The Log-Sobolev Inequality for a Submanifold in Manifolds With Asymptotic Non-Negative Intermediate Ricci Curvature

Abstract

We prove a sharp Log-Sobolev inequality for submanifolds of a complete non-compact Riemannian manifold with asymptotic non-negative intermediate Ricci curvature and Euclidean volume growth. Our work extends a result of Dong et al. (Acta Math. Sci. Ser. B (Engl. Ed.), 44(1):189–194 (2024)) which already generalizes Yi and Zheng (To appear in Chin. Ann. Math., (2023)) and Brendle (Comm. Pure Appl. Math. 75(3), 449–454 (2022)).

1 Introduction

Log-Sobolev inequalities have been already studied in mathematical physics and information theory in the 50 s ([14, 29]), then it has been extensively studied in analysis and geometry (see e.g., [10, 13, 16, 17, 22, 24,25,26]), with recent developments including their applications to heat kernel and Hamilton-Jacobi equations, inequalities in convex geometry, and their role in optimal transport. Brendle [5] obtained a sharp log-Sobolev inequality for submanifolds in Euclidean space, improving [13], using the Alexandroff-Bakelman-Pucci (ABP) method and optimal transport. Later Yi and Zheng [34] extended this result to Riemannian manifolds with non-negative sectional curvature and Euclidean volume growth conditions. Later Dong-Lin-Lu [12] further extended this to non-negative asymptotic sectional curvature. In this paper, we prove that the logarithmic Sobolev inequality holds under non-negative intermediate Ricci curvature. Furthermore, we extend our result to the asymptotic non-negative intermediate Ricci curvature by introducing error terms that depends only on the curvature’s decay at infinity, as in [12].

Let $(M^n,g)$ be a complete non-compact Riemannian manifold, for $1 \le k\le n-1$ we consider a k-dimensional subspace P of a tangent space $T_xM$ at $x\in M$. Given any tangent vector $v \in T_xM$ with $v \bot P$, the k-intermediate Ricci curvature (kth-Ricci curvature) with respect to P in the direction of v is defined as

$$\begin{aligned} {{\,\textrm{Ric}\,}}_k (P,v) = \sum _{i=1}^k \textrm{Sec}(v, e_i) |v|^2, \end{aligned}$$

where $\{e_1, \ldots , e_k\}$ is an orthonormal basis of P. The k-intermediate Ricci curvature interpolates between Ricci and sectional curvature. A Riemannian manifold has non-negative k-Ricci curvature if ${{\,\textrm{Ric}\,}}_k(P,v) \ge 0$ for any $x\in M$, k-dimensional subspace $P\subset T_xM$, and a unit tangent vector $v\in T_xM$ perpendicular to P (note that some papers [21, 30] require the stronger condition that this holds for all v, not necessary just the perpendicular ones). This condition is denoted by ${{\,\textrm{Ric}\,}}_k \ge 0$. In particular, it exhibits a monotonicity property: if $n \le m$, then ${{\,\textrm{Ric}\,}}_n \ge 0$ implies ${{\,\textrm{Ric}\,}}_m \ge 0$. This curvature condition has been well studied to explore the gap of the global results with sectional curvature bounds and Ricci curvature bounds, see for example: [9, 18, 19, 21, 23, 27, 28, 30, 32]. Our work follows this spirit and generalizes Yi-Zheng’s result with sectional curvature lower bound (Theorem 1.1, [34]) and the corresponding asymptotic extension by Dong-Lin-Lu (Theorem 1.1, [12]). Ketterer-Mondino [21] noted that it is possible to characterize lower bounds of k-intermediate Ricci curvature via optimal transport. We will adopt this approach, developed in [23] and [30], to prove our main result.

Let M be an n-dimensional non-compact Riemmanian manifold with non-negative Ricci curvature. The asymptotic volume ratio of M is defined as

$$\begin{aligned} \theta := \lim _{r \rightarrow \infty }\frac{|B_r(p)|}{\omega _nr^n}, \end{aligned}$$

where p is some fixed point in M, $\omega _n$ is the volume of the unit ball in Euclidean space $\mathbb {R}^n$, and $|B_r(p)|$ is the volume of a ball of radius r centered at p in M. The Bishop-Gromov volume comparison theorem ensures that the limit exists, it does not depend on the choice of p and that $\theta \le 1$. We say that M has Euclidean volume growth if $\theta >0$.

In this paper we are using the definition of asymptotically non negative sectional (resp. Ricci) curvature given in Abresch [1] (see also [2, 20, 35]). Zhu (see theorem 2.1 in [36]) proved the equivalent of the classical Bishop-Gromov volume comparison Theorem with ${\text {Ric}}\ge 0$ by replacing $\mathbb {R}^n$ with a different model space. We will need to define the equivalent of the usual asymptotic volume ratio.

Definition 1.1

(Abresch [1]) An n-dimensional complete non-compact Riemannian manifold (M, g) with base point o has asymptotically non-negative Sectional curvature (Ricci curvature, respectively) if and only if there exists any non-negative, non-increasing function $\lambda : [0,\infty )\rightarrow [0,\infty )$ such that the following holds:

• (1) $b_0(\lambda )=\int _0^\infty t\lambda (t)dt < \infty$

• (2) ${\text {Sec}}\ge -\lambda (d(o,p))$ at each point $p\in M$. $\left( {\text {Ric}}\ge -(n-1)\lambda (d(o,p)), \text { respectively} \right)$

The first condition also implies that $b_1(\lambda )=\int _0^\infty \lambda (t)dt < \infty$. To extend this notion of asymptotically non-negative sectional curvature to intermediate k-Ricci curvature, one can simply replace the second condition with:

$$\begin{aligned} {\text {Ric}}_k\ge -k\lambda (d(o,p)) \text { at each point } p \in M. \end{aligned}$$

More precisely, ${{\,\textrm{Ric}\,}}_k(P,v) \ge - k \lambda (d(o,p))$ for any k dimensional subspace $P \subset T_pM$ and unit tangent vector v perpendicular to P.

The usual non-negative sectional curvature (Ricci or ${{\,\textrm{Ric}\,}}_k$ respectively) condition is equivalent to requiring $\lambda \equiv 0$. It’s immediately evident from the definition to see that this class of manifolds includes those with non-negative sectional curvature (Ricci or ${\text {Ric}}_k$ respectively) outside a compact set and asymptotically flat manifolds as well. Furthermore, the monotonicity of the intermediate curvature still holds: if $k_1 \le k_2$, then ${{\,\textrm{Ric}\,}}_{k_1} \ge - k_1\lambda ( d(o,p))$ implies ${{\,\textrm{Ric}\,}}_{k_2} \ge - k_2 \lambda ( d(o,p))$.

