Eigenvalue inequalities of elliptic operators in weighted divergence form on smooth metric measure spaces

Abstract

In this paper, we study the eigenvalue problem of elliptic operators in weighted divergence form on smooth metric measure spaces. First of all, we give a general inequality for eigenvalues of the eigenvalue problem of elliptic operators in weighted divergence form on compact smooth metric measure space with boundary (possibly empty). Then applying this general inequality, we get some universal inequalities of Payne-Pólya-Weinberger-Yang type for the eigenvalues of elliptic operators in weighted divergence form on a connected bounded domain in the smooth metric measure spaces, the Gaussian shrinking solitons, and the general product solitons, respectively.

1 Introduction

A smooth metric measure space is actually a Riemannian manifold equipped with some measure which is absolutely continuous with respect to the usual Riemannian measure. More precisely, for a given complete n-dimensional Riemannian manifold $(M,\langle,\rangle)$ with the metric $\langle,\rangle$, the triple $(M,\langle,\rangle,e^{-f}\,d\nu )$ is called a smooth metric measure space, where f is a smooth real-valued function on M and dν is the Riemannian volume element related to $\langle,\rangle$ (sometimes, we also call dν the volume density). Let Ω be a bounded domain in a smooth metric measure space $(M,\langle,\rangle,e^{-f}\,d\nu )$, and let $A: \Omega \rightarrow\operatorname{End}(T\Omega )$ be a smooth symmetric and positive definite section of the bundle of all endomorphisms of TΩ, we can define the elliptic operator in weighted divergence form as

$$\begin{aligned} \mathfrak{L}_{f}=-\operatorname{div}_{f} A\nabla, \end{aligned}$$

(1.1)

where $\operatorname{div}_{f}X=e^{f}\operatorname{div}(e^{-f}X)$ is the weighted divergence of vector fields X, and ∇ is the gradient operator. When A is an identity map, $-\mathfrak{L}_{f}$ becomes the drifting Laplacian $\Delta_{f}$, for the drifting Laplacian, some universal inequalities have been given in [1–5]. When f is a constant, $\mathfrak{L}_{f}$ becomes the elliptic operator in divergence form, for some recent developments about universal inequalities of the eigenvalue of elliptic operator in divergence form on Riemannian manifolds, we refer to [6–10] and the references therein. As briefly mentioned above, it is a natural problem how to get the universal inequalities of the eigenvalues of elliptic operator in weighted divergence form. Actually, in this paper, we first consider the eigenvalue problem as follows:

$$\begin{aligned} \left \{ \textstyle\begin{array}{@{}l@{\quad}l} (\mathfrak{L}_{f}+V)u=\lambda\rho u,& \mbox{in } \Omega , \\ u=0,& \mbox{on } \partial \Omega , \end{array}\displaystyle \right . \end{aligned}$$

(1.2)

where Ω is a bounded domain in a complete smooth metric measure space $(M,\langle,\rangle,e^{-f}\,d\nu )$, V is a non-negative continuous function on M, and ρ is a weight function which is positive and continuous on M. For the eigenvalues of (1.2), we can give the following universal inequalities.

Theorem 1.1

Let Ω be a connected bounded domain in an n-dimensional complete smooth metric measure space $(M,\langle,\rangle, e^{-f}\,d\nu )$. Assume that $\xi_{1} I\leq A, tr(A)\leq n\xi_{2}$ throughout Ω, and $\rho_{1} \leq\rho(x) \leq\rho_{2}, |\nabla f|(x)\leq C_{0}, \forall x \in \Omega$, here I is the identity map, $\xi_{1}, \xi_{2},\rho_{1},\rho_{2}, C_{0}$ are positive constants and $tr(A)$ denotes the trace of A. Let $\lambda_{i}$ be the ith eigenvalue of the eigenvalue problem (1.2), then we have

$$\begin{aligned} &\sum_{i=1}^{k}( \lambda_{k+1}-\lambda_{i})^{2} \\ &\quad\leq\frac{4\xi_{2}\rho_{2}^{2}}{n\rho_{1}^{2}}\sum_{i=1}^{k}( \lambda_{k+1}-\lambda_{i}) \biggl\{ \frac {\lambda_{i}-\rho^{-1}_{2} V_{0}}{\xi_{1}}+C_{0} \biggl( \frac{\lambda_{i}-\rho ^{-1}_{2} V_{0}}{\xi_{1}} \biggr)^{\frac{1}{2}}+ \frac{n^{2}H_{0}^{2}+C_{0}^{2}}{4\rho _{1}} \biggr\} , \end{aligned}$$

(1.3)

where $H_{0}=\sup_{x\in \Omega }\mathbf{|H|}(x), V_{0}=\min_{x\in \Omega }V(x)$, and H is the mean curvature vector field of M in a Euclidean space $\mathbb {R}^{m}$.

Remark 1.2

From inequality (1.3), we can get some results which are given in [5, 8], for example, if f is a constant, then $C_{0}=0$, (1.3) becomes (3.2) in [8].

For the fourth-order elliptic operator in weighted divergence, we can consider the following eigenvalue problem:

$$\begin{aligned} \left \{ \textstyle\begin{array}{@{}l@{\quad}l} \mathfrak{L}_{f}^{2}u =\Lambda u,& \mbox{in } \Omega , \\ u=\frac{\partial u}{\partial\nu}=0,& \mbox{on } \partial \Omega , \end{array}\displaystyle \right . \end{aligned}$$

(1.4)

we also give some universal inequalities for the eigenvalues of (1.4) as follows.

Theorem 1.3

Let Ω be a connected bounded domain in an n-dimensional complete smooth metric measure space $(M,\langle,\rangle, e^{-f}\,d\nu )$. Assume that $\xi_{1} I\leq A\leq\xi_{2} I$ throughout Ω, and $|\nabla f|(x)\leq C_{0}, \forall x \in \Omega$, here I is the identity map, $\xi_{1}, \xi_{2}, C_{0}$ are positive constants. Let $\Lambda_{i}$ be the ith eigenvalue of the eigenvalue problem (1.4), then we have

$$\begin{aligned} \sum_{i=1}^{k}( \Lambda_{k+1}-\Lambda_{i})^{2} \leq{}& \frac{\xi_{2}}{n\xi_{1}} \Biggl\{ \sum_{i=1}^{k}( \Lambda_{k+1}-\Lambda_{i})^{2} \bigl((2n+4) \Lambda_{i}^{\frac{1}{2}}+4C_{0} \xi_{2}^{\frac{1}{2}} \Lambda_{i}^{\frac{1}{4}}+\xi _{2} \bigl(n^{2}H_{0}^{2}+C_{0}^{2} \bigr) \bigr) \\ &{}\times\sum_{i=1}^{k}( \Lambda_{k+1}-\Lambda_{i}) \bigl(4 \Lambda_{i}^{\frac{1}{2}}+4C_{0} \xi_{1}^{\frac{1}{2}} \Lambda_{i}^{\frac {1}{4}}+ \xi_{1} \bigl(n^{2}H_{0}^{2}+C_{0}^{2} \bigr) \bigr) \Biggr\} ^{\frac{1}{2}}, \end{aligned}$$

(1.5)

where $H_{0}=\sup_{x\in \Omega }|\mathbf{H}|(x)$, and H is the mean curvature vector field of M in a Euclidean space $\mathbb {R}^{m}$.

