Myers-type compactness theorem with the Bakry-Emery Ricci tensor

TL;DR

This paper extends classical Myers' compactness theorem to smooth metric measure spaces with Bakry-Emery Ricci tensor bounds, providing new comparison results and relaxing conditions on the weight function derivative.

Contribution

It introduces a generalized Myers-type theorem for Bakry-Emery Ricci curvature, including a new $f$-mean curvature comparison and weaker derivative conditions.

Findings

Established $f$-mean curvature comparison under Ricci bounds

Proved a Myers-type compactness theorem for Bakry-Emery Ricci tensor

Improved previous results with weaker derivative conditions on $f'$

Abstract

In this paper, we first prove the $f$-mean curvature comparison in a smooth metric measure space when the Bakry-Emery Ricci tensor is bounded from below and $|f|$ is bounded. Based on this, we define a Myers-type compactness theorem by generalizing the results of Cheeger, Gromov, and Taylor and of Wan for the Bakry-Emery Ricci tensor. Moreover, we improve a result from Soylu by using a weaker condition on a derivative $f'(t)$.

Equations

$${\int }_0^{\infty }Ric({\gamma }^{\prime }(s),{\gamma }^{\prime }(s))ds=\infty .$$

$$Ric(x)\ge (n-1)\frac{(\frac{1}{4}+{\nu }^2)}{r^2(x)}$$

$$Ric(x)\ge \frac{C(n,b,r_0)}{(r_0+r(x){)}^b}$$

$$Ri{c}_V^k:=Ric+\frac{1}{2}{\mathcal{L}}_{V}g-\frac{1}{k}{V}^{\ast }\otimes V^{\ast }$$

$$Ri{c}_V:=Ri{c}_V^{\infty }=Ric+\frac{1}{2}{\mathcal{L}}_{V}g.$$

$$Ri{c}_V=\lambda g$$

$$Ri{c}_f^k:=Ric+\text{mbox}\mathrm{Hess}f-\frac{1}{k}df\otimes df,$$

$$Ri{c}_f:=Ric+\mathrm{Hess}f,$$

$$Ri{c}_V(x)\ge {\delta }_{2}h(r(x)),$$

$$Ri{c}_f(x)\ge C_1(h,n,\delta ,ϵ)h(r(x)),$$

$$Ri{c}_f(x)\ge \frac{C_2(n,b,r_0,\delta ,ϵ)}{(r_0+r(x){)}^b}$$

$$Ri{c}_f(x)\ge C_3(h,n,a)h(r(x)),$$

$$Ri{c}_f(x)\ge \frac{C_4(n,b,r_0,a)}{(r_0+r(x){)}^b}$$

$$Ri{c}_f^k(x)\ge C_5(h,n+k)h(r(x)),$$

$$Ri{c}_f^k(x)\ge \frac{C_6(n+k,b,r_0)}{(r_0+r(x){)}^b}$$

$${\int }_0^{\infty }Ri{c}_f({\gamma }^{\prime }(t),{\gamma }^{\prime }(t))dt=\infty ,$$

$$m_f(t)\le m_H^{n+4\delta (t+1)}(t)$$

$$-\frac{(\Delta r{)}^2}{n-1}\ge \frac{\partial }{\partial r}(\Delta r)+Ric(\nabla r,\nabla r).$$

$$m^{\prime }(t)\le -\frac{m^2(t)}{n-1}-Ric({\gamma }^{\prime }(t),{\gamma }^{\prime }(t)).$$

$$m_H^{\prime }(t)=-\frac{m_H^2(t)}{n-1}-(n-1)H.$$

$$(m(t)-m_H(t){)}^{\prime }\le -\frac{m^2(t)-m_H^2(t)}{n-1}+\nabla \nabla f({\gamma }^{\prime }(t),{\gamma }^{\prime }(t)).$$

$$m_H(t)=(n-1)\frac{s{n}_H^{\prime }(t)}{s{n}_H(t)}.$$

$$(s{n}_H^2(m-m_H){)}^{\prime }=2s{n}_{H}s{n}_H^{\prime }(m-m_H)+s{n}_H^2(m-m_H{)}^{\prime }.$$

$$(s{n}_H^2(m-m_H){)}^{\prime }\le s{n}_H^2\left(-\frac{(m-m_H{)}^2}{n-1}+\nabla \nabla f\right)\le s{n}_H^2\nabla \nabla f.$$