In this setting, we also replace $\theta$, the asymptotic volume ratio of M, with another similar quantity that will keep track of the geometry at infinity. To achieve this, we define h(t) as the unique solution of the following ODE:

$$\begin{aligned} {\left\{ \begin{array}{ll} h''(t) = \lambda (t) h(t),\\ h(0) = 0, h'(0)=1. \end{array}\right. } \end{aligned}$$

We now define the asymptotic volume ratio of M with respect to h by

$$\begin{aligned} \theta _h=\lim _{r\rightarrow \infty }\frac{|B_r(o)|}{n\omega _n\int _0^{r} h^{n-1}(t)dt}, \end{aligned}$$

In the above definition, $|B_r(o)|$ represents the volume of the ball of radius r centered at o in M and $\omega _n$ represents the volume of the unit ball in $\mathbb {R}^n$. In [36], to prove the corresponding version of the Bishop-Gromov volume comparison in this setting, Zhu noted that the following function of r is non-increasing:

$$\begin{aligned} \frac{|B_r(o)|}{n\omega _n\int _0^rh^{n-1}(t)dt}. \end{aligned}$$

This ensures that the limit exists and $\theta _h$ is well-defined. Similarly, we say that M has Euclidean volume growth if $\theta _h>0$. In particular, when a manifold has non-negative intermediate Ricci curvature, $\lambda (t) \equiv 0$, $h(t) = t$, and $\theta _h$ is the usual asymptotic volume ratio $\theta$. In the second section of this paper we will show a connection between $\theta _h$ and an integral of a certain Guassian function. To state our theorem, we first define, for any non-negative t, a decreasing function P(t):

$$\begin{aligned} P(t)=(4 \pi )^{-\frac{n}{2}}\int _{\mathbb {R}^N} e^{-\frac{(|x|+t)^2}{4}}dx \end{aligned}$$

Notice that $P(0)=1$. Below is our main result.

Theorem 1.2

Let $M^{n+m}$ be a complete, non-compact manifold of dimension $n+m$ with asymptotically non-negative $\textrm{Ric}_{k}$, where $k=\min \{n-1,m-1\}$ and Euclidean volume growth. Assume that $\Sigma ^n$ is an n-dimensional compact submanifold without a boundary. Then, for any positive smooth function f, we have

$$\begin{aligned} \int _\Sigma&f \left( \log f + \log P(4b_1 n ) \theta _h + \frac{n}{2}\log (4 \pi ) +n + 4b_1^2 n^2 + (n+m-1) \log \frac{1+b_0}{e^{2r_0b_1 + b_0}}\right) d V (x) \nonumber \\&- \int _\Sigma \frac{|\nabla ^\Sigma f|^2}{f} - \int _\Sigma f |H|^2 \le \int _\Sigma f d V (x) \left( \log \int _\Sigma f(x) dV (x)\right) , \end{aligned}$$

(1)

where $\theta _h$ is the asymptotic volume ratio of M with respect to h, H is the mean curvature vector of $\Sigma$, and $r_0={\text {max}}_{x\in \Sigma }d(o,x)$.

Similar to the Michael-Simon-Sobolev inequality, the inequality (1) contains a term that involves the mean curvature and depends on the asymptotic volume ratio of M. If $\lambda \equiv 0$, the usual non-negative ${\text {Ric}}_k$ condition, in other words $b_0=b_1=0$, then our Theorem 1.2 becomes:

Corollary 1.3

Let $M^{n+m}$ be a complete, non-compact manifold of dimension $n+m$ with Euclidean volume growth and $\textrm{Ric}_{k} \ge 0$, where $k=\min \{n-1,m-1\}$. Let $\Sigma ^n$ be a n-dimensional compact submanifold without a boundary. Then, for any positive smooth function f, we have

$$\begin{aligned} \int _\Sigma&f\left( \log f + n+\frac{n}{2} \log (4\pi ) + \log \theta \right) dV - \int _\Sigma \frac{|\nabla ^\Sigma f|^2}{f} dV - \int _\Sigma f|H|^2 dV \nonumber \\&\le \left( \int _\Sigma f dV\right) \log \left( \int _\Sigma f dV\right) , \end{aligned}$$

(2)

where $\theta$ is the asymptotic volume ratio of M and H is the Mean curvature vector of $\Sigma$.

This gives the following two consequences by choosing $f\equiv 1$ in (2) and apply the argument in Section 4 of [34].

Corollary 1.4

Let $M^{n+m}$ be a complete, non-compact manifold of dimension $n+m$ with non-negative intermediate k-Ricci curvature, where $k = \min \{n-1,m-1\}$. Suppose M has a Euclidean volume growth. Then, there is no closed minimal submanifold of dimension n in M.

Corollary 1.5

Let $M^{n+m}$ be a complete, non-compact manifold of dimension $n+m$ with non-negative intermediate k-Ricci curvature, where $k = \min \{n-1,m-1\}$. Suppose there is a closed minimal submanifold of dimension n in M. Then, the asymptotic volume ratio of M is zero.

In the case of any hypersurface ($m=1$) under non-negative Ricci curvature condition, these two corollaries have been proved by Agostiniani-Fogagnolo-Mazzieri (see Theorem 1.6 in [3]). We do not recover this case in our result since $k= \min \{n-1,m-1\}$ would be zero. On the other hand, for higher dimension and codimension, these corollaries are also new in the usual non-negative k-intermediate Ricci curvature setting. When $k=1$, the standard ${\text {Ric}}_1\ge 0$ is equivalent to the non-negative sectional curvature condition. Thus, we recover the corollaries in Yi-Zheng (Corollaries 1.1 and 1.2 in [34]).

The same argument (Section 4, [34]) cannot be applied in the case of asymptotic non-negative k-intermediate Ricci curvature setting because the error term we have introduced is not scale-invariant and will blow up if we blow down the metric as in [34].

Note that the previous two corollaries do not hold for ${{\,\textrm{Ric}\,}}\ge 0$ total dimension 4; a counterexample is the Eguchi-Hanson metric on $TS^2$, which is Ricci flat with Euclidean volume growth and has a totally geodesic submanifold $S^2$. See, for example, [4, Page 270].

The proof of Theorem 1.2 relies on the Alexandroff-Bakelman-Pucci (ABP) method, a standard technique in Euclidean space [15]. Cabrè [7] extended this method to manifolds. The classical estimate in $\mathbb {R}^n$ uses affine functions, but on manifolds, hyperplanes are replaced with paraboloids interpreted as graphs of the distance squared relative to a point. If a lower bound condition on the curvature is added, this method has been successfully used to prove geometric inequalities (see for example, [31, 33]). In particular, Brendle [6] used this method to prove a monotonicity result in terms of the Jacobian of a certain transport map, while Wang [30] and Ma-Wu [23] independently generalized this method to k-intermediate Ricci curvature. It will be a key ingredient in our proof.

2 Asymptotic Estimate

In this section we recall an estimate for $\theta _h$, which will be used in the final steps of the proof in the next section. We first recall:

Lemma 2.1

(Lemma 2.2, [12]) Let M be a complete non-compact Riemannian manifold of dimension N with non-negative Ricci curvature. Then,

$$\begin{aligned} \theta = \lim _{r\rightarrow \infty } \left( \frac{(4 \pi )^{-\frac{N}{2}}}{r^N}\int _M e^{-\frac{d(x,p)^2}{4r^2}} d V (x)\right) \end{aligned}$$

for any $p \in M$.