Remark 1.4

From inequality (1.5), we can get some results which are given in [3, 11], for example, if A is an identity map, then $\xi_{1}=\xi_{2}=1$, (1.5) becomes (1.2) in [3].

On smooth metric measure spaces, we can also define the so-called weighted Ricci curvature $\operatorname{Ric}^{f}$ given by

$$\begin{aligned} \operatorname{Ric}^{f}=\operatorname{Ric}+\operatorname{Hess}f, \end{aligned}$$

which is also called the ∞-Bakry-Émery Ricci tensor. The equation $\operatorname{Ric}^{f}=\kappa\langle,\rangle$ for some constant κ is just the gradient Ricci soliton equation, which plays an important role in the study of Ricci flow. We refer the reader to [12] for some recent progress about Ricci solitons. For $\kappa>0, \kappa= 0, \operatorname{or } \kappa< 0$, the gradient Ricci soliton $(M, \langle,\rangle, e^{-f}\,d\nu , \kappa)$ is called shrinking, steady, or expanding, respectively. In the following, we would like to give two examples of Ricci solitons.

Example 1

The Gaussian shrinking soliton $(\mathbb {R}^{n},\langle,\rangle_{\mathrm{can}},e^{-\frac{1}{4}|x|^{2}}\,d\nu ,\frac{1}{2})$, where $\langle,\rangle_{\mathrm{can}}$ is the standard Euclidean metric on $\mathbb {R}^{n}$, $f=\frac{1}{4}|x|^{2}, x\in \mathbb {R}^{n}$, and $\operatorname{Ric}^{f}=\frac{1}{2}\langle,\rangle_{\mathrm{can}}$.

Example 2

More generally, consider the Riemannian product $\Sigma\times \mathbb {R}^{n}$, where $(\Sigma; \langle,\rangle_{\Sigma})$ is an Einstein manifold satisfying $\operatorname{Ric}_{\Sigma}= \kappa \langle,\rangle_{\Sigma}$. Define the smooth function $f : \Sigma\times \mathbb {R}^{n} \rightarrow \mathbb {R}$ by setting $f(p; x) = \frac{\kappa}{2} |x|^{2} + x \cdot a + b$, for any $a \in \mathbb {R}^{n}\setminus \{0\}$ and $b \in \mathbb {R}$. Then $(\Sigma\times \mathbb {R}^{n}, \langle,\rangle_{\Sigma} +\langle,\rangle_{\mathrm{can}}, e^{-f}\,d\nu _{\Sigma}\otimes\,d\nu _{\mathbb {R}}^{n}, \kappa)$ is a non-trivial gradient Ricci soliton. Similarly, one could even construct gradient Ricci solitons with a warped product structure. More details of the product solitons can be found in the Remark 3.2 in [2].

In the following, we will give some universal inequalities for the Dirichlet eigenvalues in a connected bounded domain on the Gaussian shrinking solitons and general product solitons.

Theorem 1.5

Let Ω be a connected bounded domain in the Gaussian shrinking soliton $(\mathbb {R}^{n},\langle,\rangle_{\mathrm{can}},e^{-\frac{1}{4}|x|^{2}}\,d\nu ,\frac {1}{2} )$, and assume that $\xi_{1} I\leq A, tr(A)\leq n\xi_{2}$ throughout Ω, here I is the identity map, $\xi_{1}, \xi_{2}$ are positive constants and $tr(A)$ denotes the trace of A. Let $\lambda_{i}$ be the ith eigenvalue of the Dirichlet problem

$$\begin{aligned} \left \{ \textstyle\begin{array}{@{}l@{\quad}l} \mathfrak{L}_{f}u =\lambda u,& \textit{in } \Omega , \\ u=0,& \textit{on } \partial \Omega , \end{array}\displaystyle \right . \end{aligned}$$

(1.6)

where $f=\frac{1}{4}|x|^{2}$, then we have

$$\begin{aligned} \sum_{i=1}^{k}( \lambda_{k+1}-\lambda_{i})^{2}\leq \frac{4\xi_{2}}{n\xi_{1}}\sum _{i=1}^{k}(\lambda_{k+1}-\lambda_{i}) \biggl\{ \lambda_{i}+ \frac{\xi_{1}}{16} \Bigl(4n-\min_{x\in\overline{\Omega }} \bigl\{ |x|^{2} \bigr\} \Bigr) \biggr\} . \end{aligned}$$

(1.7)

Remark 1.6

(i) For a self-shrinker, the drifting Laplacian $\Delta_{f}$ with $f=\frac{|x|^{2}}{4}$ is actually the operator $\mathfrak{L}:=\Delta -\frac{1}{2}\langle x, \nabla (\cdot)\rangle$, which was introduced by Colding-Minicozzi [13] to study self-shrinker hypersurfaces. For the Dirichlet problem of the operator $\mathfrak{L}$, some universal inequalities have been obtained by Cheng and Peng [14]. In this case, our results can be regarded as conclusions for the Dirichlet problem of the elliptic operator in weighted divergence form.

(ii) Let $b=\frac{\xi_{2}}{\xi_{1}}$, using the recursive formula in Cheng and Yang [15], we can infer from (1.7) that

$$\begin{aligned} \lambda _{k+1}+\frac{\xi_{1}}{16} \Bigl(4n-\min_{x\in\overline{\Omega }} \bigl\{ |x|^{2} \bigr\} \Bigr)\leq C_{0}(n,k)k^{\frac{2b}{n}} \biggl( \lambda _{1}+\frac{\xi_{1}}{16} \Bigl(4n-\min_{x\in \overline{\Omega }} \bigl\{ |x|^{2} \bigr\} \Bigr) \biggr), \end{aligned}$$

(1.8)

where $C_{0}(n,k)\leq1+\frac{4b}{n}$ is a constant (see [15]).

Theorem 1.7

Let Ω be a connected bounded domain in the gradient product Ricci soliton $(\Sigma\times \mathbb {R}, \langle,\rangle, e^{-\frac{\kappa t^{2}}{2}}\,d\nu ,\kappa )$, where Σ is an Einstein manifold with constant Ricci curvature κ. Set $\overline{x}=(x,t)\in \Omega$, where $x\in\Sigma, t\in \mathbb {R}$, and assume that $\xi_{1} I\leq A\leq\xi_{2}I$ throughout Ω, here I is the identity map, $\xi_{1}, \xi_{2}$ are positive constants. Let $\lambda_{i}$ be the ith eigenvalue of the eigenvalue problem (1.6), where $f=\frac{\kappa t^{2}}{2}$, then we have

$$\begin{aligned} \sum_{i=1}^{k}( \lambda_{k+1}-\lambda_{i})^{2}\leq \frac{\xi_{2}}{\xi_{1}}\sum _{i=1}^{k}(\lambda_{k+1}-\lambda_{i}) \Bigl\{ 4 \lambda_{i}+4 \xi_{1}-\kappa^{2}\xi_{1} \min_{(x,t)\in\overline {\Omega }} \bigl\{ |t|^{2} \bigr\} \Bigr\} . \end{aligned}$$

(1.9)

2 A general inequality

In this section, we will prove a general inequality, which will play a key role in the proof of our main results which are listed in Section 1.