$$s{n}_H^2(t)m(t)\le s{n}_H^2(t)m_H(t)+{\int }_0^{t}s{n}_H^2(s)\frac{\partial }{\partial s}⟨\nabla f,\gamma ⟩(s)ds.$$

$$s{n}_H^2(t)m(t)$$

$$s{n}_H^2(t)m_f(t)\le s{n}_H^2(t)m_H(t)-{\int }_0^t(s{n}_H^2(s){)}^{\prime }\frac{\partial }{\partial s}(f(\gamma (s)))ds.$$

$$s{n}_H^2(t)m_f(t)\le s{n}_H^2(t)m_H(t)-f(\gamma (t))(s{n}_H^2(t){)}^{\prime }+{\int }_0^t(s{n}_H^2(s){)}^{\prime \prime }f(\gamma (s))ds.$$

$$s{n}_H^2(t)m_f(t)$$

Full text

1. Introduction

The Myers theorem is a fundamental result in Riemannian geometry; it states that if an $n$-dimensional complete Riemannian manifold $(M,g)$ satisfies $Ric\geq(n-1)H$ with $H>0$, then $M$ is compact and ${\rm diam}(M)\leq\frac{\pi}{\sqrt{H}}$. Here, $Ric$ is the Ricci curvature of the metric $g$. This result has been generalized by different approaches ([1], [4], [3], and [9]). Herein, we recall two of them. The first was derived by W. Ambrose.

Theorem 1.1** ([1]).**

Suppose there exists a point $p$ in an $n$-dimensional complete Riemannian manifold $M$ for which every geodesic $\gamma(t)$ emanating from $p$ satisfies

[TABLE]

Then $M$ is compact.

The second theorem was derived by J. Cheeger, M. Gromov, and M. Taylor.

Theorem 1.2** ([3]).**

Let $M$ be an $n$-dimensional complete Riemannian manifold. If there exists a fixed point $p\in M$ and $r_{0},\nu>0$ such that

[TABLE]

holds for all $r(x)\geq r_{0}>0$, where $r(x)$ is a distance function defined with respect to a fixed point $p\in M$, i.e., $r(x)=d(p,x)$, then $M$ is compact and $\mbox{\rm diam}(M)<r_{0}e^{\frac{\pi}{\nu}}$.

Theorem 1.2 was improved by J. Wan as follows:

Theorem 1.3** ([11]).**

Let $M$ be an $n$-dimensional complete Riemannian manifold. If there exists a fixed point $p\in M$, $b\geq 2$, and $r_{0}>0$ such that

[TABLE]

for all $r(x)\geq 0$, where $r(x)=d(p,x)$ and $C(n,b,r_{0})$ is a constant depending on $n$, $b$, and $r_{0}$, then $M$ is compact. In particular, $C(n,b,r_{0})$ can be chosen to be equal to $(n-1)\frac{(b-1)^{b}}{(b-2)^{b-2}}r^{b-2}_{0}$ for $b>2$ and $(n-1)(1+\frac{r_{0}}{\epsilon})$, $\epsilon>0$ for $b=2$.

Now, we consider the $k$-Bakry–Emery Ricci tensor as follows:

[TABLE]

for some number $k>0$, where $\mathcal{L}_{V}$ denotes the Lie derivative in the direction of $V$, and $V^{*}$ is the metric dual of $V$. When $k=\infty$,

[TABLE]

In particular, if

[TABLE]

for some $\lambda\in\mathbb{R}$, then $(M,g)$ is a Ricci soliton, which is a self-similar solution to the Ricci flow. It is classified as expanding, steady, or shrinking when $\lambda<0$, $\lambda=0$, or $\lambda>0$, respectively. When $V=\nabla f$ for some $f\in C^{\infty}(M)$, the $k$-Bakry–Emery Ricci tensor becomes

[TABLE]

and the Ricci soliton becomes a gradient Ricci soliton

[TABLE]

where ${\rm Hess}\,f$ is the Hessian of $f$.

G. Wei and W. Wylie have generalized the Myers theorem by using $Ric_{f}$ in [12]. Several works have attempted to generalize this result ([13], [5], [10], and [14]). Note that H. Qiu generalized Theorem 1.3 to Theorem 1.4 by using $Ric_{\,V}$.