The proof is based on the Bishop-Gromov volume comparison. However, in our asymptotic case, we utilize the inequality $r\le h(r)$ instead. The upper bound for h, as provided in [36], is $h(r) \le e^{b_0}r$. Generally, the presence of $b_0\ne 0$ precludes us from replacing $r^N$ with $h^N$ in the above lemma. Following Dong-Lin-Lu [12], it is preferable to replace $r^N$ with $rh^{N-1}$, doing so, we first recall the definition of P(t):

$$\begin{aligned} P(t)=(4 \pi )^{-\frac{N}{2}}\int _{\mathbb {R}^N} e^{-\frac{(|x|+t)^2}{4}}dx. \end{aligned}$$

Lemma 2.2

(Lemma 2.2, [12]) Let M be a complete non-compact Riemannian manifold of dimension N with asymptotic non-negative Ricci curvature. Then

$$\begin{aligned} P(t)\theta _h =\lim _{r \rightarrow \infty }\frac{(4 \pi )^{-\frac{N}{2}}}{rh(r)^{N-1}} \int _M e^{- \frac{\left( \frac{d(x,o)}{r}+t\right) ^2}{4}} dV (x), \end{aligned}$$

(3)

where o is the base point of M.

This was proved in [12], we are giving the same proof of this statement also here for completeness:

Proof

First notice that we can apply De L’Hopital’s rule to the limit of $\theta _h$ and obtain

$$\begin{aligned} \theta _h=\lim _{r\rightarrow \infty }\frac{|B_r(o)|}{N\omega _N\int _0^{r} h^{N-1}(t)dt}=\lim _{r\rightarrow \infty }\frac{|\partial B_r(o)|}{N\omega _Nh^{N-1}(r)} \end{aligned}$$

(4)

Now we can compute

$$\begin{aligned}&\lim _{r \rightarrow \infty }\frac{(4 \pi )^{-\frac{N}{2}}}{rh(r)^{N-1}} \int _M e^{- \frac{\left( \frac{d(x,o)}{r}+t\right) ^2}{4}} dV (x)\\&\quad = \lim _{r \rightarrow \infty }(4 \pi )^{-\frac{N}{2}}\int _0^\infty \frac{|\partial B_{rs}(o)|}{h^{N-1}(r)}e^{-\frac{(s+t)^2}{4}}ds\\&\quad = \lim _{r \rightarrow \infty }(4 \pi )^{-\frac{N}{2}}\int _0^\infty \frac{|\partial B_{rs}(o)|}{h^{N-1}(rs)} \frac{h^{N-1}(rs)}{h^{N-1}(r)} e^{-\frac{(s+t)^2}{4}}ds\\&\quad =(4 \pi )^{-\frac{N}{2}}N \omega _N \theta _h \int _0^\infty s^{N-1}e^{-\frac{(s+t)^2}{4}}ds\\&\quad = \theta _h (4 \pi )^{-\frac{N}{2}}\int _{\mathbb {R}^n}e^{-\frac{(|x|+t)^2}{4}}dx\\&\quad =P(t)\theta _h, \end{aligned}$$

where we used the monotone convergence theorem and the identity (4). $\square$

For the next lemma, we aim to substitute the base point o on the right-hand side of inequality (3) with $\psi (x)$, where $\psi$ is a Borel map whose image is compact.

Lemma 2.3

Let M be a complete non-compact Riemannian manifold of dimension n with asymptotic non-negative intermediate Ricci curvature. Then we have

$$\begin{aligned} \lim _{r \rightarrow \infty }\frac{(4 \pi )^{-\frac{N}{2}}}{rh(r)^{N-1}} \int _M e^{- \frac{\left( \frac{d(x,\psi (x))}{r}+t\right) ^2}{4}} dV (x)=\lim _{r \rightarrow \infty }\frac{(4 \pi )^{-\frac{N}{2}}}{rh(r)^{N-1}} \int _M e^{- \frac{\left( \frac{d(x,o)}{r}+t\right) ^2}{4}} dV (x), \end{aligned}$$

where $\psi :M\rightarrow K$ is any Borel map where $K\subset M$ a compact subset.

The above lemma is proven, similar to the proof of Lemma 2.2 in [34], by utilizing the triangle inequality.

3 Proof of Theorem 1.2

In this section, we prove log Sobolev inequality with non-negative asymptotic intermediate Ricci curvature by following the papers [11, 12, 34].

Let $M^{n+m}$ be a complete non-compact manifold of dimension $n+m$ with asymptotically non-negative $\textrm{Ric}_{k}$ and Euclidean volume growth, where $k= \min \{n-1, m-1\}$. Assume that $\Sigma ^n$ is an n-dimensional compact submanifold without a boundary and let f be any positive smooth function.

We first assume that $\Sigma$ is connected, which is needed for the existence of a solution to a differential equation. The inequality (1) is invariant when we scale a function f. Thus, by scaling, we may assume that

$$\begin{aligned} \int _\Sigma f \log f dV - \int _\Sigma \frac{|\nabla ^\Sigma f|^2}{f} dV - \int _\Sigma f |H|^2 d V = 0. \end{aligned}$$

(5)

Indeed, we can scale the function f to satisfy the equation (5) because $\log (cf)$ can be any real number by choosing a suitable constant c. The left-hand-side of equation (5) comes from the inequality (1). Thus, it suffices to show that

$$\begin{aligned} \int _\Sigma&f\left( \log P(4b_1 n ) \theta _h + \frac{n}{2}\log (4 \pi ) +n + 4b_1^2 n^2 + (n+m-1) \log \frac{1+b_0}{e^{2r_0b_1 + b_0}}\right) dV\\&\le \left( \int _\Sigma f dV\right) \log \left( \int _\Sigma f dV\right) . \end{aligned}$$

In order to prove Isoperimetric or Sobolev inequalities using the ABP method (see [8] and related literature), we need to consider a suitable PDE on a submanifold $\Sigma$. To apply our assumption (5), we consider a differential equation as follows:

$$\begin{aligned} \textrm{div}_\Sigma (f \nabla ^\Sigma u) = f \log f - \frac{|\nabla ^\Sigma f |^2 }{f} - f | H|^2 \text{ on } \Sigma , \end{aligned}$$

(6)

where $\nabla ^\Sigma$ is the induced Levi-Civita connection on $\Sigma$. We do not need t o specify a boundary condition since $\partial \Sigma = \emptyset$. Using standard PDE theory (see [15] Chapter 6), since f is a positive function, the operator $u \mapsto \textrm{div}_\Sigma (f \nabla ^\Sigma u)$ has Fredholm index 0. Hence, we can find a smooth solution $u: \Sigma \rightarrow \mathbb {R}$.

We define a contact set $A_r$ as the set of all points $({\bar{x}}, {\bar{y}}) \in T^\bot \Sigma$ satisfying

$$\begin{aligned} r u(x) + \frac{1}{2} d(x, \exp _{{\bar{x}}} ( r \nabla ^\Sigma u ({\bar{x}}) + r{\bar{y}}))^2 \ge ru ({\bar{x}}) + \frac{1}{2} r^2(|\nabla ^\Sigma u({\bar{x}} )|^2 + |{\bar{y}}|^2) \end{aligned}$$

(7)

for all $x \in \Sigma$ as introduced in [6]. This is a natural extension of the Cabrè’s idea [7] to the submanifold case.

Let $\Phi _t: T^\bot \Sigma \rightarrow M$ be a map defined by

$$\begin{aligned} \Phi _t (x,y) = \exp _x (t \nabla ^\Sigma u (x) + t y). \end{aligned}$$

Then we have the following lemma, for which we will recall the proof for the reader’s convenience.

Lemma 3.1

(Lemma 3.2, [34]) For each $r \in (0, \infty )$, we have $\Phi _r (A_r) = M$.