Lemma 2.1

Let $(M,\langle,\rangle, e^{-f}\,d\nu )$ be an n-dimensional compact smooth metric measure space with boundary ∂M (possibly empty), and let $a, b$ be the random non-negative constants and $a+b \neq0$. Let $\lambda_{i}$ be the ith eigenvalue of the eigenvalue problem of the fourth-order elliptic operator in weighted divergence form with weight ρ such that

$$\begin{aligned} \left \{ \textstyle\begin{array}{@{}l@{\quad}l} (a\mathfrak{L}_{f}^{2}+b\mathfrak{L}_{f}+V)u=\lambda\rho u,& \textit{in } M, \\ u=\frac{\partial u}{\partial\nu}=0,& \textit{on } \partial M, \end{array}\displaystyle \right . \end{aligned}$$

and $u_{i}$ be the orthonormal eigenfunction corresponding to $\lambda_{i}$, that is,

$$\begin{aligned} \left \{ \textstyle\begin{array}{@{}l@{\quad}l} (a\mathfrak{L}_{f}^{2}+b\mathfrak{L}_{f}+V)u_{i}=\lambda_{i} \rho u_{i},& \textit{in } M, \\ u_{i}=\frac{\partial u_{i}}{\partial\nu}=0,& \textit{on } \partial M,\\ \int_{M} \rho u_{i} u_{j}\,d\mu =\delta_{ij},& \forall i, j=1, 2, \ldots, \end{array}\displaystyle \right . \end{aligned}$$

where $d\mu =e^{-f}\,d\nu$. Then, for any $h\in C^{4}(\overline{M})$, we have

$$\begin{aligned} &\sum_{i=1}^{k}( \lambda_{k+1}-\lambda_{i})^{2} \int_{M} u_{i}^{2}|\nabla h|^{2}\,d \mu \\ &\quad\leq\sum_{i=1}^{k}\delta( \lambda_{k+1}-\lambda_{i})^{2} \int_{M} h u_{i} p_{i}\,d\mu \\ &\qquad{} +\sum _{i=1}^{k}\frac{\lambda_{k+1}-\lambda_{i}}{\delta} \int_{M} \frac{1}{\rho} \biggl(\langle \nabla h, \nabla u_{i}\rangle+\frac{u_{i}\Delta _{f} h}{2} \biggr)^{2}\,d\mu , \end{aligned}$$

(2.1)

where δ is any positive constant and

$$\begin{aligned} p_{i} =&-2a \bigl\langle \nabla h, A\nabla (\mathfrak{L}_{f}u_{i}) \bigr\rangle +a \mathfrak{L}_{f}h \mathfrak{L}_{f}u_{i}-2a\mathfrak{L}_{f}\bigl(\langle \nabla h, A\nabla u_{i}\rangle \bigr) \\ &{}+a \mathfrak{L}_{f}(u_{i} \mathfrak{L}_{f}h)-2b\langle \nabla h, A\nabla u_{i} \rangle+bu_{i} \mathfrak{L}_{f}h. \end{aligned}$$

Proof

Let $\varphi_{i}=hu_{i}-\sum_{j=1}^{k} a_{ij}u_{j}$, here $k\geq 1$ is any integer and $a_{ij}=\sum_{j=1}^{k}\int_{M}\rho hu_{i} u_{j}\,d\mu =a_{ji}$. Then we have

$$\varphi_{i}|_{\partial M}=\frac{\partial\varphi_{i}}{\partial\nu}\bigg|_{\partial M}=0, \quad \mbox{and}\quad \int_{M} \rho\varphi_{i} u_{j}\,d\mu =0, \quad \forall i, j=1,\ldots, k. $$

By the Rayleigh-Ritz inequality, we get

$$\begin{aligned} \lambda_{k+1} \int_{M} \rho\varphi_{i}^{2}\,d\mu \leq \int_{M} \varphi_{i} \bigl(a\mathfrak{L}_{f}^{2}+b\mathfrak{L}_{f}+V \bigr)\varphi _{i}\,d\mu . \end{aligned}$$

(2.2)

From the definition of $\mathfrak{L}_{f}$, we have

$$\begin{aligned} \mathfrak{L}_{f}(hu_{i}) =&-\operatorname{div}_{f}\bigl(A\nabla (hu_{i}) \bigr) \\ =&-e^{f}\operatorname{div} \bigl( e^{-f} \bigl(A(h\nabla u_{i}+u_{i}\nabla h) \bigr) \bigr) \\ =&-\operatorname{div} \bigl( A(h\nabla u_{i}+u_{i}\nabla h) \bigr)- \bigl\langle \nabla f,A(h\nabla u_{i}+u_{i}\nabla h) \bigr\rangle \\ =&-h\operatorname{div}_{f}(A\nabla u_{i})-\langle \nabla h, A\nabla u_{i} \rangle-u_{i}\operatorname{div}_{f}(A\nabla h)-\langle \nabla u_{i}, A\nabla h\rangle \\ =&h \mathfrak{L}_{f}u_{i} -2\langle \nabla h, A\nabla u_{i} \rangle+u_{i} \mathfrak{L}_{f}h \end{aligned}$$

(2.3)

and

$$\begin{aligned} \mathfrak{L}_{f}^{2}(hu_{i}) =&\mathfrak{L}_{f}\bigl(h \mathfrak{L}_{f}u_{i}-2\langle \nabla h, A\nabla u_{i}\rangle+u_{i} \mathfrak{L}_{f}h \bigr) \\ =&h \mathfrak{L}_{f}^{2} u_{i}-2 \bigl\langle \nabla h, A\nabla (\mathfrak{L}_{f}u_{i}) \bigr\rangle +\mathfrak{L}_{f}h \mathfrak{L}_{f}u_{i} \\ &{}-2\mathfrak{L}_{f}\bigl(\langle \nabla h, A\nabla u_{i}\rangle \bigr)+ \mathfrak{L}_{f}(u_{i} \mathfrak{L}_{f}h). \end{aligned}$$

(2.4)

It follows from (2.3) and (2.4) that

$$\begin{aligned} \bigl(a\mathfrak{L}_{f}^{2}+b \mathfrak{L}_{f}+V \bigr) (hu_{i})=\lambda_{i}\rho h u_{i}+p_{i}, \end{aligned}$$

(2.5)

where $p_{i}$ is defined by

$$\begin{aligned} p_{i} =&-2a \bigl\langle \nabla h, A\nabla \bigl(\mathfrak{L}_{f}(u_{i}) \bigr) \bigr\rangle +a\mathfrak{L}_{f}h \mathfrak{L}_{f}u_{i}-2a\mathfrak{L}_{f}\bigl(\langle \nabla h, A\nabla u_{i} \rangle \bigr) \\ &{}+a \mathfrak{L}_{f}\bigl(u_{i} L(h) \bigr)-2b\langle \nabla h, A\nabla u_{i} \rangle+bu_{i} \mathfrak{L}_{f}h. \end{aligned}$$

Let us compute

$$\begin{aligned} &\int_{M} \varphi_{i} \bigl(a\mathfrak{L}_{f}^{2}+b\mathfrak{L}_{f}+V \bigr) \varphi_{i}\,d\mu \\ &\quad= \int_{M} \varphi_{i} \bigl(a\mathfrak{L}_{f}^{2}+b \mathfrak{L}_{f}+V \bigr) (hu_{i})\,d\mu \\ &\quad=\lambda_{i}\int_{M} \varphi_{i}\rho h u_{i}\,d\mu + \int_{M} \varphi_{i} p_{i}\,d\mu \\ &\quad=\lambda_{i}\int_{M} \rho\varphi_{i}^{2}\,d\mu + \int_{M} h u_{i} p_{i}\,d\mu -\sum _{j=1}^{k} a_{ij}b_{ij}, \end{aligned}$$

(2.6)

where $b_{ij}$ is defined by $b_{ij}=\int_{M} p_{i} u_{j}\,d\mu$.