Theorem 1.4** ([7]).**

Let $M$ be an $n$-dimensional complete Riemannian manifold, $V$ be a smooth vector field on $M$, and $h:[0,\infty)\rightarrow(0,\infty)$ be a continuous function. Let $r(x)=d(p,x)$ be the distance function from $p\in M$. Assume that $\langle V,\nabla r\rangle\leq\delta_{1}$ along a minimal geodesic from each point $\tilde{p}\in M$; here, $\delta_{1}$ is a constant. Suppose

[TABLE]

where $\delta_{2}$ is a positive constant depending only on $h$, $n$, and $\delta_{1}$, then $M$ is compact. In particular, $\delta_{2}$ can be chosen as $\left(\frac{n-1}{\epsilon}+2\delta_{1}\right)\cdot\left(\frac{1}{\int^{\infty}_{\epsilon}h(s)ds}\right)+\epsilon_{1}$ ($\epsilon$ and $\epsilon_{1}$ are arbitrary positive constants).

In this paper, we improve Theorem 1.4 by weakening the condition in the case of $Ric_{f}$ (Theorem 1.5 and 1.7). Here, instead of the condition $|f|\leq k$, our conditions are $|f|\leq\delta(d(p,x)+1)$ ($\delta>0$) for Theorem 1.5 and $\partial_{t}f\geq-a$ ($a\geq 0$) for Theorem 1.7. Additionally, we prove Theorem 1.9 in the case of $Ric_{f}^{\,k}$.

Theorem 1.5**.**

Let $M$ be an $n$-dimensional complete Riemannian manifold and $h:[0,\infty)\rightarrow(0,\infty)$ be a continuous function. If there exists $p\in M$ and $|f|\leq\delta(d(p,x)+1)$ for some constant $\delta>0$ such that

[TABLE]

*where $r(x)=d(p,x)$ and $C_{1}$ is a positive constant depending only on $h$, $n$, $\delta$, and $\epsilon$, then $M$ is compact. In particular, $C_{1}$ can be chosen as

$\left(4\delta+\frac{n+4\delta(\epsilon+1)-1}{\epsilon}\right)\cdot\left(\frac{1}{\int^{\infty}_{\epsilon}h(s)ds}\right)+\epsilon_{1}$ ($\epsilon$ and $\epsilon_{1}$ are arbitrary positive constants).*

As a corollary, we make the following improvement to Theorem 1.3.

Corollary 1.6**.**

Let $M$ be an $n$-dimensional complete Riemannian manifold. If there exists $p\in M$, $b\in\mathbb{R}$, $r_{0}>0$, and $|f|\leq\delta(d(p,x)+1)$ for some constant $\delta>0$ such that

[TABLE]

for all $r(x)\geq 0$, where $r(x)=d(p,x)$ and $C_{2}$ is a constant depending on $n$, $b$, $r_{0}$, $\delta$, and $\epsilon$, then $M$ is compact. In particular, $C_{2}$ can be chosen as $\left(4\delta+\frac{n+4\delta(\epsilon+1)-1}{\epsilon}\right)\cdot\left((b-1)(r_{0}+\epsilon)^{b-1}\right)+\epsilon_{1}$ for $b>1$, and $C_{2}$ can be chosen as $\epsilon_{1}$ for $b\leq 1$. Here, $\epsilon$ and $\epsilon_{1}$ are arbitrary positive constants.

Furthermore, we have

Theorem 1.7**.**

Let $M$ be an $n$-dimensional complete Riemannian manifold and $h:[0,\infty)\rightarrow(0,\infty)$ be a continuous function. If there exists $p\in M$ and $\partial_{t}f\geq-a\,(a\geq 0)$ such that

[TABLE]

where $r(x)=d(p,x)$ and $C_{3}$ is a positive constant depending on $h$, $n$, and $a$, then $M$ is compact. In particular, $C_{3}$ can be chosen as $\left(a+\frac{n-1}{\epsilon}\right)\cdot\left(\frac{1}{\int^{\infty}_{\epsilon}h(s)ds}\right)+\epsilon_{1}$ ($\epsilon$ and $\epsilon_{1}$ are arbitrary positive constants).

As a corollary, we obtain the following result, similar to Corollary 1.6.