Proof

Take any $p \in M$. Since $\Sigma$ is compact, there is ${\bar{x}} \in \Sigma$ where the function $x \mapsto ru(x) + \frac{1}{2} d(x,p)^2$ attains its minimum. Let ${\bar{\gamma }}$ be a minimizing geodesic such that ${\bar{\gamma }}(0) = {\bar{x}}$ and ${\bar{\gamma }} (r) = p$. Then the geodesic ${\bar{\gamma }}$ minimizes the functional $u(\gamma (0)) + \frac{1}{2} \int _0^r |\gamma '(t)|^2 dt$ among all smooth curves $\gamma$ with $\gamma (0) \in \Sigma$ and $\gamma (r) =p$. By the first variational formula, we have

$$\begin{aligned} \nabla ^\Sigma u( {\bar{x}}) - {\bar{\gamma }}'(0) \in T_{{\bar{x}}}^\bot \Sigma . \end{aligned}$$

That is, there is ${\bar{y}} \in T_{x}^\bot \Sigma$ satisfying

$$\begin{aligned} \nabla ^\Sigma u ({\bar{x}} ) + {\bar{y}} = {\bar{\gamma }}'(0). \end{aligned}$$

We can easily check that $\Phi _r(\bar{x}, \bar{y}) = p$ and $(\bar{x}, \bar{y}) \in A_r$ using the geodesic ${\bar{\gamma }}$, which implies that $p \in \Phi _r(A_r)$. $\square$

As we see in the above proof, we can define a map $\psi :M \rightarrow \Sigma$ as $\psi (p)$ is the point where the function $ru(x) + \frac{1}{2} d(x,p)^2$ attains its minimum. Then by Lemma 3.1, we have

$$\begin{aligned}&\int _M e^{-\left( \frac{d (\psi (p),p)}{2r}+2b_1n\right) ^2} \, dV (p)\nonumber \\&\le \int _M \left( \int _{\{(x,y)\in A_r|\Phi _r(x,y) = p\}} e^{-\left( \frac{d(x, \Phi _r(x,y))}{2r} + 2b_1n \right) ^2}d \mathcal {H}^0\right) d V (p), \end{aligned}$$

(8)

where $\mathcal {H}^0$ is the counting measure. The left hand side of the inequality (8) is related to the asymptotic volume ratio $\theta _h$ as we showed in Lemma 2.2. Also, we apply Area formula with the map $\Phi _r: A_r \rightarrow M$ to the right hand side of (8). That is,

$$\begin{aligned} \int _M&\left( \int _{\{(x,y)\in A_r|\Phi _r(x,y) = p\}} e^{-\left( \frac{d(x, \Phi _r(x,y))}{2r} + 2b_1n \right) ^2}d \mathcal {H}^0\right) d V (p)\nonumber \\&= \int _{A_r} e^{-\left( \frac{d(x, \Phi _r(x,y))}{2r} + 2b_1n \right) ^2} |\textrm{det} D\Phi _r(x,y)| d V (x,y)\nonumber \\&= \int _\Sigma \left( \int _{T_x^\bot \Sigma } e^{-\left( \frac{d(x, \Phi _r(x,y))}{2r} + 2b_1n \right) ^2} |\textrm{det} D\Phi _r(x,y)| \chi _{A_r}(x,y) \, dy\,\right) d V (x). \end{aligned}$$

(9)

We now want to estimate the integrand in the right hand side of (9). Take any $r >0$ and $(\bar{x},{\bar{y}} )$ in $A_r$. Define a geodesic ${\bar{\gamma }}(t) = \exp _{{\bar{x}}}(t \nabla ^\Sigma u({\bar{x}}) + t {\bar{y}})$ on [0, r]. Let $\{e_1,\ldots , e_n\}$ be an orthonormal basis of $T_{{\bar{x}}}\Sigma$ such that $\textrm{Hess}_\Sigma u (e_i, e_j) - \langle II(e_i, e_j), \bar{y}\rangle$ is diangonal. Let us denote $E_i$ by a parallel transport vector field of $e_i$ along ${\bar{\gamma }}$ for each $1 \le i\le n$. Take any orthonormal frame $\{e_{n+1},\ldots , e_{n+m}\}$ of $T^\bot \Sigma$ near ${\bar{x}}$ such that $\langle \nabla _{e_i} e_\alpha , e_\beta \rangle = 0$ at ${\bar{x}}$ for all $1 \le i \le n$ and $n+1 \le \alpha , \beta \le n+m$.

For each $1 \le i \le n$, let us consider a Jacobi field $X_i(t)$ along ${\bar{\gamma }}(t)$ with the initial conditions

$$\begin{aligned} {\left\{ \begin{array}{ll} X_i(0) = e_i\\ \langle X_i' (0), e_j \rangle = \textrm{Hess}_\Sigma u (e_i, e_j) - \langle II(e_i, e_j), {\bar{y}} \rangle \quad \text{ for } \text{ all } 1 \le j \le n\\ \langle X_i'(0), e_\alpha \rangle = \langle II (e_i, \nabla ^\Sigma u ({\bar{x}} )), e_\alpha \rangle \quad \text{ for } \text{ all } n+1 \le \alpha \le n+m. \end{array}\right. } \end{aligned}$$

For each $n+1 \le \alpha \le n+m$, consider a Jacobi field $X_\alpha (t)$ along ${\bar{\gamma }}(t)$ with the initial conditions

$$\begin{aligned} X_\alpha (0) = 0 \quad \text{ and } \quad X_\alpha ' (0) = e_\alpha . \end{aligned}$$

Define $(n+m)\times (n+m)$ matrices P(t) and S(t) satisfying $P_{ij}(t) = \langle X_i (t), E_j(t) \rangle$ and $S_{ij}(t) = \bar{R} ({\bar{\gamma }} '(t), E_i(t),{\bar{\gamma }}'(t),E_j(t))$ for $1 \le i, j \le n+m$. Since $X_i(t)$ is a Jacobi field along ${\bar{\gamma }} (t)$, we have $P''(t)= - P(t) S (t)$. Define $Q(t) = P^{-1}(t)P'(t)$. Then

$$\begin{aligned} Q'(t) = - (Q(t))^2 -S(t). \end{aligned}$$

Let $tr_n Q(t) = \sum _{i=1}^{n}Q_{ii}(t)$ and $tr_m Q(t) = \sum _{\alpha =n+1}^{n+m} Q_{\alpha \alpha }(t)$. Then

$$\begin{aligned} (tr_n Q(t))' + \frac{1}{n} (tr_n Q(t))^2 \le - tr_n S(t) \end{aligned}$$

and

$$\begin{aligned} (tr_m Q(t))' + \frac{1}{m} (tr_m Q(t))^2 \le - tr_m S(t), \end{aligned}$$

where $tr_n S(t)$ and $tr_m S(t)$ are defined as a partial trace of S(t). Let us note that

$$\begin{aligned} tr_nS(t) = \sum _{i=1}^n {\bar{R}} ({\bar{\gamma }}'(t), E_i(t), {\bar{\gamma }}'(t), E_i(t)). \end{aligned}$$

However, we cannot apply curvature assumption directly because $\bar{\gamma }'(t)$ may not be perpendicular to the plane spanned by $E_1(t), \ldots , E_n(t)$. Since $E_i(t)$ is a parallel vector field along ${\bar{\gamma }}(t),$ it is enough to consider the angle at $t =0$, denoted by a, between the vector ${\bar{\gamma }}'(0)$ and the tangent plane $T_{{\bar{x}} } \Sigma$. Since ${\bar{\gamma }}'(0) = \nabla ^\Sigma u ({\bar{x}} ) + {\bar{y}}$, depending on the vectors $\nabla ^\Sigma u ({\bar{x}} )$ and ${\bar{y}}$ the angle is determined.

