Global Lower Bounds on the First Eigenvalue for a Diffusion Operator

Abstract

We derive global lower bounds for the first eigenvalue of a symmetric diffusion $\Delta _X:=\Delta -\nabla _X$ on Riemannian manifolds with the Bakry–Émery–Ricci curvature bounded from below.

1 Introduction

Let $(M^n,g)$ be an n-dimensional Riemannian manifold and X be a smooth vector field on $M^n$. The diffusion operator

$$\begin{aligned} \Delta _X:=\Delta +\nabla _X \end{aligned}$$

(1.1)

is an important generalization of the Laplacian operator $\Delta$, in particular, the Witten–Laplacian

$$\begin{aligned} \Delta _f:=\Delta -\nabla _{\nabla f} \end{aligned}$$

(1.2)

is a special case of (1.1) by taking $X=-\nabla f$ for some $f\in C^\infty (M^n)$.

As in [1, 3, 9], the m-dimensional Bakry–Émery–Ricci curvature of the diffusion operator $\Delta _X$ is defined as

$$\begin{aligned} Ric_X^m:=Ric-\frac{1}{2}L_Xg-\frac{X\otimes X}{m-n} \end{aligned}$$

(1.3)

for any number $m\in (n,\infty )$, where $L_X$ stands for the Lie derivative along the direction X. In particular, the m-dimensional Bakry–Émery–Ricci curvature of the Witten–Laplacian operator $\Delta _f$ is defined as

$$\begin{aligned} Ric_f^m:=Ric+Hess f-\frac{d f\otimes d f}{m-n}, \end{aligned}$$

(1.4)

where Hessf is the Hessian of f.

For m-dimensional Bakry–Émery–Ricci curvatures, we can allow m to be infinite:

$$\begin{aligned} Ric_X^\infty :=Ric-\frac{1}{2}L_Xg, \end{aligned}$$

(1.5)

and

$$\begin{aligned} Ric_f^\infty :=Ric+Hess f, \end{aligned}$$

(1.6)

which are called the $\infty$-dimensional Bakry–Émery–Ricci curvature of the diffusion operator $\Delta _X$ (of the Witten–Laplacian operator), respectively. We refer the readers to [5, 6, 11, 12, 15, 16] for applications of Bakry–Émery–Ricci curvatures.

There are several well-known results on lower bound estimates for the first eigenvalue of Laplacian operator on closed Riemannian manifolds (see Section 5 of [7] for a summary): Lichnerowicz [10] (see also [14]) showed that the first nonzero eigenvalue of the Laplacian on a closed manifold must satisfy $\lambda _1\ge mK$ if the Ricci curvature is bounded from below by $(m-1)K$. When the Ricci curvature is nonnegative, Li–Yau [8] proved that $\lambda _1\ge \frac{\pi ^2}{(1+a)d^2}$, where $0\le a<1$ is a constant and d is the diameter of the underlying closed manifold. More generally, they [8] also derived a lower bound estimate that depends on the lower bound of the Ricci curvature, the upper bound of the diameter, and the dimension of the manifold alone. In this note, we will prove that these results are still valid for the first eigenvalue of the diffusion operator $\Delta _X$ on the closed manifold $M^n$ under the condition of the Barky–Émery–Ricci curvature bounded from below.

Wu [17, 18] established upper bounded first nontrivial eigenvalue for the Witten–Laplacian under the condition that the m-dimensional ($\infty$-dimensional) Barky–Émery–Ricci curvature bounded below, respectively. We will consider global lower bounds in the case of the diffusion operator (1.1) via the m-dimensional ($\infty$-dimensional) Bakry–Émery–Ricci curvature of (1.1).

The main theorems only consider m-dimensional Bakry–Émery–Ricci curvatures since the $\infty$-dimensional cases can be obtained by similar proof. On the closed Riemannian manifold $(M^n,g)$, we derive a series lower bounds for the first eigenvalue of the diffusion operator $\Delta _X$.

Theorem 1.1

Let $(M^n,g)$ be an n-dimensional closed manifold with $Ric_X^m\ge -(m-1)Kg$ for some constant $K>0$. Then, the first nonzero eigenvalue of the diffusion operator $\Delta _X$ on $M^n$ must satisfy

$$\begin{aligned} \lambda _1\ge mK. \end{aligned}$$

Theorem 1.2

Let $(M^n,g)$ be an n-dimensional closed manifold with $Ric_X^m\ge 0$ and d be the diameter of $M^n$. Then,

$$\begin{aligned} \lambda _1\ge \frac{\pi ^2}{(1-a)d^2} \end{aligned}$$

for some constant $0\le a<1$.

Theorem 1.3

Let $(M^n,g)$ be an n-dimensional closed manifold with $Ric_f^m\ge -(m-1)Kg$ for some constant $K>0$ and d be the diameter of $M^n$. Then, there exist positive constants $A_1(m)$ and $A_2(m)$ depending only on m so that the first nonzero eigenvalue of $\Delta _f$ satisfies

$$\begin{aligned} \lambda _1\ge \frac{A_1}{d^2}\exp \{-A_2d\sqrt{K}\}. \end{aligned}$$

We also prove global lower bound estimate on the first eigenvalue for the diffusion operator $\Delta _X$ on a complete noncompact Riemannian manifold.