Corollary 1.8**.**

Let $M$ be an $n$-dimensional complete Riemannian manifold. If there exists $p\in M$, $b\in\mathbb{R}$, $r_{0}>0$, and $\partial_{t}f\geq-a\,(a\geq 0)$ such that

[TABLE]

for all $r(x)\geq 0$, where $r(x)=d(p,x)$ and $C_{4}$ is a constant depending only on $n$, $b$, $r_{0}$, and $a$, then $M$ is compact. In particular, $C_{4}$ can be chosen as $(2ar_{0}+(n-1)(b-2))r^{b-2}_{0}\frac{(b-1)^{b}}{(b-2)^{b-1}}$ for $b>2$, $C_{4}$ can be chosen as $(2a+\frac{n-1}{\epsilon})(b-1)(r_{0}+\epsilon)^{b-1}$ for $1<b\leq 2$, and $C_{4}$ can be chosen as $\epsilon_{1}$ for $b\leq 1$. Here, $\epsilon$ and $\epsilon_{1}$ are arbitrary positive constants.

For the case of $Ric_{f}^{\,k}$, we have

Theorem 1.9**.**

Let $M$ be an $n$-dimensional complete Riemannian manifold and $h:[0,\infty)\rightarrow(0,\infty)$ be a continuous function. If there exists $p\in M$ such that

[TABLE]

*where $k\in(0,\infty)$, $r(x)=d(p,x)$, and $C_{5}$ is a positive constant depending on $h$ and $n+k$, then $M$ is compact. In particular, $C_{5}$ can be chosen as

$\left(\frac{n+k-1}{\epsilon}\right)\cdot\left(\frac{1}{\int^{\infty}_{\epsilon}h(s)ds}\right)+\epsilon_{1}$ ($\epsilon$ and $\epsilon_{1}$ are arbitrary positive constants).*

Corollary 1.10**.**

Let $M$ be an $n$-dimensional complete Riemannian manifold. If there exists $p\in M$, $b\in\mathbb{R}$, and $r_{0}>0$ such that

[TABLE]

*for all $r(x)\geq 0$, where $k\in(0,\infty)$, $r(x)=d(p,x)$, and $C_{6}$ is a constant depending on $n+k$, $b$, and $r_{0}$, then $M$ is compact. In particular, $C_{6}$ can be chosen as $(n+k-1)\frac{(b-1)^{b}}{(b-2)^{b-2}}r_{0}^{b-2}$ for $b>2$, $C_{6}$ can be chosen as

$(\frac{n+k-1}{\epsilon})(b-1)(r_{0}+\epsilon)^{b-1}$ for $1<b\leq 2$, and $C_{6}$ can be chosen as $\epsilon_{1}$ for $b\leq 1$. Here, $\epsilon$ and $\epsilon_{1}$ are arbitrary positive constants.*

Remark 1.11**.**

If $\int^{\infty}_{\epsilon}h(s)ds=\infty$, then the constants $C_{1}$, $C_{3}$, and $C_{5}$ can be chosen as arbitrary positive real numbers.

Moreover, we obtain an Ambrose-type result. Theorem 1.1 was generalized by S. Zhang [14] using the Bakry–Emery Ricci tensor. Another generalization was derived by M.P. Cavalcate, J.Q. Oliveira, and M.S. Santos [2] with the condition $f$: $\frac{df}{dt}\leq 0$. The following is a further improvement made by Y. Soylu [8].

Theorem 1.12** ([8]).**

Let $M$ be an $n$-dimensional complete Riemannian manifold, where $n\geq 2$. Suppose there exists a point $p\in M$ such that every geodesic $\gamma(t)$ emanating from $p$ satisfies

[TABLE]

and $f^{\prime}(t)\leq\frac{1}{4}(1-\frac{1}{t})$ for all $t\geq 1$. Then, $M$ is compact.

In this paper, we improve the above result by finding a weaker condition for the derivative $f^{\prime}(t)$:

Theorem 1.13**.**

Let $M$ be an $n$-dimensional complete Riemannian manifold, where $n\geq 2$. Suppose there exists a point $p\in M$ such that every geodesic $\gamma(t)$ emanating from $p$ satisfies

[TABLE]

*and $f^{\prime}(t)\leq C(1-\frac{1}{t^{\alpha}})$ for some constants $C>0$, $\alpha>1$, and for all $t\geq 1$.

Then, $M$ is compact.*

This paper is organized as follows. In Section 2, we prove the $f$-mean curvature comparison theorem. In Section 3, we prove Theorems 1.5, 1.7, and 1.9. In Section 4, we prove Corollaries 1.6, 1.8, and 1.10. In the final section, we prove Theorem 1.13.

