /2\sqrt{L}, \pi /2\sqrt{L})$ by

  $$\begin{aligned} L_{R,l}v(x)=v''(x)-(l-1)\sqrt{L}\tan (\sqrt{L}x)v'(x); \end{aligned}$$

• (2) If $R<0$, $L_{R,l}$ defined on $(-\infty , \infty )$ by

  $$\begin{aligned} L_{R,l}v(x)=v''(x)-(l-1)\sqrt{-L}\tanh (\sqrt{-L}x)v'(x); \end{aligned}$$

• (3) If $R=0$, $L_{R,l}$ defined on $(-\infty , \infty )$ by

  $$\begin{aligned} L_{R,l}v(x)=v''(x). \end{aligned}$$

Next we will apply Corollary 2.7 to eigenfunctions and prove the following comparison theorem on the gradient of the eigenfunctions, which is an extension of Kröger’s comparison result in [24].

Theorem 3.3

Let (X, d, m) be a compact $RCD^*(K,N)$-space, and let $\lambda _1$ be the first eigenvalue on X. Let $l\in \mathbb {R}$ and $l\geqslant N$, and let f be an eigenfunction with respect to $\lambda _1$. Suppose $\lambda _1>\max \left\{ 0,\frac{lK}{l-1}\right\} .$ Let v be a Neumann eigenfunction of $L_{K,l}$ with respect to the same eigenvalue $\lambda _1$ on some interval. If $[\min f, \max f] \subset [\min v, \max v]$, then

$$\begin{aligned} \Gamma (f) \leqslant (v'\circ v^{-1})^2(f)\text { m-a.e.}. \end{aligned}$$

Proof

Without loss of generality, we may assume that $[\min f,\max f] \subset (\min v, \max v)$.

Denote by T(x) the function such that

$$\begin{aligned} L_{K,l}(v)=v''-Tv. \end{aligned}$$

As in Corollary 3 in Section 4 of [6], we can choose a smooth bounded function $h_1$ on $[\min f,\max f]$ such that

$$\begin{aligned} h'_1<\min \{Q_1(h_1),Q_2(h_1)\}, \end{aligned}$$

where $Q_1, Q_2$ are given by following

$$\begin{aligned} Q_1(h_1):=-(h_1-T)\left( h_1-\frac{2l}{l-1}T+\frac{2\lambda _1 v}{v'}\right) , \end{aligned}$$

$$\begin{aligned} Q_2(h_1):=-h_1\left( \frac{l-2}{2(l-1)}h_1-T+\frac{\lambda _1 v}{v'}\right) . \end{aligned}$$

We can then take a smooth function g on $[\min f,\max f]$, $g\leqslant 0$ and $g'=-\frac{h_1}{v'}\circ v^{-1}.$

According to [3, Theorem6.5] (see also [20, Theorem1.1]), we have that f is Lipschitz continuous. Notice that $\Delta f=-\lambda _1f\in \mathbb V$. Hence $f\in D_{\mathbb {V}}(\Delta )\cap Lip(X)$.

Now define a function F on X by

$$\begin{aligned} \psi (f)F=\Gamma (f)-\phi (f), \end{aligned}$$

where $\psi (f):=e^{-g(f)}$ and $\phi (f)~:=(v'\circ v^{-1})^2(f)$. Since $f\in D_{\mathbb {V}}(\Delta )\cap Lip(X)$, by Theorem 2.6, we have $\Gamma (f)\in \mathbb M_\infty$. According to Lemma 2.4 and Remark 2.5, we have $\psi (f), \phi (f) \in D(\Delta )\cap Lip(X)$ and $F\in \mathbb M_\infty$. Moreover

$$\begin{aligned} \Delta ^*F=\frac{1}{\psi }\Delta ^*\Gamma +\frac{1}{\psi }\big (-2\Gamma (\psi ,F)-\Delta \psi F-\Delta \phi \big )\cdot m, \end{aligned}$$

where and in the sequel, we denote by $\Gamma =\Gamma (f)$ and $\phi =\phi (f),\psi =\psi (f).$ By using Theorem 2.6 again, we have $\Delta ^sF\geqslant 0$ on X and

$$\begin{aligned} \Delta ^{ab}F=\frac{1}{\psi }\left( \Delta ^{ab}\Gamma -2\Gamma (\psi ,F)-\Delta \psi F-\Delta \phi \right) \quad m\mathrm{-}a.e.\ \mathrm{in}\ X. \end{aligned}$$

Since $l\geqslant N$, the (X, d, m) satisfies also $RCD^*(K,l)$ condition. Applying inequality (2.2) to f and using $\Delta f =-\lambda _1 f$, we have, for m-a.e. $x\in \{x: \Gamma (x)>0\}$,

$$\begin{aligned} \Delta ^{ab}\Gamma \geqslant -2\lambda _1\Gamma +\frac{2{\lambda _1}^2}{l}f^2+2K\Gamma +\frac{2l}{l-1}\left( \frac{\lambda _1f}{l}+\frac{\Gamma (f,\Gamma )}{2\Gamma }\right) ^2. \end{aligned}$$

Fix arbitrarily a constant $\epsilon _0>0$. We want to show $F \leqslant \epsilon _0$ m-a.e. in X.