Let us consider the case $\nabla ^\Sigma u ( {\bar{x}} ) \ne 0$ and ${\bar{y}} \ne 0$. In this case, ${\bar{\gamma }}'(0)$ is not perpendicular to the tangent plane. After we change the basis as in [23], we get

$$\begin{aligned} tr_n S(t)&= \sin ^2 (a)|{\bar{\gamma }}'(0)|^2 Ric_n\left( \frac{\bar{\gamma }'(t)}{|{\bar{\gamma }}'(t)|}, P_1(t)\right) \nonumber \\&\quad + \cos ^2(a)|\bar{\gamma }'(0)|^2 Ric_{n-1}\left( \frac{{\bar{\gamma }}'(t)}{|{\bar{\gamma }}'(t)|}, P_2(t)\right) \nonumber \\&\ge -n\sin ^2 (a) |{\bar{\gamma }} '(0)|^2 \lambda (d(o, {\bar{\gamma }}(t))) - (n-1) \cos ^2 (a) |{\bar{\gamma }}'(0)|^2 \lambda (d( o, {\bar{\gamma }}(t)))\nonumber \\&= (\cos ^2 ( a) - n) |{\bar{\gamma }}'(0)|^2\lambda (d(o, \bar{\gamma }(t))), \end{aligned}$$

(10)

where $P_1(t)$ is an n-dimensional subspace spanned by $E_2(t), \ldots , E_{n+1}(t)$ and $P_2(t)$ is an $(n-1)$-dimensional subspace spanned by $E_2(t),\ldots , E_n(t)$. Similarly, we have

$$\begin{aligned} tr_mS(t) \ge (\sin ^2 (a) - m) |{\bar{\gamma }}'(0)|^2 \lambda (d(o,{\bar{\gamma }}(t))). \end{aligned}$$

(11)

Indeed, the inequalities (10) and (11) hold for the case $\nabla ^\Sigma u ({\bar{x}}) =0$ or ${\bar{y}} =0$. More specifically, for the case $\nabla ^\Sigma u ({\bar{x}}) = 0$, since $\bar{\gamma }' ( 0 ) = {\bar{y}} \in T_{{\bar{x}}}^\bot \Sigma$, the angle $a = \frac{\pi }{2}$ and we have

$$\begin{aligned} tr_n S(t)&= |{\bar{\gamma }}'(0)|^2Ric_n(P_3(t), {\bar{\gamma }}'(t))\\&\ge -n |{\bar{\gamma }}'(0)|^2\lambda (d(o, {\bar{\gamma }}(t))) \\&= (\cos ^2 ( a) - n)|{\bar{\gamma }}'(0)|^2 \lambda (d(o, \bar{\gamma }(t))) \end{aligned}$$

and

$$\begin{aligned} tr_m S(t)&= \sum _{\alpha = n+1}^{n+m} {\bar{R}} ({\bar{\gamma }}'(t) , E_\alpha (t), {\bar{\gamma }}'(t) , E_\alpha (t))\\&= \sum _{\alpha =n+2}^{n+m} {\bar{R}} ({\bar{\gamma }}'(t) , E_\alpha (t), {\bar{\gamma }}'(t) , E_\alpha (t)) \\&= |{\bar{\gamma }}'(0)|^2 Ric_{m-1}(P_4 (t), {\bar{\gamma }}'(t)) \\&\ge -(m-1) |{\bar{\gamma }}'(0)|^2\lambda (d(o, {\bar{\gamma }}'(t)))\\&\ge (\sin ^2 (a) - m) |{\bar{\gamma }}'(0)|^2 \lambda (d(o,\bar{\gamma }(t))), \end{aligned}$$

where $P_3(t) = Span (E_1(t), \ldots , E_n(t))$ and $P_4 (t) = Span (E_{n+2}(t), \ldots , E_{n+m}(t))$ with the assumption $E_{n+1} = \frac{{\bar{y}}}{|{\bar{y}}|}$. In the same manner, for the case ${\bar{y}} = 0$, since ${\bar{\gamma }}'(0) = \nabla ^\Sigma u({\bar{x}}) \in T_{{\bar{x}}} \Sigma$, by assuming $E_{1}(t) = \frac{{\bar{\gamma }}'(t)}{|\bar{\gamma }'(t)|}$, we can verify that the same inequalities (10) and (11) hold. In all, for $({\bar{x}}, {\bar{y}}) \in A_r {\setminus } \{(0,0)\}$, we obtain

$$\begin{aligned} {\left\{ \begin{array}{ll} (tr_nQ(t))' + \frac{1}{n}(tr_nQ(t))^2 \le (n-\cos ^2 (a) ) |{\bar{\gamma }}'(0)|^2\lambda (d(o,{\bar{\gamma }}(t)))\\ (tr_mQ(t))' + \frac{1}{m}(tr_mQ(t))^2 \le (m-\sin ^2 (a))|\bar{\gamma }'(0)|^2\lambda (d(o,{\bar{\gamma }}(t))) \end{array}\right. } \end{aligned}$$

by combining inequalities (10) and (11) with Riccati equation. It follows from the triangle inequality $d(o, {\bar{\gamma }}(t)) \ge |d(o, {\bar{x}} ) - d( {\bar{x}}, {\bar{\gamma }}(t))|$ and the definition of $A_r$ that

$$\begin{aligned} {\left\{ \begin{array}{ll} (tr_nQ(t))' + \frac{1}{n}(tr_nQ(t))^2 \le (n-\cos ^2(a)) |\bar{\gamma }'(0)|^2 \lambda \left( \left| d ( o, {\bar{x}} ) - t \sqrt{|\nabla ^\Sigma u ( {\bar{x}} ) |^2 + |{\bar{y}} |^2}\right| \right) \\ (tr_mQ(t))' + \frac{1}{m}(tr_mQ(t))^2 \le (m-\sin ^2(a))|\bar{\gamma }'(0)|^2\lambda \left( \left| d ( o, {\bar{x}} ) - t \sqrt{|\nabla ^\Sigma u ( {\bar{x}} ) |^2 + |{\bar{y}} |^2}\right| \right) . \end{array}\right. } \end{aligned}$$

Let $\phi (t) = e^{\frac{1}{n}\int _0^t \sum _{i=1}^n Q_{ii}(\tau ) d \tau }$. Then

$$\begin{aligned} {\left\{ \begin{array}{ll} \phi ''(t) \le \frac{n-\cos ^2 (a)}{n}|{\bar{\gamma }}'(0)|^2\lambda \left( \left| d ( o, {\bar{x}} ) - t \sqrt{|\nabla ^\Sigma u ( {\bar{x}} ) |^2 + |{\bar{y}} |^2}\right| \right) \,\phi (t)\\ \phi '(0) = \frac{1}{n} ( \Delta _{\Sigma } u ({\bar{x}}) - \langle H ({\bar{x}}), {\bar{y}} \rangle )\\ \phi (0) =1. \end{array}\right. } \end{aligned}$$