On the other hand, by (2.2) and (2.6), we have

$$\begin{aligned} (\lambda_{k+1}-\lambda_{i}) \int_{M} \rho\varphi_{i}^{2}\,d\mu \leq \int_{M} h u_{i} p_{i}\,d\mu -\sum _{j=1}^{k} a_{ij}b_{ij}. \end{aligned}$$

(2.7)

By a similar computation to (2.8)-(2.12) in [6], we have

$$\begin{aligned} &2 \int_{M} \mathfrak{L}_{f}u_{j}\langle A\nabla h, \nabla u_{i} \rangle \,d\mu -2 \int_{M} \mathfrak{L}_{f}u_{i}\langle A\nabla h, \nabla u_{j} \rangle\,d\mu \\ &\quad=- \int_{M} h u_{i}\mathfrak{L}_{f}^{2} u_{j} \,d\mu + \int_{M} h u_{j}\mathfrak{L}_{f}^{2} u_{i} \,d\mu + \int_{M} u_{i}\mathfrak{L}_{f}u_{j}\mathfrak{L}_{f}h\,d\mu - \int_{M} u_{j}\mathfrak{L}_{f}u_{i}\mathfrak{L}_{f}h\,d\mu , \end{aligned}$$

(2.8)

$$\begin{aligned} & \int_{M} u_{j} \mathfrak{L}_{f}\langle \nabla h, A\nabla u_{i} \rangle \,d\mu + \int_{M} u_{j} \bigl\langle \nabla h, A\nabla (\mathfrak{L}_{f}u_{i}) \bigr\rangle \,d\mu \\ &\quad= \int_{M} \mathfrak{L}_{f}u_{j} \langle \nabla h, A\nabla u_{i} \rangle\,d\mu - \int_{M} \mathfrak{L}_{f}u_{i} \langle \nabla h, A\nabla u_{j} \rangle\,d\mu + \int_{M} u_{j}\mathfrak{L}_{f}u_{i}\mathfrak{L}_{f}h\,d\mu , \end{aligned}$$

(2.9)

$$\begin{aligned} & \int_{M} u_{j}\mathfrak{L}_{f}(u_{i}\mathfrak{L}_{f}h)\,d\mu = \int_{M} u_{i}\mathfrak{L}_{f}u_{j}\mathfrak{L}_{f}h\,d\mu , \end{aligned}$$

(2.10)

and

$$\begin{aligned} \int_{M} u_{j} \bigl\{ -2\langle \nabla h, A\nabla u_{i} \rangle+u_{i} \mathfrak{L}_{f}h \bigr\} \,d\mu = \int_{M} hu_{i}\mathfrak{L}_{f}u_{j} \,d\mu - \int_{M} hu_{j}\mathfrak{L}_{f}u_{i}\,d\mu . \end{aligned}$$

(2.11)

Combining (2.8)-(2.11) and a similar calculation to (2.13) in [6], we get

$$\begin{aligned} b_{ij}= \int_{M} p_{i} u_{j}\,d\mu =(\lambda_{j}- \lambda_{i})a_{ij}. \end{aligned}$$

(2.12)

We infer from (2.7) and (2.12) that

$$\begin{aligned} (\lambda_{k+1}-\lambda_{i}) \int_{M} \rho\varphi_{i}^{2}\,d\mu \leq \int_{M} h u_{i} p_{i}\,d\mu -\sum _{j=1}^{k}(\lambda_{j}-\lambda_{i}) a^{2}_{ij}. \end{aligned}$$

(2.13)

Setting $t_{ij}=\int_{M} u_{j}(\langle \nabla h, \nabla u_{i}\rangle+\frac{u_{i}\Delta _{f} h}{2})\,d\mu$, thus $c_{ij}=-c_{ji}$ and

$$\begin{aligned} & \int_{M} -2\varphi_{i} \biggl(\langle \nabla h, \nabla u_{i}\rangle+\frac{u_{i}\Delta _{f} h}{2} \biggr)\,d\mu \\ &\quad= \int_{M} \bigl(-2hu_{i}\langle \nabla h, \nabla u_{i} \rangle-hu^{2}_{i}\Delta _{f} h \bigr) \,d\mu +2\sum_{j=1}^{k} a_{ij}c_{ij} \\ &\quad= \int_{M} u_{i}^{2}|\nabla h|^{2}\,d \mu +2\sum_{j=1}^{k} a_{ij}c_{ij}. \end{aligned}$$

(2.14)

Using (2.13), (2.14), and the Schwarz inequality, we can get

$$\begin{aligned} &(\lambda_{k+1}-\lambda_{i})^{2} \Biggl( \int_{M} u_{i}^{2}|\nabla h|^{2}\,d \mu +2\sum_{j=1}^{k} a_{ij}c_{ij} \Biggr) \\ &\quad=(\lambda_{k+1}-\lambda_{i})^{2} \int_{M} -2\sqrt{\rho} \varphi_{i} \Biggl( \frac{1}{\sqrt{\rho}} \biggl(\langle \nabla h, \nabla u_{i}\rangle+ \frac{u_{i}\Delta _{f} h}{2} \biggr)-\sum_{j=1}^{k} c_{ij}\sqrt{ \rho}u_{j} \Biggr)\,d\mu \\ &\quad\leq\delta(\lambda_{k+1}-\lambda_{i})^{3} \int_{M} \rho \varphi_{i}^{2}\,d\mu \\ &\qquad{} + \frac{\lambda_{k+1}-\lambda_{i}}{\delta} \int_{M} \Biggl( \frac{1}{\sqrt{\rho}} \biggl(\langle \nabla h, \nabla u_{i}\rangle+ \frac{u_{i}\Delta _{f} h}{2} \biggr)-\sum _{j=1}^{k} c_{ij}\sqrt{\rho}u_{j} \Biggr)^{2}\,d\mu \\ &\quad\leq\delta(\lambda_{k+1}-\lambda_{i})^{2} \Biggl( \int_{M} h u_{i} p_{i}\,d\mu -\sum _{j=1}^{k}(\lambda_{j}-\lambda_{i}) a^{2}_{ij} \Biggr) \\ &\qquad{}+\frac{\lambda_{k+1}-\lambda_{i}}{\delta} \Biggl( \int_{M} \frac{1}{\rho} \biggl(\langle \nabla h, \nabla u_{i}\rangle+\frac{u_{i}\Delta _{f} h}{2} \biggr)^{2}-\sum _{j=1}^{k} c_{ij}^{2} \Biggr), \end{aligned}$$