Theorem 1.4

Let $(M^n,g)$ be an n-dimensional complete noncompact Riemannian manifold with $Ric_X^m\ge -(m-1)Kg$ for some constant $K\ge 0$. Then, there is a global lower bound estimate on the first eigenvalue for the diffusion operator $\Delta _X$

$$\begin{aligned} \lambda _1\ge -\frac{(m-1)^4K^2}{4}. \end{aligned}$$

(1.7)

In the rest of this paper, Theorems 1.1–1.3 are proved in Sect. 2, while Theorem 1.4 is established in Sect. 3.

2 The Closed Case

In this section, we will prove Theorems 1.1–1.3 of lower bound estimates for the first eigenvalue of the diffusion operator $\Delta _X$ on the closed manifold $M^n$, which are generalizations of [8, 10, 14].

Theorem 2.1

Let $(M^n,g)$ be an n-dimensional closed manifold with $Ric_X^m\ge -(m-1)Kg$ for some constant $K>0$. Then, the first nonzero eigenvalue of the diffusion operator $\Delta _X$ on $M^n$ must satisfy

$$\begin{aligned} \lambda _1\ge mK. \end{aligned}$$

(2.1)

Proof

This result generalizes the lower bound result by Lichnerowicz [10] (see also Obata [14]). Let u be a nonconstant eigenfunction satisfying

$$\begin{aligned} \Delta _Xu=\lambda u \end{aligned}$$

(2.2)

with $\lambda >0$.

Consider the function $A=|\nabla u|^2-\frac{\lambda }{m}u^2$. A direct computation implies that

$$\begin{aligned} \frac{1}{2}\Delta _XA= & {} \nabla _i\left( \nabla _i\nabla _ju\nabla _ju -\frac{\lambda }{m}u\nabla _i u\right) +X^i\left( \nabla _i\nabla _ju\nabla _ju -\frac{\lambda }{m}u\nabla _i u\right) \nonumber \\= & {} |Hess u|^2+\langle \Delta \nabla u,\nabla u\rangle -\frac{\lambda }{m}|\nabla u|^2-\frac{\lambda }{m}u\Delta u\nonumber \\&+X^i\nabla _i\nabla _ju\nabla _ju-\frac{\lambda }{m}u\nabla _X u\nonumber \\= & {} |Hess u|^2+\langle \nabla \Delta u,\nabla u\rangle +Ric(\nabla u,\nabla u)\nonumber \\&+X^i\nabla _i\nabla _ju\nabla _ju-\frac{\lambda }{m}|\nabla u|^2 -\frac{\lambda }{m}u\Delta _X u\nonumber \\= & {} |Hess u|^2+\langle \nabla \Delta _X u,\nabla u\rangle +Ric_X^m(\nabla u,\nabla u)\nonumber \\&+\frac{(Xu)^2}{m-n}-\frac{\lambda }{m}|\nabla u|^2-\frac{\lambda }{m}u\Delta _X u, \end{aligned}$$

(2.3)

where Hessu is the Hessian of u, and we used the second Bianchi identity in the third equality.

Since $m>n$, we have

$$\begin{aligned} 0\le & {} \left( \sqrt{\frac{m-n}{mn}}\Delta u-\sqrt{\frac{n}{m(m-n)}}\nabla _Xu\right) ^2\nonumber \\\le & {} \frac{(\Delta u)^2}{n}-\frac{1}{m}[(\Delta u)^2+2\Delta u\nabla _Xu+(\nabla _Xu)^2]+\frac{(\nabla _Xu)^2}{m-n}\nonumber \\\le & {} |Hess u|^2-\frac{1}{m}(\Delta _Xu)^2+\frac{(\nabla _Xu)^2}{m-n}, \end{aligned}$$

i.e.,

$$\begin{aligned} |Hess u|^2+\frac{(\nabla _Xu)^2}{m-n}\ge \frac{1}{m}(\Delta _Xu)^2. \end{aligned}$$

(2.4)

Plugging the fact of $Ric_X^m\ge -(m-1)Kg$, (2.2) and (2.4) into (2.3), we get

$$\begin{aligned} \frac{1}{2}\Delta _XA\ge & {} \frac{\lambda ^2u^2}{m}+\lambda |\nabla u|^2-(m-1)K|\nabla u|^2-\frac{\lambda }{m}|\nabla u|^2-\frac{\lambda ^2u^2}{m}\nonumber \\= & {} (m-1)\bigg (\frac{\lambda }{m}-K\bigg )|\nabla u|^2. \end{aligned}$$

(2.5)

If $\lambda \le mK$, then $\Delta _XA\le 0$ on $M^n$. By the compactness of $M^n$ and the strong maximum principle, A must be identically constant. In particular, the right hand side of (2.5) must be identically 0, i.e., $\lambda \equiv mK$ since u is nonconstant.