To use comparison theorems with $\phi (t)$, let us define $\psi _1 (t)$ and $\psi _2(t)$ as a solution to each PDE similarly to [11]:

$$\begin{aligned} {\left\{ \begin{array}{ll} \psi _1''(t) = \frac{n-\cos ^2 (a)}{n}|{\bar{\gamma }}'(0)|^2\lambda \left( \left| d ( o, {\bar{x}} ) - t \sqrt{|\nabla ^\Sigma u ( {\bar{x}} ) |^2 + |{\bar{y}} |^2}\right| \right) \,\psi _1(t)\\ \psi _1(0) = 0, \psi _1'(0) = 1 \end{array}\right. } \end{aligned}$$

and

$$\begin{aligned} {\left\{ \begin{array}{ll} \psi _2''(t) = \frac{n-\cos ^2 (a)}{n}|{\bar{\gamma }}'(0)|^2\lambda \left( \left| d ( o, {\bar{x}} ) - t \sqrt{|\nabla ^\Sigma u ( {\bar{x}} ) |^2 + |{\bar{y}} |^2}\right| \right) \,\psi _2(t)\\ \psi _2(0) = 1, \psi _2'(0) = 0. \end{array}\right. } \end{aligned}$$

Then we see that a function $\psi (t) = \psi _2(t) + \frac{1}{n} ( \Delta _{\Sigma } u ({\bar{x}}) - \langle H ({\bar{x}}), {\bar{y}} \rangle ) \psi _1(t)$ satisfies the following PDE.

$$\begin{aligned} {\left\{ \begin{array}{ll} \psi ''(t) = \frac{n-\cos ^2 (a)}{n}|{\bar{\gamma }}'(0)|^2\lambda \left( \left| d ( o, {\bar{x}} ) - t \sqrt{|\nabla ^\Sigma u ( {\bar{x}} ) |^2 + |{\bar{y}} |^2}\right| \right) \,\psi (t)\\ \psi '(0) = \frac{1}{n} ( \Delta _{\Sigma } u ({\bar{x}}) - \langle H ({\bar{x}}), {\bar{y}} \rangle )\\ \psi (0) =1. \end{array}\right. } \end{aligned}$$

By comparing $\frac{\phi '(t)}{\phi (t)}$ with $\frac{\psi '(t)}{\psi (t)}$, we have

$$\begin{aligned} tr_n Q(t) = n \cdot \frac{\phi '(t)}{\phi (t)} \le n \cdot \frac{\psi '(t)}{\psi (t)} = \frac{\psi _2' +\frac{1}{n} ( \Delta _{\Sigma } u ({\bar{x}}) - \langle H ({\bar{x}}), {\bar{y}} \rangle ) \psi _1' }{\psi _2 + \frac{1}{n} ( \Delta _{\Sigma } u ({\bar{x}}) - \langle H ({\bar{x}}), {\bar{y}} \rangle ) \psi _1}. \end{aligned}$$

Define

$$\begin{aligned} {\bar{\phi }}(t) = t e^{\frac{1}{m} \int _0^t \sum _{\alpha =n+1}^{n+m} (Q_{\alpha \alpha }(t) - \frac{1}{t} ) d\tau }. \end{aligned}$$

Then

$$\begin{aligned} {\left\{ \begin{array}{ll} {\bar{\phi }}'' \le \frac{m- \sin ^2 (a)}{m}|{\bar{\gamma }}'(0)|^2\lambda \left( \left| d(o,{\bar{x}} ) - t \sqrt{|\nabla ^\Sigma u({\bar{x}})|^2 + |{\bar{y}} |^2})\right| \right) \, {\bar{\phi }}\\ {\bar{\phi }}(0) = 0, \quad {\bar{\phi }}'(0) = 1. \end{array}\right. } \end{aligned}$$

To estimate $\frac{{\bar{\phi }}'}{{\bar{\phi }}}$, let us introduce a $C^2$ solution ${\bar{\psi }}$ to the PDE

$$\begin{aligned} {\left\{ \begin{array}{ll} {\bar{\psi }}'' = \frac{m- \sin ^2 (a)}{m}|{\bar{\gamma }}'(0)|^2\lambda \left( \left| d(o,{\bar{x}} ) - t \sqrt{|\nabla ^\Sigma u({\bar{x}})|^2 + |{\bar{y}} |^2})\right| \right) \, {\bar{\psi }}\\ {\bar{\psi }}(0) = 0, \quad {\bar{\psi }}'(0) = 1. \end{array}\right. } \end{aligned}$$

Thus,

$$\begin{aligned} tr_mQ(t) = m \cdot \frac{{\bar{\phi }}'}{{\bar{\phi }}} \le m \cdot \frac{{\bar{\psi }}'}{{\bar{\psi }}}. \end{aligned}$$

Since $\frac{d}{dt} (\det P(t)) = \det P(t) {{\,\textrm{tr}\,}}Q(t),$ we have

$$\begin{aligned} \frac{d}{dt} \log (\det P(t))= & {} tr Q(t) \le n\cdot \frac{\psi '}{\psi } + m \cdot \frac{{\bar{\psi }}' }{{\bar{\psi }}} \\= & {} \frac{d}{dt} (n\log (\psi ) + m \log ({\bar{\psi }})) = \frac{d}{dt} \log (\psi ^n {\bar{\psi }}^m). \end{aligned}$$

By integrating the above inequality from $\epsilon$ to t and $\epsilon \rightarrow 0$, we get

$$\begin{aligned} \det P(t)&\le \psi ^n(t) {\bar{\psi }}^m(t) = \left( \psi _2(t) + \frac{1}{n} ( \Delta _{\Sigma } u ({\bar{x}}) - \langle H ({\bar{x}}) , {\bar{y}} \rangle ) \psi _1(t)\right) ^n {\bar{\psi }}^m(t)\\&= \left( \frac{\psi _2(t)}{\psi _1(t)} + \frac{1}{n} (\Delta _{\Sigma } u ({\bar{x}}) - \langle H ({\bar{x}}) , {\bar{y}} \rangle ) \right) ^n \psi _1^n(t) {\bar{\psi }}^m (t) \end{aligned}$$

Moreover, comparison theorems imply those inequalities

$$\begin{aligned} {\left\{ \begin{array}{ll} \frac{\psi _2}{\psi _1} \le \frac{2b_1(n-\cos ^2(a))}{n} \sqrt{|\nabla ^\Sigma u ( {\bar{x}} ) |^2 + |{\bar{y}} |^2} + \frac{1}{t}\\ \psi _1(t) \le t e^{\frac{n- \cos ^2 (a)}{n}(2b_1 d(o, {\bar{x}} ) + b_0)}\\ {\bar{\psi }} (t) \le t e^{\frac{m- \sin ^2 (a)}{m}(2b_1 d(o, {\bar{x}} ) + b_0)}. \end{array}\right. } \end{aligned}$$