(2.15)

where δ is any positive constant. Summing over i from 1 to k in (2.15) and noticing $a_{ij}=a_{ji}, c_{ij}=-c_{ji}$, we have

$$\begin{aligned} &\sum_{i=1}^{k}( \lambda_{k+1}-\lambda_{i})^{2} \int_{M} u_{i}^{2}|\nabla h|^{2}\,d \mu -2\sum_{i,j=1}^{k}(\lambda_{k+1}- \lambda_{i}) (\lambda_{i}-\lambda_{j}) a_{ij}c_{ij} \\ &\quad\leq\sum_{i=1}^{k}\delta( \lambda_{k+1}-\lambda_{i})^{2} \int_{M} h u_{i} p_{i}\,d\mu +\sum _{i=1}^{k}\frac{\lambda_{k+1}-\lambda_{i}}{\delta} \int_{M} \frac{1}{\rho} \biggl(\langle \nabla h, \nabla u_{i}\rangle+\frac{u_{i}\Delta _{f} h}{2} \biggr)^{2}\,d\mu \\ &\qquad{}-\sum_{i,j=1}^{k}\delta( \lambda_{k+1}-\lambda_{i}) (\lambda_{j}-\lambda_{i})^{2} a^{2}_{ij}- \sum_{i,j=1}^{k}\frac{\lambda_{k+1}-\lambda_{i}}{\delta}c_{ij}^{2}, \end{aligned}$$

(2.16)

which implies that

$$\begin{aligned} &\sum_{i=1}^{k}(\lambda_{k+1}- \lambda_{i})^{2} \int_{M} u_{i}^{2}| \nabla h|^{2} \,d\mu \\ &\quad\leq\sum_{i=1}^{k}\delta( \lambda_{k+1}-\lambda_{i})^{2} \int_{M} h u_{i} p_{i}\,d\mu +\sum _{i=1}^{k}\frac{\lambda_{k+1}-\lambda_{i}}{\delta} \int_{M} \frac{1}{\rho} \biggl(\langle \nabla h, \nabla u_{i}\rangle+\frac{u_{i}\Delta _{f} h}{2} \biggr)^{2}\,d\mu . \end{aligned}$$

This completes the proof of Lemma 2.1. □

3 Proof of Theorem 1.1 and Theorem 1.3

In this section, we will give the proof of Theorem 1.1 and Theorem 1.3 by using Lemma 2.1.

Proof of Theorem 1.1

From the Nash embedding theorem, we know that there exists an isometric immersion from a complete Riemannian manifold M into a Euclidean space $\mathbb {R}^{m}$. Thus, M can be considered as an n-dimensional complete isometrically immersed submanifold in $\mathbb {R}^{m}$. Let $y_{1}, y_{2}, \ldots, y_{m}$ be the standard coordinate functions of $\mathbb {R}^{m}$. Then we have

$$\begin{aligned}& \sum_{\alpha=1}^{m}|\nabla y_{\alpha }|^{2}=n, \end{aligned}$$

(3.1)

$$\begin{aligned}& \Delta (y_{1}, y_{2},\ldots, y_{m})=(\Delta y_{1}, \Delta y_{2}, \ldots, \Delta y_{m})=n \mathbf{H}, \end{aligned}$$

(3.2)

$$\begin{aligned}& \sum_{\alpha=1}^{m}\langle \nabla y_{\alpha }, \nabla f\rangle^{2}=\sum_{\alpha=1}^{m} \bigl(\nabla f(y_{\alpha }) \bigr)^{2}=|\nabla f|^{2}, \end{aligned}$$

(3.3)

$$\begin{aligned}& \sum_{\alpha=1}^{m}\langle \nabla y_{\alpha }, \nabla u_{i}\rangle^{2}=\sum_{\alpha=1}^{m} \bigl( \nabla u_{i}(y_{\alpha }) \bigr)^{2}=|\nabla u_{i}|^{2}, \end{aligned}$$

(3.4)

$$\begin{aligned}& \sum_{\alpha=1}^{m}\langle \nabla y_{\alpha }, \nabla u_{i}\rangle \langle \nabla y_{\alpha },\nabla f\rangle=\sum _{\alpha=1}^{m}\nabla u_{i}(y_{\alpha })\nabla f(y_{\alpha })=\langle \nabla u_{i},\nabla f\rangle, \end{aligned}$$

(3.5)

$$\begin{aligned}& \sum_{\alpha=1}^{m}\Delta y_{\alpha }\langle \nabla y_{\alpha },\nabla u_{i}\rangle=\sum_{\alpha=1}^{m} \Delta y_{\alpha }\nabla u_{i}(y_{\alpha })=\langle n\mathbf{H},\nabla u_{i}\rangle=0, \end{aligned}$$

(3.6)

and

$$\begin{aligned} \sum_{\alpha=1}^{m}\Delta y_{\alpha }\langle \nabla y_{\alpha },\nabla f\rangle=\sum _{\alpha=1}^{m}\Delta y_{\alpha }\nabla f(y_{\alpha })= \langle n\mathbf{H},\nabla f\rangle=0. \end{aligned}$$

(3.7)

Then we infer from (3.3)-(3.7) that

$$\begin{aligned} \sum_{\alpha=1}^{m} \Delta _{f} y_{\alpha }\langle \nabla y_{\alpha },\nabla u_{i} \rangle=\sum_{\alpha=1}^{m} \bigl(\Delta y_{\alpha }-\langle \nabla y_{\alpha }, \nabla f\rangle \bigr)\langle \nabla y_{\alpha },\nabla u_{i}\rangle=\langle \nabla u_{i},\nabla f \rangle \end{aligned}$$

(3.8)

and

$$\begin{aligned} \sum_{\alpha=1}^{m}( \Delta _{f} y_{\alpha })^{2} =&\sum _{\alpha=1}^{m} \bigl(\Delta y_{\alpha }-\langle \nabla y_{\alpha },\nabla f\rangle \bigr)^{2} \\ =&\sum_{\alpha=1}^{m} \bigl((\Delta y_{\alpha })^{2}-2\Delta y_{\alpha }\langle \nabla y_{\alpha },\nabla f\rangle +\langle \nabla y_{\alpha },\nabla f \rangle^{2} \bigr) \\ =&n^{2}|\mathbf{H}|^{2}+|\nabla f|^{2}. \end{aligned}$$

(3.9)

Let $a=0, b=1$ in (2.1), then taking $h=y_{\alpha }$ and summing over α, and noticing $\rho^{-1}_{2}\leq\|u_{i}\|^{2}\leq\rho^{-1}_{1}$, we get

$$\begin{aligned} n\rho^{-1}_{2}\sum _{i=1}^{k}(\lambda_{k+1}- \lambda_{i})^{2} \leq&\sum_{i=1}^{k}\delta( \lambda_{k+1}-\lambda_{i})^{2} \int_{\Omega}\sum_{\alpha=1}^{m} y_{\alpha }u_{i} p_{\alpha i}\,d\mu \\ &{}+\sum_{i=1}^{k}\frac{\lambda_{k+1}-\lambda_{i}}{\delta} \int_{\Omega}\frac{1}{\rho}\sum_{\alpha=1}^{m} \biggl(\langle \nabla y_{\alpha }, \nabla u_{i}\rangle+\frac{u_{i}\Delta _{f} y_{\alpha }}{2} \biggr)^{2}\,d\mu , \end{aligned}$$