2. $f$-Mean curvature comparison

We first recall a few definitions. Let $(M,g,e^{-f}dv_{g})$ be a smooth metric measure space on an $n$-dimensional complete Riemannian manifold $M$. For the measure $e^{-f}dv_{g}$, the $f$-mean curvature is defined by $m_{f}:=m-\partial_{r}f$, where $m$ is the mean curvature of the geodesic sphere with an inward-pointing normal vector. Then, the $f$-Laplacian is defined by $\Delta_{f}:=\Delta-\langle\nabla f,\nabla\rangle$. Note that $m_{f}(r)=\Delta_{f}(r)$ and $m(r)=\Delta(r)$, where $r$ is the distance function. Now, we prove the $f$-mean curvature comparison theorems.

Theorem 2.1**.**

Let $M$ be an $n$-dimensional complete Riemannian manifold with $Ric_{f}(\gamma^{\prime},\gamma^{\prime})\geq(n-1)H$, $H\in\mathbb{R}$. Fix $p\in M$. If $|f|\leq\delta(d(p,x)+1)$ for some constant $\delta>0$ and $d(p,x)$ is the distance function from $p$ to $x$, then

[TABLE]

*(when $H>0$, assume t $\leq\frac{\pi}{4\sqrt{H}}$).

Here, $m^{n+4\delta(t+1)}_{H}(t)$ is the mean curvature of the geodesic sphere in $M^{n+4\delta(t+1)}_{H}$. The simply connected model space of dimension $n+4\delta(t+1)$ has a constant curvature $H$, and $m_{H}$ is the mean curvature of the model space of dimension $n$.*

Proof. By the Bochner formula and Schwarz inequality, the distance function $r$ satisfies the Riccati inequality,

[TABLE]

i.e.,

[TABLE]

The equality holds if and only if the radial sectional curvatures are constant. Therefore, the mean curvature of the model space $m_{H}$ satisfies

[TABLE]

Since $Ric_{f}(\gamma^{\prime},\gamma^{\prime})=Ric(\gamma^{\prime},\gamma^{\prime})+{\rm Hess}\,f(\gamma^{\prime},\gamma^{\prime})$, we obtain

[TABLE]

Let $sn_{H}(t)$ be the solution to $sn^{\prime\prime}_{H}(t)+Hsn_{H}(t)=0$ such that $sn_{H}(0)=0$ and $sn^{\prime}_{H}(0)=1$, then

[TABLE]

We compute

[TABLE]

By inequality (2.2), we have

[TABLE]

Since $\nabla\nabla f(\gamma^{\prime}(t),\gamma^{\prime}(t))=\frac{\partial}{\partial t}\langle\nabla f,\gamma^{\prime}\rangle(t)$, integrating the above inequality from [math] to $t$ yields

[TABLE]

By performing integration by parts on the last term, we obtain

[TABLE]

Since $\langle\nabla f,\gamma^{\prime}\rangle(t)=\frac{\partial}{\partial t}(f(\gamma(t)))$ and $m_{f}(t)=m(t)-\frac{\partial}{\partial t}f(\gamma(t))$, we have

[TABLE]

Conducting integration by parts on the last term again, we obtain

[TABLE]

If $|f|\leq\delta(d(p,x)+1)$ and $t\in(0,\frac{\pi}{4\sqrt{H}}]$ when $H>0$, then $(sn_{H}^{2}(t))^{\prime}\geq 0$ and $(sn_{H}^{2}(t))^{\prime\prime}\geq 0$. Therefore, we have

[TABLE]

From (2.3), we compute

[TABLE]

Thus, we can see that

[TABLE]

Therefore,

[TABLE]

Theorem 2.2** ([12]).**

Let $M$ be an $n$-dimensional complete Riemannian manifold with $Ric_{f}(\gamma^{\prime},\gamma^{\prime})\geq(n-1)H$, $H\in\mathbb{R}$. Fix $p\in M$. If $\partial_{t}f\geq-a$ for some constant $a\geq 0$, then

[TABLE]

(when $H>0$ assume $t\leq\frac{\pi}{2\sqrt{H}}$).

The proof of Theorem 2.2 is similar to the above proof. For the detailed proof, see Theorem 1.1 of [12].