Since $F\leqslant e^g\cdot \Gamma \leqslant \Gamma$, we have $\{x: F(x)\geqslant {\epsilon _0}\}\subset \{x: \Gamma (x)>0\}$. Following the argument from line 29 on page 1182 to line 10 on page 1183 of [32], we get:

$$\begin{aligned} \Delta ^{ab}F \geqslant \psi T_1\cdot F^2+ T_2\cdot F+T_3 \Gamma (f,F) \quad m\mathrm{-}a.e. \text { on } \{x: F(x)\geqslant \epsilon _0\}, \end{aligned}$$

(3.5)

where

$$\begin{aligned} {v'}^2T_1=Q_2(h_1)-h'_1, \quad T_2=Q_1(h_1)-h'_1, \end{aligned}$$

and

$$\begin{aligned} T_3=\frac{2l}{l-1}\left( -\frac{g'}{2}+\frac{1}{2\Gamma }\left( \frac{2\lambda _1f}{l}+\phi '+\phi g'\right) \right) +2g'. \end{aligned}$$

Note that both $T_1$ and $T_2$ are positive, $\Gamma$ is bounded on X and $T_3$ is bounded on $\{x: F(x)\geqslant \epsilon _0\}$. It follows from (3.5) that

$$\begin{aligned} \Delta ^{ab}F \geqslant c_1\cdot F-c_2\cdot \sqrt{\Gamma (F)} \quad a.e. \text { on } \{x: F(x)\geqslant \epsilon _0\} \end{aligned}$$

(3.6)

for some constant $c_2>0$ and $c_1=\min _{s\in [\min f,\max f]} T_2(s)>0$. By combining with $\Delta ^sF\geqslant 0$ on X and Proposition 3.1, we conclude that $F \leqslant \epsilon _0$ m-a.e. in X.

At last, by the arbitrariness of $\epsilon _0$, we have $F \leqslant 0$ m-a.e. in X. This completes the proof of Theorem 3.3. $\square$

Let $v_{R,l}$ be the solution of the equation

$$\begin{aligned} L_{R,l}v=-\lambda _1 v \end{aligned}$$

with initial value $v(a)=-1$ and $v'(a)=0$, where

$$\begin{aligned} a= \left\{ \begin{array}{ll} -\frac{\pi }{2\sqrt{R/(l-1)}}&{} \qquad \text { if }R>0,\\ 0&{}\qquad \text { if } R \leqslant 0. \end{array} \right. \end{aligned}$$

We denote

$$\begin{aligned} b= \inf \{x>a:~v'_{R,l}(x)=0 \} \end{aligned}$$

and

$$\begin{aligned} m_{R,l}=v_{R,l}(b) \end{aligned}$$

Note that $v_{R,l}$ is non-decreasing on [a, b].

Next we show the following comparison theorem on the maximum of eigenfunctions.

Theorem 3.4

Let (X, d, m) be a compact $RCD^*(K,N)$-space, and let f be an eigenfunction with respect to the first eigenvalue $\lambda _1$ on X. Suppose $\min f=-1, \max f \leqslant 1$. Then we have

$$\begin{aligned} \max f \geqslant m_{K,N}. \end{aligned}$$

Proof

We argue by contradiction. Suppose $\max f<m_{K,N}$. Since $m_{K,l}$ is continuous on l, we can find some real number $l>N$ such that

$$\begin{aligned} \max f \leqslant m_{K,l}\text { and }\lambda _1>\max \{0,\frac{lK}{l-1}\}. \end{aligned}$$

Then following the proof of Proposition 5 in [6], we obtain that the ratio

$$\begin{aligned} R(s)=-\frac{\int _X f 1_{\{f\leqslant v(s)\}}\mathrm {d}m}{\rho (s)v'(s)} \end{aligned}$$

is increasing on $[a,v^{-1}(0)]$ and decreasing on $[v^{-1}(0),b]$, where the function $\rho$ is

$$\begin{aligned} \rho (s)~:=\left\{ \begin{array}{ll} \cos ^{l-1}(\sqrt{L}s)&{} \quad \text { if } L>0\\ s^{l-1}&{} \quad \text { if } L=0\\ \sinh ^{l-1}(\sqrt{-L}s)&{} \quad \text { if } L<0 \end{array} \right. \end{aligned}$$

and $L=K/(l-1)$. It follows that for any $s\in [a,v^{-1}(-1/2)]$, since $v(s)\leqslant -\frac{1}{2}$, we have

$$\begin{aligned} m(\{f\leqslant v(s)\})\leqslant -2\int _X f 1_{\{f\leqslant v(s)\}} \mathrm {d}m \leqslant 2C \rho (s) v'(s), \end{aligned}$$

(3.7)

where $C=R(v^{-1}(0))$.