(3.10)

where $p_{\alpha i}=-2\langle \nabla y_{\alpha }, A\nabla u_{i}\rangle+u_{i} \mathfrak{L}_{f}y_{\alpha }$. Since

$$-2 \int_{\Omega}y_{\alpha }u_{i}\langle \nabla y_{\alpha }, A\nabla u_{i}\rangle\,d\mu = \int_{\Omega}u_{i}^{2}\langle \nabla y_{\alpha },A \nabla y_{\alpha }\rangle\,d\mu - \int_{\Omega}y_{\alpha }u_{i}^{2}\mathfrak{L}_{f}y_{\alpha }\,d\mu , $$

we infer from above equality and $\sum_{\alpha=1}^{m}\langle \nabla y_{\alpha },A \nabla y_{\alpha }\rangle =tr(A)\leq n\xi_{2}$ that

$$\begin{aligned} \int_{\Omega}\sum_{\alpha=1}^{m} y_{\alpha }u_{i} p_{\alpha i}\,d\mu =& \int_{\Omega}\sum_{\alpha=1}^{m} y_{\alpha }u_{i} \bigl(-2\langle \nabla y_{\alpha }, A\nabla u_{i}\rangle+u_{i} \mathfrak{L}_{f}y_{\alpha }\bigr)\,d\mu \\ =& \int_{\Omega}\sum_{\alpha=1}^{m} u_{i}^{2}\langle \nabla y_{\alpha },A \nabla y_{\alpha }\rangle\,d\mu \leq n\xi_{2}\|u_{i} \|^{2}\leq n \xi_{2}\rho^{-1}_{1}. \end{aligned}$$

(3.11)

From $\rho^{-1}_{2}\leq\|u_{i}\|^{2}\leq\rho^{-1}_{1}$ and $A\geq\xi_{1} I$, we have

$$\begin{aligned} \lambda_{i} =& \int_{\Omega}u_{i}(\mathfrak{L}_{f}+V) u_{i}\,d\mu = \int_{\Omega}-u_{i}\operatorname{div}_{f}(A\nabla u_{i})\,d\mu + \int_{\Omega}V u_{i}^{2}\,d\mu \\ =& \int_{\Omega}\langle \nabla u_{i},A\nabla u_{i} \rangle\,d\mu + \int_{\Omega}V u_{i}^{2}\,d\mu \geq \xi_{1}\|\nabla u_{i}\|^{2}+ \rho^{-1}_{2} V_{0}, \end{aligned}$$

which implies

$$\begin{aligned} \|\nabla u_{i}\|^{2}\leq\frac{\lambda_{i}-\rho^{-1}_{2} V_{0}}{\xi_{1}}. \end{aligned}$$

(3.12)

Using the Schwarz inequality and the above inequality, we have

$$\begin{aligned} \int_{\Omega}\langle \nabla f, \nabla u_{i}\rangle\,d\mu \leq \int_{\Omega}|\nabla f||\nabla u_{i}|\,d\mu \leq C_{0} \bigl\{ \|\nabla u_{i}\| ^{2} \bigr\} ^{\frac{1}{2}}\leq C_{0} \biggl( \frac{\lambda_{i}-\rho^{-1}_{2} V_{0}}{\xi_{1}} \biggr)^{\frac{1}{2}}. \end{aligned}$$

(3.13)

Combining (3.4), (3.8), (3.9), (3.12), and (3.13), we have

$$\begin{aligned} & \int_{\Omega}\frac{1}{\rho}\sum_{\alpha=1}^{m} \biggl(\langle \nabla y_{\alpha }, \nabla u_{i}\rangle+\frac{u_{i}\Delta _{f} y_{\alpha }}{2} \biggr)^{2}\,d\mu \\ &\quad= \int_{\Omega}\frac{1}{\rho}\sum_{\alpha=1}^{m} \biggl(\langle \nabla y_{\alpha }, \nabla u_{i}\rangle^{2}+u_{i} \Delta _{f} y_{\alpha }\langle \nabla y_{\alpha }, \nabla u_{i} \rangle+ \frac{u_{i}^{2}(\Delta _{f} y_{\alpha })^{2}}{4} \biggr)\,d\mu \\ &\quad= \int_{\Omega}\frac{1}{\rho} \biggl(|\nabla u_{i}|^{2}+ \langle \nabla f, \nabla u_{i}\rangle+\frac{u_{i}^{2}}{4} \bigl(n^{2}| \mathbf{H}|^{2}+|\nabla f|^{2} \bigr) \biggr)\,d\mu \\ &\quad\leq\frac{1}{\rho_{1}} \biggl\{ \frac{\lambda_{i}-\rho^{-1}_{2} V_{0}}{\xi _{1}}+C_{0} \biggl( \frac{\lambda_{i}-\rho^{-1}_{2} V_{0}}{\xi_{1}} \biggr)^{\frac {1}{2}}+\frac{1}{4\rho_{1}} \bigl(n^{2}H_{0}^{2}+C_{0}^{2} \bigr) \biggr\} . \end{aligned}$$

(3.14)

Substituting (3.11) and (3.14) into (3.10), we have

$$\begin{aligned} \sum_{i=1}^{k}(\lambda_{k+1}- \lambda_{i})^{2} \leq&\sum_{i=1}^{k} \delta( \lambda_{k+1}-\lambda_{i})^{2}\frac{\rho_{2}\xi_{2}}{\rho_{1}} \\ &{}+\sum_{i=1}^{k} \frac{\lambda_{k+1}-\lambda_{i}}{\delta} \frac{\rho_{2}}{n\rho_{1}} \biggl\{ \frac{\lambda_{i}-\rho^{-1}_{2} V_{0}}{\xi _{1}}+C_{0} \biggl( \frac{\lambda_{i}-\rho^{-1}_{2} V_{0}}{\xi_{1}} \biggr)^{\frac {1}{2}}\\ &{}+\frac{n^{2}H_{0}^{2}+C_{0}^{2}}{4\rho_{1}} \biggr\} . \end{aligned}$$

Taking

$$\begin{aligned} \delta= \biggl\{ \frac{\sum_{i=1}^{k}(\lambda_{k+1}-\lambda_{i}) \{\frac{\lambda _{i}-\rho^{-1}_{2} V_{0}}{\xi_{1}}+C_{0} ( \frac{\lambda_{i}-\rho^{-1}_{2} V_{0}}{\xi_{1}} )^{\frac{1}{2}}+\frac{n^{2}H_{0}^{2}+C_{0}^{2}}{4\rho_{1}} \} }{\sum_{i=1}^{k}(\lambda_{k+1}-\lambda_{i})^{2} n\xi_{2}} \biggr\} ^{\frac{1}{2}}, \end{aligned}$$

we can get (3.1). This completes the proof of Theorem 1.1. □

Proof of Theorem 1.3

Let $a=1,b=0,V\equiv0,\rho \equiv1$ in (2.1), then taking $h=y_{\alpha }$ and summing over α, where $\{y_{\alpha }\}_{\alpha =1}^{m}$ are defined as above, we get

$$\begin{aligned} n\sum_{i=1}^{k}( \Lambda_{k+1}-\Lambda_{i})^{2} \leq&\sum _{i=1}^{k}\delta(\Lambda_{k+1}- \Lambda_{i})^{2} \int_{\Omega}\sum_{\alpha=1}^{m} y_{\alpha }u_{i} p_{\alpha i}\,d\mu \\ &{}+\sum_{i=1}^{k}\frac{\Lambda_{k+1}-\Lambda_{i}}{\delta} \int_{\Omega}\sum_{\alpha=1}^{m} \biggl(\langle \nabla y_{\alpha }, \nabla u_{i}\rangle+\frac{u_{i}\Delta _{f} y_{\alpha }}{2} \biggr)^{2}\,d\mu , \end{aligned}$$