To conclude, the first nonzero eigenvalue of the diffusion operator $\Delta _X$ on $M^n$ is no less than mK. $\square$

When the $\infty$-dimensional Bakry–Émery–Ricci curvature of the diffusion operator $\Delta _X$ is bounded below, we have

Corollary 2.2

Let $(M^n,g)$ be an n-dimensional closed manifold with $Ric_X^\infty \ge -nKg$ for some constant $K>0$ and $|X|^2\le \alpha$ for some constant $\alpha \ge 0$ . Then, the first nonzero eigenvalue of the diffusion operator $\Delta _X$ on $M^n$ must satisfy

$$\begin{aligned} \lambda _1\ge (n+1)\left( K+\frac{\alpha }{n}\right) . \end{aligned}$$

(2.6)

Proof

Note that $(Xh)^2\le \alpha |\nabla h|^2$. Taking $m=n+1$ in the proof of Theorem 2.1, (2.5) becomes

$$\begin{aligned} \frac{1}{2}\Delta _XA\ge & {} \frac{\lambda ^2u^2}{n+1}+\lambda |\nabla u|^2-(nK+\alpha )|\nabla u|^2 -\frac{\lambda }{n+1}|\nabla u|^2-\frac{\lambda ^2u^2}{n+1}\nonumber \\= & {} n\big (\frac{\lambda }{n+1}-K-\frac{\alpha }{n}\big )|\nabla u|^2. \end{aligned}$$

Similarly, $\lambda \equiv (n+1)(K+\frac{\alpha }{n})$ if $\lambda \le (n+1)(K+\frac{\alpha }{n})$. To conclude, the first nonzero eigenvalue of the diffusion operator $\Delta _X$ on $M^n$ is no less than $(n+1)(K+\frac{\alpha }{n})$. $\square$

Then, we generalize two lower bound results by Li–Yau [8]. Let $\lambda _1$ be the least nontrivial eigenvalue of the diffusion operator $\Delta _X$ on the closed manifold $M^n$ and let u be the corresponding eigenfunction. By multiplying with a constant, it is possible to exist a positive constant $a\in [0,1)$ so that

$$\begin{aligned} a=\inf _{M^n}u+1=\sup _{M^n}u-1. \end{aligned}$$

Theorem 2.3

Let $(M^n,g)$ be an n-dimensional closed manifold with $Ric_X^m\ge 0$ and d be the diameter of $M^n$. Then,

$$\begin{aligned} \lambda _1\ge \frac{\pi ^2}{(1-a)d^2}. \end{aligned}$$

(2.7)

Proof

Set $\lambda =\lambda _1$ and $v=u-a$. Then, the equation becomes

$$\begin{aligned} \Delta _Xv=\lambda (v+a). \end{aligned}$$

Let $B:=|\nabla v|^2-cv^2$ with $c=\lambda (1-a)>0$ and $x_0$ be a maximum point of B. Choose a frame $\{e_i\}_{i=1}^n$ so that $v_1(x_0)=\nabla _{e_1}v(x_0)=|\nabla v(x_0)|$ if $|\nabla v(x_0)|\ne 0$. Note that

$$\begin{aligned} \frac{1}{2}\nabla _iB=\nabla _i\nabla _jv\nabla _jv-cv\nabla _iv, \end{aligned}$$

(2.8)

so at $x_0$

$$\begin{aligned} 0=v_1(v_{11}-cv), \end{aligned}$$

(2.9)

and the Hessian of v satisfies

$$\begin{aligned} |Hess v|^2\ge v^2_{11}=c^2v^2. \end{aligned}$$

(2.10)

Covariant differentiating (2.8) with respect to $e_i$ again, then using the Bochner formula and (2.10), we can get at $x_0$

$$\begin{aligned} 0\ge & {} \frac{1}{2}\Delta _X B\nonumber \\= & {} |Hess v|^2+\nabla _i\nabla _i\nabla _jv\nabla _jv-c|\nabla v|^2-cv\Delta v\nonumber \\&+X^i(\nabla _i\nabla _jv\nabla _jv-cv\nabla _iv)\nonumber \\= & {} |Hess v|^2+\langle \nabla v,\nabla \Delta v\rangle +Ric(\nabla v,\nabla v)\nonumber \\&-c|\nabla v|^2-cv\Delta _X v+X^i\nabla _i\nabla _jv\nabla _jv\nonumber \\\ge & {} c^2v^2+\langle \nabla v,\nabla \Delta _Xv\rangle +Ric_X^m(\nabla v,\nabla v)\nonumber \\&+\frac{(Xv)^2}{m-n}-cv_1^2-c\lambda v(v+a)\nonumber \\\ge & {} c^2v^2+\lambda v_1^2-cv_1^2-c\lambda v(v+a)\nonumber \\= & {} (\lambda -c)(v_1^2-cv^2)-ac\lambda v\nonumber \\\ge & {} a\lambda B(x_0)-ac\lambda . \end{aligned}$$

(2.11)

Hence, for all $x\in M^n$,

$$\begin{aligned} |\nabla v(x)|^2+cv^2(x)=B(x)\le B(x_0)\le c, \end{aligned}$$

i.e.,

$$\begin{aligned} |\nabla v(x)|^2\le \lambda (1-a)(1-v^2(x)). \end{aligned}$$

(2.12)

Also (2.12) is trivially satisfied if $|\nabla v(x_0)|=0$. Let $\gamma$ be the shortest geodesic from the minimizing point of v to the maximizing point. The length of $\gamma$ is at most d. Integrating the gradient estimate (2.12) along this segment with respect to arclength, we obtain

$$\begin{aligned} \sqrt{\lambda (1-a)}\cdot d\ge \sqrt{\lambda (1-a)}\int _\gamma \hbox {d}s\ge \int _\gamma \frac{|\nabla v|\hbox {d}s}{\sqrt{1-v^2}}\ge \int _{-1}^1\frac{\hbox {d}u}{\sqrt{1-v^2}}=\pi . \end{aligned}$$