Take $p \in X$ with $f(p)=-1$. By

$$\begin{aligned} f-f(p) \geqslant 0,\quad \text { and } \quad \Delta (f-f(p))=-\lambda _1 f \leqslant \lambda _1 \end{aligned}$$

The mean value inequality (2.4) implies that

$$\begin{aligned} \int \!\!\!\!\!\!-_{B_r(p)} (f-f(p)) \mathrm {d}m \leqslant C \lambda _1 r^2 \end{aligned}$$

for all $r>0$ such that $B_r(x)\subset X$. Denote $C_1= C \lambda _1$. Let $A(r)=\{f-f(p)>2C_1 r^2\} \cap B_p(r)$. Then

$$\begin{aligned} \frac{m (A(r))}{m (B_p(r))} \leqslant \frac{\int _{B_r(p)}(f-f(p)) \mathrm {d}m}{2C_1r^2 m(B_r(p))} \leqslant \frac{1}{2}. \end{aligned}$$

Hence

$$\begin{aligned} \frac{1}{2} m(B_r(p))&\leqslant m(B_r(p)\backslash A(r)) \\&\leqslant m(\{f-f(p) \leqslant 2C_1 r^2\})\\&= m(\{f \leqslant -1+ 2C_1 r^2\}). \end{aligned}$$

By using (3.7) and following the argument from line 1 on page 1186 to line 3 on page 1187 of [32], one can get that there exists a constant $C_2>0$ such that

$$\begin{aligned} m(B_p(r)) \leqslant C_2 r^l \end{aligned}$$

for all sufficiently small $r>0$.

Fix $r_0>0$. By Bishop-Gromov inequality (2.3), we have

$$\begin{aligned} m(B_p(r)) \geqslant \frac{m(B_p(r_0))}{\int _0^{r_0} \mathfrak {s}_{\frac{K}{N-1}}(t)^{N-1} \mathrm {d}t}\int _0^r \mathfrak {s}_{\frac{K}{N-1}}(t)^{N-1} \mathrm {d}t \geqslant C_3 r^N \end{aligned}$$

for any $0<r<r_0$. The combination of the above two inequalities implies that $C_2 r^{l-N} \geqslant C_3$ holds for any sufficiently small r. Hence, we have $l \leqslant N$, which contradicts to the assumption $l >N$. Therefore, the Proof of Theorem 3.4 is finished. $\square$

Now we are in the position to prove the main result—Theorem 1.2.

Proof of Theorem 1.2

Let $\lambda _1$ and f denote respectively the first non-zero eigenvalue and a corresponding eigenfunction with $\min f=-1$ and $\max f \leqslant 1$. By Theorem 4.22 of [16], we have $\lambda _1 \geqslant NK/(N-1)$ if $K>0$ and $N>1$. Now fix any $R<K$, we have

$$\begin{aligned} \lambda _1>\max \left\{ \frac{NR}{N-1},0\right\} . \end{aligned}$$

Then we may use the results of Sections 3 and 6 of [6], we can find an interval [a, b] such that the one-dimensional model operator $L_{R,N}$ has the first Neumann eigenvalue $\lambda _1$ and a corresponding eigenfunction v with $v(a)=\min v=-1$ and $v(b)=\max v=\max f.$ By Theorem 13 in Section 7 of [6], we have

$$\begin{aligned} \lambda _1 \geqslant \hat{\lambda }(R,N,b-a), \end{aligned}$$

(3.8)

where $\hat{\lambda }(R,N,b-a)$ is the first non-zero Neumann eigenvalue of $L_{R,N}$ on the symmetric interval $(-\frac{b-a}{2}, \frac{b-a}{2})$. Note that f is continuous, we take two points x and y in X such that $f(x)=-1$ and $f(y)=\max f$. Let $g=v^{-1}\circ f$, then $g(x)=a$, $g(y)=b$ and, by Theorem 3.3, $\Gamma (g) \leqslant 1$ m-a.e. in X. Hence, we have

$$\begin{aligned} b-a=g(y)-g(x) \leqslant d(x,y) \leqslant \max _{z_1,z_2 \in X}d(z_1,z_2):=d, \end{aligned}$$

where d is the diameter of X. Together with (3.8) and the fact that the function $\hat{\lambda }(R,N,s)$ decreases with s, we conclude

$$\begin{aligned} \lambda _1 \geqslant \hat{\lambda }(R,N,d). \end{aligned}$$

By the arbitrariness of R, we finally prove the theorem. $\square$