Thus,

$$\begin{aligned} \det P(t)&\le \left( \frac{\psi _2(t)}{\psi _1(t)} + \frac{1}{n} (\Delta _{\Sigma } u ({\bar{x}}) - \langle H ({\bar{x}}) , {\bar{y}} \rangle ) \right) ^n \psi _1^n(t) {\bar{\psi }}^m (t)\nonumber \\&\le \left( \frac{2b_1(n - \cos ^2 (a))}{n} \sqrt{|\nabla ^\Sigma u ( {\bar{x}} ) |^2 + |{\bar{y}} |^2} + \frac{1}{t}+ \frac{1}{n} (\Delta _{\Sigma } u ({\bar{x}}) - \langle H ({\bar{x}}) , {\bar{y}} \rangle ) \right) ^n \nonumber \\&\quad \cdot t^n e^{(n- \cos ^2 (a))(2b_1 d(o, {\bar{x}} ) + b_0)} \cdot t^m e^{(m- \sin ^2 (a))(2b_1 d(o, {\bar{x}} ) + b_0)}\nonumber \\&\le \left( 2b_1 \sqrt{|\nabla ^\Sigma u ( {\bar{x}} ) |^2 + |{\bar{y}} |^2} + \frac{1}{t}+ \frac{1}{n} (\Delta _{\Sigma } u ({\bar{x}}) - \langle H ({\bar{x}}) , {\bar{y}} \rangle ) \right) ^n\nonumber \\&\quad t^{n+m}e^{(n+m-1)(2b_1 d( o,{\bar{x}})+b_0)}\nonumber \\&\le \left( 2b_1\sqrt{|\nabla ^\Sigma u ( {\bar{x}} ) |^2 + |{\bar{y}} |^2} + \frac{1}{t}+ \frac{1}{n} (\Delta _{\Sigma } u ({\bar{x}}) - \langle H ({\bar{x}}) , {\bar{y}} \rangle ) \right) ^n \nonumber \\&\quad t^{n+m}e^{(n+m-1)(2b_1 r_0+b_0)}, \end{aligned}$$

(12)

The following lemma can be also found in the proof of Lemma 3.7 in [34].

Lemma 3.2

For $(x,y) \in A_r$, we have:

$$\begin{aligned} \Delta _\Sigma u (x) - \langle H(x), y \rangle \le \log f(x)+ \frac{|\nabla ^\Sigma u|^2 + |y|^2}{4}- \frac{|2H(x)+y|^2}{4}. \end{aligned}$$

Proof

Since $\textrm{div}_\Sigma (f \nabla ^\Sigma u) = f \log f - \frac{|\nabla ^\Sigma f |^2 }{f} - f | H|^2$, we get

$$\begin{aligned} f \Delta _\Sigma u + \langle \nabla ^\Sigma f, \nabla ^\Sigma u \rangle = f \log f - \frac{|\nabla ^\Sigma f |^2 }{f} - f | H|^2. \end{aligned}$$

Since $f(x) \ne 0$ for all $x \in \Sigma$, we have

$$\begin{aligned} \Delta _\Sigma u - \langle H(x) , y \rangle = \log f - \frac{|\nabla ^\Sigma f |^2 }{f^2} - | H(x)|^2 -\frac{\langle \nabla ^\Sigma f , \nabla ^\Sigma u \rangle }{f} - \langle H(x) , y\rangle . \end{aligned}$$

(13)

Note that

$$\begin{aligned} \frac{|\nabla ^\Sigma f |^2}{f^2} + \frac{\langle \nabla ^\Sigma f , \nabla ^\Sigma u \rangle }{f}&= \frac{4 |\nabla ^\Sigma f |^2 + 4 \langle \nabla ^\Sigma f, f \nabla ^\Sigma u \rangle }{4f^2} \nonumber \\&= \frac{4 |\nabla ^\Sigma f|^2 + 4 \langle \nabla ^\Sigma f, f \nabla ^\Sigma u \rangle + |f \nabla ^\Sigma u |^2}{4f^2} -\frac{|f \nabla ^\Sigma u|^2}{4f^2}\nonumber \\&= \frac{| 2 \nabla ^\Sigma f+ f \nabla ^\Sigma u |^2 }{4f^2} - \frac{|\nabla ^\Sigma u|^2}{4} \end{aligned}$$

(14)

and

$$\begin{aligned} |H(x)|^2 + \langle H(x), y\rangle&= \frac{|2H(x)|^2 + \langle 4H(x), y \rangle + |y|^2}{4} - \frac{|y|^2}{4}\nonumber \\&= \frac{|2H(x) + y|^2}{4} - \frac{|y|^2}{4}. \end{aligned}$$

(15)

Combining identities (13), (14), and (15), we obtain

$$\begin{aligned} \Delta _\Sigma u (x) - \langle H(x) , y \rangle&= \log f - \frac{|\nabla ^\Sigma f |^2 }{f^2} - | H|^2 -\frac{\langle \nabla ^\Sigma f , \nabla ^\Sigma u \rangle }{f} - \langle H(x) , y\rangle \\&= \log f + \frac{|\nabla ^\Sigma u (x)|^2 + |y|^2}{4} - \frac{|2 \nabla ^\Sigma f + f \nabla ^\Sigma u |^2}{ 4f^2} \nonumber \\&\quad - \frac{|2H(x) + y |^2}{4}\\&\le \log f + \frac{|\nabla ^\Sigma u(x) |^2 + |y|^2}{4} - \frac{|2H(x)+y|^2}{4}. \end{aligned}$$

$\square$

Lemma 3.3

For $({\bar{x}},{\bar{y}}) \in A_r \setminus \{(0,0)\}$, we have:

$$\begin{aligned} |\textrm{det} D\Phi _t ({\bar{x}},{\bar{y}})|\le & {} e^{\alpha + \frac{n}{t} -n - 4b_1^2 n^2} t^{m+n} f \exp \left( \left( \frac{d({\bar{x}}, \Phi _t({\bar{x}},{\bar{y}}))}{2t} + 2b_1n\right) ^2\right) \\{} & {} \exp \left( - \frac{|2H+{\bar{y}}|^2}{4} \right) , \end{aligned}$$

where $\alpha = (n+m-1)(2r_0b_1+b_0)$.

Proof

Combining inequality (12) and Lemma 3.2, we obtain

$$\begin{aligned} \det P(t)&\le \left( 2b_1 \sqrt{|\nabla ^\Sigma u ( {\bar{x}} ) |^2 + |{\bar{y}} |^2} + \frac{1}{t}+ \frac{1}{n} (\Delta _{\Sigma } u ({\bar{x}}) - \langle H ({\bar{x}}) , {\bar{y}} \rangle ) \right) ^n t^{n+m}e^{\alpha }\\&\le \left( 2b_1\sqrt{|\nabla ^\Sigma u ( {\bar{x}} ) |^2 + |{\bar{y}} |^2} + \frac{1}{t}\right. \\&\left. \quad + \frac{1}{n} \left( \log f+ \frac{|\nabla ^\Sigma u|^2 + |{\bar{y}}|^2}{4}- \frac{|2H({\bar{x}})+{\bar{y}}|^2}{4} \right) \right) ^n t^{n+m}e^{\alpha } \end{aligned}$$