(2.7) follows immediately. $\square$

Trivially, Theorem 2.3 holds for $Ric_X^\infty \ge 0$:

Corollary 2.4

Let $(M^n,g)$ be an n-dimensional closed manifold with $Ric_X^\infty \ge 0$ and d be the diameter of $M^n$. Then,

$$\begin{aligned} \lambda _1\ge \frac{\pi ^2}{(1-a)d^2}. \end{aligned}$$

(2.13)

Under the condition of the m-dimensional Bakry–Émery–Ricci curvature of the Witten–Laplacian operator bounded below, we have

Theorem 2.5

Let $(M^n,g)$ be an n-dimensional closed manifold with $Ric_f^m\ge -(m-1)Kg$ for some constant $K>0$ and d be the diameter of $M^n$. Then, there exist positive constants $A_1(m)$ and $A_2(m)$ depending only on m so that the first nonzero eigenvalue of $\Delta _f$ satisfies

$$\begin{aligned} \lambda _1\ge \frac{A_1}{d^2}\exp \{-A_2d\sqrt{K}\}. \end{aligned}$$

Proof

Let u be a nonconstant eigenfunction satisfying

$$\begin{aligned} \Delta _fu=\lambda u. \end{aligned}$$

By the fact that

$$\begin{aligned} \lambda \int _{M^n}ue^{-f}=\int _{M^n}\Delta _fue^{-f}=0, \end{aligned}$$

u must change sign. Hence, we may normalize u to satisfy $\min u=-1$ and $\max u\le 1$. Let us consider the function

$$\begin{aligned} w=\log (a+u) \end{aligned}$$

for some constant $a>1$. The function w satisfies

$$\begin{aligned} \Delta _fw= & {} \frac{\Delta _fu}{a+u}-\frac{|\nabla u|^2}{(a+u)^2}\nonumber \\= & {} \frac{\lambda u}{a+u}-|\nabla w|^2. \end{aligned}$$

(2.14)

Calculating directly, we get

$$\begin{aligned} \frac{1}{2}\Delta _f|\nabla w|^2= & {} |Hess w|^2+\nabla _i\nabla _i\nabla _jw\nabla _jw-\nabla _i\nabla _jw\nabla _jw\nabla _if\nonumber \\= & {} |Hess w|^2+\langle \nabla w,\nabla \Delta w\rangle +Ric(\nabla w,\nabla w)-\nabla _i\nabla _jw\nabla _jw\nabla _if\nonumber \\= & {} |Hess w|^2+\langle \nabla w,\nabla \Delta _f w\rangle +Ric_f^m(\nabla w,\nabla w)+\frac{\langle \nabla f,\nabla w\rangle ^2}{m-n}\nonumber \\= & {} |Hess w|^2+\left\langle \nabla w,\nabla \left( \frac{\lambda u}{a+u}-|\nabla w|^2\right) \right\rangle +Ric_f^m(\nabla w,\nabla w)\nonumber \\&+\frac{\langle \nabla f,\nabla w\rangle ^2}{m-n}\nonumber \\\ge & {} |Hess w|^2+\frac{a\lambda }{a+u}|\nabla w|^2-\langle \nabla w,\nabla |\nabla w|^2\rangle -(m-1)K|\nabla w|^2\nonumber \\&+\frac{\langle \nabla f,\nabla w\rangle ^2}{m-n},\nonumber \\ \end{aligned}$$

(2.15)

where Hessw is the Hessian of w, and we used the Bochner formula in the second equality and (2.14) in the fourth.

Note that

$$\begin{aligned} 0\le & {} \left( \sqrt{\frac{m-n}{mn}}\Delta w+\sqrt{\frac{n}{m(m-n)}}\langle \nabla w,\nabla f\rangle \right) ^2\\= & {} \left( \frac{1}{n}-\frac{1}{m}\right) (\Delta w)^2+\frac{2}{m}\Delta w\langle \nabla w,\nabla f\rangle +\left( \frac{1}{m-n}-\frac{1}{m}\right) \langle \nabla w,\nabla f\rangle ^2\\\le & {} |Hess w|^2-\frac{1}{m}\big ((\Delta w)^2-2\Delta w\langle \nabla w,\nabla f\rangle +\langle \nabla w,\nabla f\rangle ^2\big )+\frac{\langle \nabla w,\nabla f\rangle ^2}{m-n}\\= & {} |Hess w|^2-\frac{(\Delta _fw)^2}{m}+\frac{\langle \nabla w,\nabla f\rangle ^2}{m-n}\nonumber \\= & {} |Hess w|^2-\frac{1}{m}\left( \frac{\lambda u}{a+u}-|\nabla w|^2\right) ^2+\frac{\langle \nabla w,\nabla f\rangle ^2}{m-n}\nonumber \\\le & {} |Hess w|^2-\frac{1}{m}\left( |\nabla w|^4-\frac{2\lambda u}{a+u}|\nabla w|^2\right) +\frac{\langle \nabla w,\nabla f\rangle ^2}{m-n}, \end{aligned}$$

where we used (2.14) in the third equality. Therefore, we have

$$\begin{aligned} |Hess w|^2+\frac{\langle \nabla w,\nabla f\rangle ^2}{m-n}\ge \frac{1}{m}\left( |\nabla w|^4-\frac{2\lambda u}{a+u}|\nabla w|^2\right) . \end{aligned}$$

(2.16)