(3.15)

where $p_{\alpha i}=-2\langle \nabla y_{\alpha }, A\nabla (\mathfrak{L}_{f}u_{i}) \rangle+\mathfrak{L}_{f}y_{\alpha }\mathfrak{L}_{f}u_{i}-2\mathfrak{L}_{f}(\langle \nabla y_{\alpha }, A\nabla u_{i}\rangle)+ \mathfrak{L}_{f}(u_{i} \mathfrak{L}_{f}y_{\alpha })$. By a direct computation, we have

$$\begin{aligned} &\int_{\Omega}y_{\alpha }u_{i}p_{\alpha i}\,d \mu \\ &\quad= \int_{\Omega}y_{\alpha }u_{i} \bigl\{ -2 \bigl\langle \nabla y_{\alpha }, A\nabla (\mathfrak{L}_{f}u_{i}) \bigr\rangle +\mathfrak{L}_{f}y_{\alpha }\mathfrak{L}_{f}u_{i}-2\mathfrak{L}_{f}\bigl( \langle \nabla y_{\alpha }, A\nabla u_{i}\rangle \bigr) \\ &\qquad{}+ \mathfrak{L}_{f}(u_{i} \mathfrak{L}_{f}y_{\alpha }) \bigr\} \,d\mu \\ &\quad= \int_{\Omega}2 \bigl\{ u_{i}\mathfrak{L}_{f}u_{i}\langle \nabla y_{\alpha }, A \nabla y_{\alpha }\rangle+y_{\alpha }\mathfrak{L}_{f}u_{i}\langle \nabla u_{i}, A\nabla y_{\alpha }\rangle- y_{\alpha }u_{i} \mathfrak{L}_{f}y_{\alpha }\mathfrak{L}_{f}u_{i} \bigr\} \,d\mu \\ &\qquad{}+ \int_{\Omega}y_{\alpha }u_{i} \mathfrak{L}_{f}y_{\alpha }\mathfrak{L}_{f}u_{i}\,d\mu \\ &\qquad{}+ \int_{\Omega}\bigl\{ \mathfrak{L}_{f}y_{\alpha }u_{i}+y_{\alpha }\mathfrak{L}_{f}u_{i}-2\langle \nabla y_{\alpha },A\nabla u_{i}\rangle \bigr\} \bigl\{ -2\langle \nabla y_{\alpha }, A\nabla u_{i}\rangle+ u_{i} \mathfrak{L}_{f}y_{\alpha }\bigr\} \,d\mu \\ &\quad= \int_{\Omega}2 u_{i}\mathfrak{L}_{f}u_{i}\langle \nabla y_{\alpha }, A\nabla y_{\alpha }\rangle\,d\mu + \int_{\Omega}4 \langle \nabla y_{\alpha }, A\nabla u_{i} \rangle^{2}\,d\mu \\ &\qquad{}- \int_{\Omega}4 u_{i} \mathfrak{L}_{f}y_{\alpha }\langle \nabla y_{\alpha }, A\nabla u_{i}\rangle\,d\mu + \int_{\Omega}(u_{i}\mathfrak{L}_{f}y_{\alpha })^{2} \,d\mu . \end{aligned}$$

(3.16)

Since $\xi_{1} I\leq A\leq\xi_{2} I$, we can infer from (3.1)-(3.9) that

$$\begin{aligned}& \begin{aligned}[b] \sum_{\alpha=1}^{m} \int_{\Omega}2 u_{i}\mathfrak{L}_{f}u_{i}\langle \nabla y_{\alpha }, A\nabla y_{\alpha }\rangle\,d\mu &\leq2n\xi_{2} \int_{\Omega}u_{i}\mathfrak{L}_{f}u_{i}\,d\mu \\ &\leq2n\xi_{2} \bigl\{ \|u_{i}\|^{2}\|\mathfrak{L}_{f}u_{i}\|^{2} \bigr\} ^{\frac{1}{2}}=2n\xi_{2} \Lambda_{i}^{\frac{1}{2}}, \end{aligned} \end{aligned}$$

(3.17)

$$\begin{aligned}& \begin{aligned}[b] \sum_{\alpha=1}^{m} \int_{\Omega}4 \langle \nabla y_{\alpha }, A\nabla u_{i} \rangle^{2}\,d\mu &=4\|A\nabla u_{i}\|^{2}\leq4 \xi_{2} \int_{\Omega}\langle \nabla u_{i}, A\nabla u_{i} \rangle \\ &=4\xi_{2} \int_{\Omega}u_{i}\mathfrak{L}_{f}u_{i}\,d\mu \leq4 \xi_{2} \Lambda_{i}^{\frac{1}{2}}, \end{aligned} \end{aligned}$$

(3.18)

$$\begin{aligned}& \begin{aligned}[b] &\sum_{\alpha=1}^{m}- \int_{\Omega}4 u_{i} \mathfrak{L}_{f}y_{\alpha }\langle \nabla y_{\alpha }, A\nabla u_{i}\rangle\,d\mu \\ &\quad\leq \biggl\vert 4 \xi_{2} \int_{\Omega}u_{i} \Delta_{f} y_{\alpha }\langle \nabla y_{\alpha }, A\nabla u_{i}\rangle\,d\mu \biggr\vert \\ &\quad= \biggl\vert 4\xi_{2} \int_{\Omega}u_{i} \langle \nabla f, A\nabla u_{i} \rangle\,d\mu \biggr\vert \\ &\quad\leq4\xi_{2} \int_{\Omega}u_{i} |\nabla f||\nabla u_{i}|\,d\mu \\ &\quad\leq4C_{0}\xi_{2} \bigl\{ \|u_{i} \|^{2}\|A\nabla u_{i}\|^{2} \bigr\} ^{\frac{1}{2}} \\ &\quad=4C_{0}\xi_{2}^{\frac{3}{2}} \biggl\{ \int_{\Omega}\langle \nabla u_{i},A\nabla u_{i} \rangle\,d\mu \biggr\} ^{\frac{1}{2}} \leq4C_{0}\xi_{2}^{\frac{3}{2}}\Lambda _{i}^{\frac{1}{4}} , \end{aligned} \end{aligned}$$

(3.19)

and

$$\begin{aligned} \sum_{\alpha=1}^{m} \int_{\Omega}(u_{i}\mathfrak{L}_{f}y_{\alpha })^{2} \,d\mu \leq&\xi_{2}^{2}\sum_{\alpha=1}^{m} \int_{\Omega}u_{i}^{2}( \Delta_{f} y_{\alpha })^{2}\,d\mu \\ =&\xi_{2}^{2} \int_{\Omega}u_{i}^{2} \bigl(n^{2}| \mathbf{H}|^{2}+| \nabla f|^{2} \bigr)\,d\mu \\ \leq& \xi _{2}^{2} \bigl(n^{2}H_{0}^{2}+C_{0}^{2} \bigr). \end{aligned}$$