3. Proof of Theorems 1.5, 1.7, and 1.9

In this section, we prove Theorem 1.5, 1.7, and 1.9. The proof uses the $f$-mean curvature comparison calculated above and Riccati inequality.

First, we prove Theorem 1.5. Suppose $M$ is non-compact. For any $p\in M$, there exists a unit speed ray $\gamma(t)$ starting from $p$. Let $r(x)=d(p,x)$ be the distance function from $p$. By the Bochner formula and Schwarz inequality, we have

[TABLE]

Integrating the above inequality from $\epsilon$ to $t$ we obtain

[TABLE]

From (1.1), the above inequality becomes

[TABLE]

Now, we claim that $-4\delta\leq m_{f}(t)\leq\frac{n+4\delta(t+1)-1}{t}$.

Since $Ric_{f}>0$, $m_{f}(t)\leq\frac{n+4\delta(t+1)-1}{t}$ holds by inequality (2.1), it suffices to show that $-4\delta\leq m_{f}(t)$.

Consider the excess function

[TABLE]

for $0\leq t\leq i$. By the triangle inequality, we have

[TABLE]

Therefore,

[TABLE]

It follows that

[TABLE]

i.e.,

[TABLE]

Thus,

[TABLE]

If $i\rightarrow\infty$, then $m_{f}(t)\geq-4\delta$. Hence, we proved the claim that

[TABLE]

From (3.1) and (3.2), we derive

[TABLE]

Let $t\rightarrow\infty$. Then, the above inequality becomes

[TABLE]

Therefore, we obtain

[TABLE]

Choosing

[TABLE]

where $\epsilon_{1}>0$. This yields a contradiction. Thus, $M$ must be compact.

Second, we prove Theorem 1.7. Its proof is similar to the previous proof; thus, the setting is identical. Suppose that $M$ is non-compact. Based on the previous proof, we know that

[TABLE]

From (1.3), we have

[TABLE]

We claim that $-a\leq m_{f}(t)\leq\frac{n-1}{t}+a$. Based on Theorem 2.2, $m_{f}(t)\leq\frac{n-1}{t}+a$ and $-a\leq m_{f}(t)$ hold according to the same argument as in the previous proof. Thus,

[TABLE]

Let $t\rightarrow\infty$. Then we obtain

[TABLE]

If we choose

[TABLE]

for some positive constant $\epsilon_{1}$, then this yields a contradiction. Thus, $M$ must be compact.

Finally, we prove Theorem 1.9. The setting is identical to that of the previous proof. Suppose that $M$ is non-compact. By the Bochner formula and Schwarz inequality for the distance function $r$, we have the following Riccati inequality (Appendix A of [12]):

[TABLE]

Integrating both sides of (3.4), we obtain

[TABLE]

From (1.5), the above inequality becomes

[TABLE]

We claim that $0\leq m_{f}(t)\leq\frac{n+k-1}{t}$. By Theorem A.1 in [12], $m_{f}(t)\leq\frac{n+k-1}{t}$ and $m_{f}(t)\geq 0$ hold by the same argument as in the previous proof. Thus

[TABLE]

Let $t\rightarrow\infty$. Then, we have

[TABLE]

Choosing

[TABLE]

where $\epsilon_{1}>0$. This is a contradiction. Thus, $M$ is compact.

4. Proof of Corollaries 1.6, 1.8, and 1.10

In this section, we prove Corollaries 1.6, 1.8, and 1.10. These corollaries generalize Theorem 1.3 by using the Bakry–Emery Ricci tensor. The settings of the corollaries are same as those of the previous theorems.

First, we prove Corollary 1.6. Suppose that $M$ is non-compact. We know that

[TABLE]

By assumption (1.2) and $-4\delta\leq m_{f}(t)\leq\frac{n+4\delta(t+1)-1}{t}$, the above inequality becomes

[TABLE]

Now, we obtain $C_{2}$. First, when $b>1$, we have

[TABLE]

If $t\rightarrow\infty$, then we obtain

[TABLE]

Choosing

[TABLE]

for any positive constant $\epsilon_{1}$ yields a contradiction.

Second, when $b=1$, inequality (4.1) becomes

[TABLE]

If $t\rightarrow\infty$ and $C_{2}=\epsilon_{1}$, the above inequality is a contradiction.

Similarly, when $b<1$, if $t\rightarrow\infty$ and $C_{2}=\epsilon_{1}$ in inequality (4.1), we obtain a contradiction. Thus, $M$ must be compact.