Applying (2.16)–(2.15), we obtain

$$\begin{aligned} \frac{1}{2}\Delta _f|\nabla w|^2\ge & {} \frac{1}{m}\left( |\nabla w|^4-\frac{2\lambda u}{a+u}|\nabla w|^2\right) +\frac{a\lambda }{a+u}|\nabla w|^2\nonumber \\&-\langle \nabla w,\nabla |\nabla w|^2\rangle -(m-1)K|\nabla w|^2. \end{aligned}$$

(2.17)

If $x_1\in M^n$ is a point where $|\nabla w|^2$ achieves its maximum, the maximum implies that at such point

$$\begin{aligned} 0\ge \frac{1}{m}|\nabla w|^4-\left( \frac{2\lambda u}{m(a+u)}-\frac{a\lambda }{a+u}+(m-1)K\right) |\nabla w|^2, \end{aligned}$$

(2.18)

i.e.,

$$\begin{aligned} |\nabla w|^2(x)\le & {} |\nabla w|^2(x_1)\nonumber \\\le & {} \frac{2\lambda u(x_1)}{a+u(x_1)}-\frac{ma\lambda }{a+u(x_1)}+m(m-1)K\nonumber \\\le & {} \frac{2\lambda }{a-1}+m(m-1)K \end{aligned}$$

(2.19)

for all $x\in M^n$. Integrating $|\nabla w|=|\nabla \log (a+u)|$ along a minimal geodesic $\gamma$ joining the points at which $u=-1$ and $u=\max u$, we have

$$\begin{aligned} \log \left( \frac{a}{a-1}\right)\le & {} \log \left( \frac{a+\max u}{a-1}\right) \nonumber \\\le & {} \int _\gamma |\nabla \log (a+u)|\nonumber \\\le & {} d\left( \frac{2\lambda }{a-1}+m(m-1)K\right) ^\frac{1}{2} \end{aligned}$$

(2.20)

for all $a>1$. Setting $t=\frac{a-1}{a}$, we have

$$\begin{aligned} 2\lambda \ge \frac{2}{a}\lambda \ge t\left( \frac{1}{d^2}\left( \log \frac{1}{t}\right) ^2-m(m-1)K\right) \end{aligned}$$

(2.21)

for all $0<t<1$. Maximizing the right hand side as a function of t by setting

$$\begin{aligned} t=\exp (-1-\sqrt{1+m(m-1)Kd^2}), \end{aligned}$$

we obtain the estimate

$$\begin{aligned} \lambda\ge & {} \frac{1}{d^2}(1+\sqrt{1+m(m-1)Kd^2})\exp (-1-\sqrt{1+m(m-1)Kd^2})\nonumber \\\ge & {} \frac{A_1}{d^2}\exp \{-A_2d\sqrt{K}\} \end{aligned}$$

(2.22)

as claimed. $\square$

As we obtained Theorem 2.5, we can derive the following result by similar arguments.

Corollary 2.6

Let $(M^n,g)$ be an n-dimensional closed manifold with $Ric_f^\infty \ge -nKg$ for some constant $K>0$, $|X|^2\le \alpha$ for some constant $\alpha \ge 0$ and d be the diameter of $M^n$. Then, there exist positive constants $A_3(n)$ and $A_4(n)$ depending only on m so that the first nonzero eigenvalue of $\Delta _f$ satisfies

$$\begin{aligned} \lambda _1\ge \frac{A_3}{d^2}\exp \{-A_4d\sqrt{K+\alpha }\}. \end{aligned}$$

Proof

Note that $(Xh)^2\le \alpha |\nabla h|^2$. Taking $m=n+1$ in the proof of Theorem 2.1, (2.17) becomes

$$\begin{aligned} \frac{1}{2}\Delta _f|\nabla w|^2\ge & {} \frac{1}{n+1}\left( |\nabla w|^4-\frac{2\lambda u}{a+u}|\nabla w|^2\right) \nonumber \\&+\frac{a\lambda }{a+u}|\nabla w|^2-\langle \nabla w,\nabla |\nabla w|^2\rangle -(nK+\alpha )|\nabla w|^2. \end{aligned}$$

(2.23)

Furthermore, we can obtain

$$\begin{aligned} \lambda\ge & {} \frac{1}{d^2}(1+\sqrt{1+(n+1)(nK+\alpha )d^2})\exp (-1-\sqrt{1+(n+1)(nK+\alpha )d^2})\nonumber \\\ge & {} \frac{A_3}{d^2}\exp \{-A_4d\sqrt{K+\alpha }\} \end{aligned}$$

(2.24)

as claimed. $\square$

3 The Complete Noncompact Case

In this section, we prove a global lower bound for the first eigenvalue of diffusion operator $\Delta _X$ by using the technique of gradient estimate as Wu [17, 18] did. First of all, we present a smooth cutoff function originated by Calabi [2] (see also [4, 13]).