By the inequality $\lambda \le e^{\lambda -1}$, we have

$$\begin{aligned} |\textrm{det} D\Phi _t ({\bar{x}},{\bar{y}})|&\le \exp \left( 2b_1n \sqrt{|\nabla ^\Sigma u ( {\bar{x}} ) |^2 + |{\bar{y}} |^2} + \frac{n}{t}+ \log f \right. \\&\left. \quad + \frac{|\nabla ^\Sigma u|^2 + |{\bar{y}}|^2}{4}- \frac{|2H+{\bar{y}}|^2}{4} -n\right) t^{n+m}e^{\alpha }\\&=e^{\alpha + \frac{n}{t} -n} t^{m+n} f\exp \left( 2b_1n \sqrt{|\nabla ^\Sigma u ( {\bar{x}} ) |^2 + |{\bar{y}} |^2} \right. \\&\left. \quad + \frac{|\nabla ^\Sigma u|^2 + |{\bar{y}}|^2}{4}- \frac{|2H+{\bar{y}}|^2}{4}\right) \\&\le e^{\alpha + \frac{n}{t} -n} t^{m+n} f({\bar{x}})\exp \left( 2b_1n \sqrt{|\nabla ^\Sigma u ( {\bar{x}} ) |^2 + |{\bar{y}} |^2} \right. \\&\left. \quad + \frac{|\nabla ^\Sigma u|^2 + |{\bar{y}}|^2}{4}- \frac{|2H({\bar{x}})+{\bar{y}}|^2}{4}\right) \\&= e^{\alpha + \frac{n}{t} -n} t^{m+n} f({\bar{x}})\exp \left( \left( \frac{\sqrt{|\nabla ^\Sigma u|^2 + |{\bar{y}}|^2}}{2} + 2b_1n\right) ^2\right) \\&\quad \exp \left( - \frac{|2H+{\bar{y}}|^2}{4} - 4 b_1^2 n^2\right) \\&= e^{\alpha + \frac{n}{t} -n-4b_1^2 n^2} t^{m+n} f({\bar{x}})\exp \left( \left( \frac{d({\bar{x}}, \Phi _t({\bar{x}},{\bar{y}}))}{2t} \right. \right. \\&\left. \left. \quad + 2b_1n\right) ^2\right) \exp \left( - \frac{|2H({\bar{x}})+{\bar{y}}|^2}{4} \right) . \end{aligned}$$

$\square$

By combining inequalities (8) and (9) and Lemma 3.3, we obtain

$$\begin{aligned} \int _M&e^{- \left( \frac{d(\psi (p), p)}{2t} + 2b_1n\right) ^2} \, d V (p) \\&\le \int _\Sigma \left( \int _{T_x^\bot \Sigma } e^{-\left( \frac{d( x, \Phi _t(x, y))}{2t} + 2b_1n\right) ^2} |\textrm{det} D\Phi _t(x,y)| \chi _{A_t}(x,y) \, dy\,\right) d V (x)\\&=\int _\Sigma \left( \int _{T_x^\bot \Sigma } e^{\alpha + \frac{n}{t} -n-4b_1^2 n^2} t^{m+n} f( x) \exp \left( - \frac{|2H( x)+ y|^2}{4} \right) dy\right) d V (x)\\&= t^{n+m} e^{\frac{n}{t} -n+\alpha -4b_1^2 n^2}(4 \pi )^\frac{m}{2}\int _\Sigma f (x)d V (x). \end{aligned}$$

Then

$$\begin{aligned}&\frac{(4\pi )^{-\frac{n+m}{2}}}{t (h(t))^{n+m-1}}\int _M e^{- \left( \frac{d(\psi (p), p)}{2t} + 2b_1n\right) ^2} \, d V (p)\nonumber \\&\quad \le \frac{(4 \pi )^{-\frac{n}{2}}t^{n+m-1}}{(h(t))^{n+m-1}} e^{\frac{n}{t} -n+\alpha -4b_1^2 n^2}\int _\Sigma f (x)d V (x) \end{aligned}$$

(16)

Let us take limit $t \rightarrow \infty$ to the both sides of the above inequality. Then by Lemma 2.2 and 2.3, we get

$$\begin{aligned} P(4b_1 n )\theta _h \le \lim _{t \rightarrow \infty }\frac{t^{n+m-1}}{(h(t ))^{n+m-1}} (4 \pi )^{-\frac{n}{2}} e^{\frac{n}{t} -n+\alpha -4b_1^2 n^2}\int _\Sigma f (x)d V (x). \end{aligned}$$

Since

$$\begin{aligned} \lim \limits _{t \rightarrow \infty } \frac{t^{n+m-1}}{(h(t))^{n+m-1}} = \left( \lim \limits _{t \rightarrow \infty } \frac{t}{h(t)}\right) ^{n+m-1} = \left( \lim \limits _{t \rightarrow \infty }\frac{1}{h'(t)}\right) ^{n+m-1} \le \frac{1}{(1+b_0)^{n+m-1}}, \end{aligned}$$

we have

$$\begin{aligned} P(4b_1 n )\theta _h \le (4 \pi )^{-\frac{n}{2}} e^{ -n+\alpha -4b_1^2 n^2}\frac{1}{(1+b_0)^{n+m-1}} \int _\Sigma f (x)d V (x). \end{aligned}$$

By taking a logarithmic function, we obtain

$$\begin{aligned}{} & {} \log P(4b_1 n ) \theta _h + \frac{n}{2}\log (4 \pi ) +n - \alpha + 4b_1^2 n^2 + (n+m-1) \log (1+b_0)\\{} & {} \le \log \int _\Sigma f(x) dV (x), \end{aligned}$$

or equivalently,

$$\begin{aligned}{} & {} \log P(4b_1 n ) \theta _h + \frac{n}{2}\log (4 \pi ) +n + 4b_1^2 n^2 + (n+m-1) \log \frac{1+b_0}{e^{2r_0b_1 + b_0}}\\{} & {} \le \log \int _\Sigma f(x) dV (x). \end{aligned}$$

Multiplying the inequality by f and integrate it over $\Sigma$, then

$$\begin{aligned} \int _\Sigma&f \left( \log P(4b_1 n ) \theta _h + \frac{n}{2}\log (4 \pi ) +n + 4b_1^2 n^2 + (n+m-1) \log \frac{1+b_0}{e^{2r_0b_1 + b_0}}\right) d V (x) \\&\le \int _\Sigma f d V (x) \left( \log \int _\Sigma f(x) dV (x)\right) . \end{aligned}$$

By adding our assumption

$$\begin{aligned} \int _\Sigma f \log f - \int _\Sigma \frac{|\nabla ^\Sigma f|^2}{f} - \int _\Sigma f |H|^2 = 0, \end{aligned}$$

we conclude that

$$\begin{aligned}&\int _\Sigma f \left( \log f + \log P(4b_1 n ) \theta _h + \frac{n}{2}\log (4 \pi ) +n + 4b_1^2 n^2 \right. \\&\left. \quad + (n+m-1) \log \frac{1+b_0}{e^{2r_0b_1 + b_0}}\right) d V (x) \\&\quad - \int _\Sigma \frac{|\nabla ^\Sigma f|^2}{f} - \int _\Sigma f |H|^2 \le \int _\Sigma f d V (x) \left( \log \int _\Sigma f(x) dV (x)\right) . \end{aligned}$$

It now remains to show the case when $\Sigma$ is disconnected. We follow Brendle [5] and notice that

$$\begin{aligned} a\log (a)+b\log (b) <a\log (a+b)+b\log (a+b)=(a+b)\log (a+b). \end{aligned}$$

The above simple inequality will take care of the right hand side when we apply inequality (1) on each connected component individually. And this completes our proof of Theorem 1.2.