Secondly, we prove Corollary 1.8. Suppose that $M$ is non-compact. By assumption (1.4) and $-a\leq m_{f}(t)\leq\frac{n-1}{t}+a$, we have

[TABLE]

Now, we obtain $C_{4}$. First, when $b>2$, we have

[TABLE]

If $t\rightarrow\infty$, then we obtain

[TABLE]

Solving

[TABLE]

it follows that

[TABLE]

When $\epsilon=\frac{r_{0}}{b-2}$, $\frac{(r_{0}+\epsilon)^{b-1}}{\epsilon}$ has a minimum value. Substituting this into (4.3), we obtain

[TABLE]

Then, we can choose $C_{4}(n,b,r_{0},a)=(2ar_{0}+(n-1)(b-2))r_{0}^{b-2}\frac{(b-1)^{b}}{(b-2)^{b-1}}$. Similarly, from inequality (4.3), we can choose $C_{4}=\left(2a+\frac{n-1}{\epsilon}\right)(b-1)(r_{0}+\epsilon)^{b-1}$ for $1<b\leq 2$. When $b\leq 1$, it is the same as Corollary 1.6. Thus, $M$ is compact.

Finally, we prove Corollary 1.10. Suppose that $M$ is a non-compact. By assumption (1.6) and $0\leq m_{f}(t)\leq\frac{n+k-1}{t}$, we obtain

[TABLE]

Now, we obtain $C_{6}$. First, for $b>2$, we have

[TABLE]

Let $t\rightarrow\infty$. We obtain

[TABLE]

If $\frac{n+k-1}{\epsilon}-\frac{C_{6}}{b-1}\left(\frac{1}{(r_{0}+\epsilon)^{b-1}}\right)\leq 0$, then we have

[TABLE]

Similar to the above proof, when $\epsilon=\frac{r_{0}}{b-2}$, we can choose $C_{6}(n+k,b,r_{0})=(n+k-1)\frac{(b-1)^{b}}{(b-2)^{b-2}}r_{0}^{b-2}$. Similarly, from inequality (4.4) and (4.5), we can choose $C_{6}=(\frac{n+k-1}{\epsilon})(b-1)(r_{0}+\epsilon)^{b-1}$ for $1<b\leq 2$ and $C_{6}=\epsilon_{1}$ for $b\leq 1$, where $\epsilon_{1}$ is an arbitrary positive constant. Thus, $M$ is compact.

5. Ambrose-type Theorem

In this section, we prove Theorem 1.13. Suppose that $M$ is non-compact. Let $\gamma(t)$ be a unit speed ray starting from $p$. For every $t>0$, let $m(t)$ be the Laplacian of the distance function from a fixed point $p\in M$. By applying the Bochner formula

[TABLE]

to the distance function $r(x)=d(x,p)$, we obtain

[TABLE]

By the Schwarz inequality, we have the Riccati inequality

[TABLE]

By adding ${\rm Hess}\,f(\gamma^{\prime},\gamma^{\prime})$ to both sides of this inequality, we obtain

[TABLE]

Integrating both sides of (5.1), we obtain

[TABLE]

From assumption

[TABLE]

we have

[TABLE]

By multiplying both sides of (5.2) with $\frac{1}{n}$, we obtain

[TABLE]

Thus, for any constant $A\geq C+2$, there exists $t_{1}>1$ such that for any $t\geq t_{1}$, we have

[TABLE]

By assumption $f^{\prime}(t)\leq C(1-\frac{1}{t^{\alpha}})$, we have

[TABLE]

In other words,

[TABLE]

By multiplying both sides of (5.3) by $n$, we obtain

[TABLE]

Now, we consider the increasing sequence $\{t_{\ell}\}$ defined by

[TABLE]

Note that $\{t_{\ell}\}$ converges to $T:=t_{1}+2$ as $\ell\rightarrow\infty$.

We claim that $-m(t)\geq 2^{\ell}n$ for all $t\geq t_{\ell}$. We prove this claim through induction. If $\ell=1$, the claim is trivially true from the inequality in (5.4). Now, for all $t\geq t_{\ell+1}$, by the inequality in (5.4), we have

[TABLE]

proving the claim. Therefore,

[TABLE]

This contradicts the smoothness of $m(t)$, which completes the proof of Theorem 1.13.