Choose a smooth function $\xi :[0,+\infty )\rightarrow [0,1]$ so that $0\le \xi (s)\le 1$, $\xi (s)=1$ for $s\le 1$ and $\xi (s)=0$ for $s\ge 2$. Moreover, for some constants $C_1>0$ and $C_2>0$,

$$\begin{aligned} -C_1\le \frac{\xi '(s)}{\sqrt{\xi (s)}}\le 0 \end{aligned}$$

and

$$\begin{aligned} \xi ''(s)\ge -C_2. \end{aligned}$$

Let $(M^n,g)$ be a Riemannian manifold with $Ric\ge -(n-1)Kg$ for some constant $K:=K(2\rho )>0$ in $B({\bar{x}},2\rho )$ and $r(x):=d(x,{\bar{x}})$ be the distance function from a fixed point ${\bar{x}}\in M^n$. For any $\rho >0$, we define the cutoff function by

$$\begin{aligned} \phi (x)=\xi \left( \frac{r(x)}{\rho }\right) . \end{aligned}$$

We can assume without loss of generality that the function $\phi$ is smooth in $B({\bar{x}},2\rho )$ by the arguments of Calabi [2] (see also [4]).

It is clear that $0\le \phi \le 1$ on $M^n$, $\phi \equiv 1$ on $\overline{B({\bar{x}},\rho )}$ and $\phi \equiv 0$ outside $B({\bar{x}},2\rho )$. Moreover, we have

$$\begin{aligned} \frac{|\nabla \phi |^2}{\phi }=\frac{(\xi ')^2}{\rho ^2\xi }\le \frac{C_1^2}{\rho ^2} \end{aligned}$$

(3.1)

in $B({\bar{x}},2\rho )$.

If $Ric_X^m\ge -(m-1)Kg$ for some constant $K:=K(2\rho )>0$ in $B({\bar{x}},2\rho )$, the generalized Laplacian comparison theorem (see Corollary 3.3 of [9]) implies that $\Delta _X r\le (m-1)(\frac{1}{r}+\sqrt{K})$.

To deal with $\Delta \phi$, we divide the arguments into two cases:

• Case 1: $r(x)<\rho$. In this case, $\phi =1$ around x. Therefore, $\Delta \phi =0$.

• Case 2: $\rho \le r(x)<2\rho$. By direct computations, we have

  $$\begin{aligned} \Delta _X \phi= & {} \frac{\xi '}{\rho }\Delta _X r+\frac{\phi ''}{\rho ^2}|\nabla r|^2\nonumber \\\ge & {} -\frac{C_1}{\rho }(m-1)\left( \frac{1}{r(x)}+\sqrt{K}\right) -\frac{C_2}{\rho ^2}\nonumber \\\ge & {} -\frac{C_1}{\rho }(m-1)\left( \frac{1}{\rho }+\sqrt{K}\right) -\frac{C_2}{\rho ^2}. \end{aligned}$$

  (3.2)

  Therefore, we obtain

  $$\begin{aligned} \Delta _X\phi \ge -\frac{(m-1)C_1(1+\rho \sqrt{K})+C_2}{\rho ^2} \end{aligned}$$

  (3.3)

  in $B({\bar{x}},2\rho )$. Then, we prove the following essential inequality.

Lemma 3.1

Let $(M^n,g)$ be an n-dimensional complete noncompact Riemannian manifold. If u is a positive function defined on the geodesic ball $B({\bar{x}},\rho )\in M^n$ satisfying $\Delta _Xu=\lambda u$ for some constant $\lambda$ and $h=\log u$, then we have

$$\begin{aligned} \frac{1}{2}\Delta _X|\nabla h|^2\ge & {} \frac{m}{4(m-1)}\cdot \frac{|\nabla |\nabla h|^2|^2}{|\nabla h|^2}+\frac{1}{m-1}(|\nabla h|^2-\lambda )^2+Ric_X^m(\nabla h,\nabla h)\nonumber \\&-\left[ \frac{\lambda }{(m-1)|\nabla h|^2}+\frac{m-2}{m-1}\right] \langle \nabla |\nabla h|^2,\nabla h\rangle . \end{aligned}$$

(3.4)

Proof

Note that $\nabla h=\frac{\nabla u}{u}$, we have

$$\begin{aligned} \Delta _Xh=\frac{\Delta _Xu}{u}-\frac{|\nabla u|^2}{u^2}=\lambda -|\nabla h|^2. \end{aligned}$$

(3.5)

Moreover,

$$\begin{aligned} \frac{1}{2}\Delta _X|\nabla h|^2= & {} |Hess h|^2+\nabla _i\nabla _i\nabla _j h\nabla _jh+X^i\nabla _j\nabla _jh\nabla _jh\nonumber \\= & {} |Hess h|^2+\langle \nabla \Delta h,\nabla h\rangle +Ric(\nabla h,\nabla h)+X^i\nabla _j\nabla _jh\nabla _jh\nonumber \\= & {} |Hess h|^2+\langle \nabla \Delta _Xh,\nabla h\rangle +Ric_X^m(\nabla h,\nabla h)+\frac{(Xh)^2}{m-n}\nonumber \\= & {} |Hess h|^2-\langle \nabla |\nabla h|^2,\nabla h\rangle +Ric_X^m(\nabla h,\nabla h)+\frac{(Xh)^2}{m-n}, \end{aligned}$$

(3.6)

where Hessh is the Hessian of h, and we used the Bochner formula in the second equality and (3.5) in the last.

As in the proof of Theorem 2.3, we choose a frame $\{e_i\}_{i=1}^n$ so that $|\nabla h|=\nabla _{e_1} h\equiv h_1$. Then, we have

$$\begin{aligned} |\nabla |\nabla h|^2|^2=4|Hess h(\nabla h,\cdot )|^2=4h_1^2\sum _{i=1}^nh_{1i}^2=4|\nabla h|^2\cdot \sum _{i=1}^nh_{1i}^2, \end{aligned}$$

i.e.,

$$\begin{aligned} \frac{|\nabla |\nabla h|^2|^2}{|\nabla h|^2}=4\sum _{i=1}^nh_{1i}^2. \end{aligned}$$

(3.7)

Similar as the estimate (2.5) in [17], we get

$$\begin{aligned} |Hess h|^2+\frac{(Xh)^2}{m-n}\ge & {} h_{11}^2+2\sum _{j=2}^nh_{1j}^2+\frac{1}{n-1}(\Delta h-h_{11})^2+\frac{(Xh)^2}{m-n}\nonumber \\= & {} h_{11}^2+2\sum _{j=2}^nh_{1j}^2+\frac{1}{n-1}(|\nabla h|^2-\lambda +h_{11}+Xh)^2+\frac{(Xh)^2}{m-n}\nonumber \\\ge & {} h_{11}^2+2\sum _{j=2}^nh_{1j}^2+\frac{(|\nabla h|^2-\lambda +h_{11})^2}{m-1}\nonumber \\\ge & {} \frac{m}{4(m-1)}\cdot \frac{|\nabla |\nabla h|^2|^2}{|\nabla h|^2}+\frac{(|\nabla h|^2-\lambda )^2}{m-1}\nonumber \\&+\frac{|\nabla h|^2-\lambda }{m-1}\cdot \frac{\langle \nabla |\nabla h|^2,\nabla h\rangle }{|\nabla h|^2},\nonumber \\ \end{aligned}$$

(3.8)

where we used Cauchy’s inequality in the first line, (3.5) in the second, the fact of $\frac{(a+b)^2}{n-1}+\frac{b^2}{m-n}\ge \frac{a^2}{m-1}$ in the third and (3.7) in the last.

(3.4) follows by applying (3.8)–(3.6). $\square$

Now we are ready to derive the global lower estimates for the first eigenvalue of $\Delta _X$ in the noncompact case.

Theorem 3.2

Let $(M^n,g)$ be an n-dimensional complete noncompact Riemannian manifold with $Ric_X^m\ge -(m-1)Kg$ for some constant $K\ge 0$. Then, there is a global lower bound estimate on the first eigenvalue for the diffusion operator $\Delta _X$

$$\begin{aligned} \lambda _1\ge -\frac{(m-1)^4K^2}{4}. \end{aligned}$$

(3.9)

Proof

Define $H=\phi |\nabla h|^2$. Then, in $B({\bar{x}},2\rho )$, we have

$$\begin{aligned} \frac{1}{2}\phi \Delta _XH\ge & {} \frac{m}{4(m-1)}\cdot \frac{\phi ^2|\nabla |\nabla h|^2|^2}{|\nabla h|^2}+\frac{1}{m-1}(H-\phi \lambda )^2-(m-1)KH\nonumber \\&-\phi ^2\left[ \frac{\lambda }{(m-1)|\nabla h|^2}+\frac{m-2}{m-1}\right] \langle \nabla |\nabla h|^2,\nabla h\rangle +\phi \langle \nabla \phi ,\nabla |\nabla h|^2\rangle \nonumber \\&+H\Delta _X\phi \nonumber \\\ge & {} \frac{m}{4(m-1)}\cdot \frac{|\nabla H|^2-|\nabla h|^2|\nabla \phi |^2}{|\nabla h|^2}+\frac{1}{m-1}(H-\phi \lambda )^2-(m-1)KH\nonumber \\&-\phi \left[ \frac{\lambda }{(m-1)|\nabla h|^2}+\frac{m-2}{m-1}\right] \langle \nabla H,\nabla h\rangle \nonumber \\&+H\left[ \frac{\lambda }{(m-1)|\nabla h|^2}+\frac{m-2}{m-1}\right] \langle \nabla \phi ,\nabla h\rangle \nonumber \\&+\langle \nabla \phi ,\nabla H\rangle -\frac{|\nabla \phi |^2}{\phi }H-\frac{(m-1)C_1(1+\rho \sqrt{K})+C_2}{\rho ^2}H, \end{aligned}$$

(3.10)

where we used Lemma 3.1 in the first inequality and (3.3) in the second.

Suppose that $x_1\in B({\bar{x}},2\rho )\subset M^n$ is a maximum of G. Applying the maximum principle to (3.10), we get

$$\begin{aligned} 0\ge & {} -\frac{m|\nabla \phi |^2}{4(m-1)}+\frac{1}{m-1}(H-\phi \lambda )^2-(m-1)KH\nonumber \\&+H\left[ \frac{\lambda }{(m-1)|\nabla h|^2}+\frac{m-2}{m-1}\right] \langle \nabla \phi ,\nabla h\rangle \nonumber \\&-\frac{|\nabla \phi |^2}{\phi }H-\frac{(m-1)C_1(1+\rho \sqrt{K})+C_2}{\rho ^2}H\nonumber \\\ge & {} -\frac{m+C_1^2}{4(m-1)\rho ^2}+\frac{H^2-2\phi \lambda +(\phi \lambda )^2}{m-1}\nonumber \\&-(m-1)KH-\frac{|\lambda |\phi C_1}{(m-1)\rho }H^\frac{1}{2}-\frac{(m-2)C_1}{(m-1)\rho }H^\frac{3}{2}\nonumber \\&-\frac{(m-1)C_1(1+\rho \sqrt{K})+C_1^2+C_2}{\rho ^2}H, \end{aligned}$$

(3.11)

where we used (3.1) and Cauchy’s inequality. Note that

$$\begin{aligned} \frac{(\phi \lambda )^2}{m-1}-\frac{|\lambda |\phi C_1}{(m-1)\rho }H^\frac{1}{2}\ge -\frac{C_1^2}{4(m-1)\rho ^2}H, \end{aligned}$$

and

$$\begin{aligned} -\frac{(m-2)C_1}{(m-1)\rho }H^\frac{3}{2}\ge -\frac{1}{2(m-1)}H^2 -\frac{(m-2)^2C_1^2}{2(m-1)\rho ^2}H. \end{aligned}$$

Hence, (3.11) reduces to

$$\begin{aligned} 0\ge & {} \frac{H^2}{2(m-1)}-\frac{2\phi \lambda }{m-1}-\frac{m+C_1^2}{4(m-1)\rho ^2}\nonumber \\&-\left[ (m-1)K+\frac{C_1^2}{4(m-1)\rho ^2}+\frac{(m-2)^2C_1^2}{2(m-1)\rho ^2}\right. \nonumber \\&\left. +\frac{(m-1)C_1(1+\rho \sqrt{K})+C_1^2+C_2}{\rho ^2}\right] H. \end{aligned}$$

(3.12)

Therefore, we get

$$\begin{aligned}&\left[ (m-1)K+\frac{C_1^2}{4(m-1)\rho ^2}+\frac{(m-2)^2C_1^2}{2(m-1)\rho ^2} +\frac{(m-1)C_1(1+\rho \sqrt{K})+C_1^2+C_2}{\rho ^2}\right] ^2\nonumber \\&\quad +\frac{2}{m-1}\cdot \left[ \frac{2\phi \lambda }{m-1}+\frac{m+C_1^2}{4(m-1)\rho ^2}\right] \ge 0. \end{aligned}$$

(3.13)

Taking $\rho \rightarrow \infty$ in (3.13), it becomes

$$\begin{aligned} (m-1)^2K^2+\frac{4\lambda }{(m-1)^2}\ge 0, \end{aligned}$$

(3.14)

i.e.,

$$\begin{aligned} \lambda \ge -\frac{(m-1)^4K^2}{4}. \end{aligned}$$

(3.15)

This completes the proof. $\square$

We have similar result via $\infty$-dimensional Bakry–Émery–Ricci curvature of the diffusion operator $\Delta _X$.

Corollary 3.3

Let $(M^n,g)$ be an n-dimensional complete noncompact Riemannian manifold with $Ric_X^\infty \ge -nKg$ for some constant $K\ge 0$ and $|X|^2\le \alpha$. Then, there is a global lower bound estimate on the first eigenvalue for the diffusion operator $\Delta _X$

$$\begin{aligned} \lambda _1\ge -\frac{n^2(nK+\alpha )^2}{4}. \end{aligned}$$

(3.16)

Proof

Taking $m=n+1$ in Lemma 3.1, we can obtain

$$\begin{aligned} \frac{1}{2}\Delta _X|\nabla h|^2\ge & {} \frac{n+1}{4n}\cdot \frac{|\nabla |\nabla h|^2|^2}{|\nabla h|^2}+\frac{1}{n}(|\nabla h|^2-\lambda )^2+Ric_X^\infty (\nabla h,\nabla h)\nonumber \\&-(Xh)^2-\left( \frac{\lambda }{n|\nabla h|^2}+\frac{n-1}{n}\right) \langle \nabla |\nabla h|^2,\nabla h\rangle . \end{aligned}$$

(3.17)

Since $Ric_X^\infty \ge -nKg$ and $|X|^2\le \alpha$, it is clear that

$$\begin{aligned} Ric_X^{n+1}\ge -(nK+\alpha )g. \end{aligned}$$

(3.18)

Then, the generalized Laplacian comparison theorem (see Corollary 3.3 of [9]) implies that $\Delta _X r\le n(\frac{1}{r}+\sqrt{K+\frac{\alpha }{n}})$.

Using the same method as in the proof of (3.3), we can get

$$\begin{aligned} \Delta _X\phi \ge -\frac{nC_1(1+\rho \sqrt{K+\frac{\alpha }{n}})+C_2}{\rho ^2} \end{aligned}$$

(3.19)

Furthermore, (3.13) becomes

$$\begin{aligned}&\left[ nK+\alpha +\frac{C_1^2}{4n\rho ^2}+\frac{(n-1)^2C_1^2}{2n\rho ^2} +\frac{(nC_1(1+\rho \sqrt{K+\frac{\alpha }{n}})+C_1^2+C_2}{\rho ^2}\right] ^2\nonumber \\&\quad +\frac{2}{n}\cdot \left[ \frac{2\phi \lambda }{n}+\frac{n+1+C_1^2}{4n\rho ^2}\right] \ge 0. \end{aligned}$$

(3.20)

Taking $\rho \rightarrow \infty$ in (3.20), we get

$$\begin{aligned} (nK+\alpha )^2+\frac{4\lambda }{n^2}\ge 0, \end{aligned}$$

(3.21)

i.e.,

$$\begin{aligned} \lambda \ge -\frac{n^2(nK+\alpha )^2}{4} \end{aligned}$$

(3.22)

as desired. $\square$