(3.20)

Combining (3.16)-(3.20), we have

$$\begin{aligned} \sum_{\alpha=1}^{m} \int_{\Omega}y_{\alpha }u_{i}p_{\alpha i}\,d \mu \leq \xi_{2} \bigl((2n+4) \Lambda_{i}^{\frac {1}{2}}+4C_{0} \xi_{2}^{\frac{1}{2}}\Lambda_{i}^{\frac{1}{4}}+ \xi_{2} \bigl(n^{2}H_{0}^{2}+C_{0}^{2} \bigr) \bigr). \end{aligned}$$

(3.21)

Since $\|\nabla u_{i}\|^{2}\leq\frac{1}{\xi_{1}}\int_{\Omega}\langle \nabla u_{i}, A\nabla u_{i}\rangle \,d\mu =\frac{1}{\xi_{1}}\int_{\Omega}u_{i}\mathfrak{L}_{f}u_{i}\,d\mu \leq\frac{\Lambda_{i}^{\frac {1}{2}}}{\xi_{1}}$, then from (3.14), we have

$$\begin{aligned} & \int_{\Omega}\sum_{\alpha=1}^{m} \biggl(\langle \nabla y_{\alpha }, \nabla u_{i} \rangle+\frac{u_{i}\Delta _{f} y_{\alpha }}{2} \biggr)^{2}\,d\mu \\ &\quad= \int_{\Omega}\biggl(|\nabla u_{i}|^{2}+\langle \nabla f, \nabla u_{i}\rangle+\frac{u_{i}^{2}}{4} \bigl(n^{2}| \mathbf{H}|^{2}+|\nabla f|^{2} \bigr) \biggr)\,d\mu \\ &\quad\leq \frac{\Lambda_{i}^{\frac{1}{2}}}{\xi_{1}} +\frac{C_{0}\Lambda_{i}^{\frac {1}{4}}}{\xi_{1}^{\frac{1}{2}}}+\frac{n^{2}H_{0}^{2}+C_{0}^{2}}{4}. \end{aligned}$$

(3.22)

Taking (3.21) and (3.22) into (3.15), we have

$$\begin{aligned} n\sum_{i=1}^{k}( \Lambda_{k+1}-\Lambda_{i})^{2} \leq&\sum _{i=1}^{k}\delta(\Lambda_{k+1}- \Lambda_{i})^{2} \xi_{2} \bigl((2n+4) \Lambda_{i}^{\frac {1}{2}}+4C_{0} \xi_{2}^{\frac{1}{2}}\Lambda_{i}^{\frac{1}{4}}+\xi _{2} \bigl(n^{2}H_{0}^{2}+C_{0}^{2} \bigr) \bigr) \\ &{}+\sum_{i=1}^{k}\frac{\Lambda_{k+1}-\Lambda_{i}}{\delta} \biggl(\frac{\Lambda_{i}^{\frac{1}{2}}}{\xi_{1}}+\frac{C_{0}\Lambda_{i}^{\frac {1}{4}}}{\xi_{1}^{\frac{1}{2}}}+ \frac{n^{2}H_{0}^{2}+C_{0}^{2}}{4} \biggr). \end{aligned}$$

(3.23)

Let

$$\begin{aligned} \delta= \biggl\{ \frac{\sum_{i=1}^{k}(\Lambda_{k+1}-\Lambda_{i}) \{\frac{\Lambda _{i}^{\frac{1}{2}}}{\xi_{1}}+\frac{C_{0}\Lambda_{i}^{\frac{1}{4}}}{\xi_{1}^{\frac {1}{2}}}+\frac{n^{2}H_{0}^{2}+C_{0}^{2}}{4} \}}{\sum_{i=1}^{k}(\Lambda_{k+1}-\Lambda_{i})^{2} \xi_{2}((2n+4) \Lambda_{i}^{\frac {1}{2}}+4C_{0}\xi_{2}^{\frac{1}{2}}\Lambda_{i}^{\frac{1}{4}}+\xi _{2}(n^{2}H_{0}^{2}+C_{0}^{2}))} \biggr\} ^{\frac{1}{2}}, \end{aligned}$$

we can infer from (3.23) that

$$\begin{aligned} \sum_{i=1}^{k}( \Lambda_{k+1}-\Lambda_{i})^{2} \leq& \frac{\xi_{2}}{n\xi_{1}} \Biggl\{ \sum_{i=1}^{k}( \Lambda_{k+1}-\Lambda_{i})^{2} \bigl((2n+4) \Lambda_{i}^{\frac{1}{2}}+4C_{0} \xi_{2}^{\frac{1}{2}}\Lambda_{i}^{\frac{1}{4}}+\xi _{2} \bigl(n^{2}H_{0}^{2}+C_{0}^{2} \bigr) \bigr) \\ &{}\times\sum_{i=1}^{k}( \Lambda_{k+1}-\Lambda_{i}) \bigl(4 \Lambda_{i}^{\frac{1}{2}}+4C_{0} \xi_{1}^{\frac{1}{2}} \Lambda_{i}^{\frac {1}{4}}+ \xi_{1} \bigl(n^{2}H_{0}^{2}+C_{0}^{2} \bigr) \bigr) \Biggr\} ^{\frac{1}{2}}, \end{aligned}$$

(3.24)

this completes the proof of Theorem 1.3. □

4 Proof of Theorem 1.5 and Theorem 1.7

In this section, applying Lemma 2.1, we will give the proof of Theorem 1.5 and Theorem 1.7.

Proof of Theorem 1.5

Let $a=0, b=1,V\equiv0, \rho \equiv1$ in (2.1), then taking $h=x_{\alpha }$ and summing over α, where $\{x_{\alpha }\}_{\alpha =1}^{n}$ are the coordinate functions of $\mathbb {R}^{n}$, we have

$$\begin{aligned} n\sum_{i=1}^{k}( \lambda_{k+1}-\lambda_{i})^{2} \leq&\sum_{i=1}^{k} \delta( \lambda_{k+1}-\lambda_{i})^{2} \int_{\Omega}\sum_{\alpha =1}^{n} x_{\alpha }u_{i} p_{\alpha i}\,d\mu \\ &{}+\sum_{i=1}^{k}\frac{\lambda_{k+1}-\lambda_{i}}{\delta} \sum _{\alpha =1}^{n} \biggl(\langle \nabla x_{\alpha }, \nabla u_{i}\rangle+\frac{u_{i}\Delta _{f} x_{\alpha }}{2} \biggr)^{2}\,d\mu , \end{aligned}$$

(4.1)

where $p_{\alpha i}=-2\langle \nabla x_{\alpha }, A\nabla u_{i}\rangle+u_{i} \mathfrak{L}_{f}x_{\alpha }$.

By a similar computation to (3.11), we have

$$\begin{aligned} \int_{\Omega}\sum_{\alpha =1}^{n} x_{\alpha }u_{i} p_{\alpha i}\,d\mu \leq n\xi_{2}. \end{aligned}$$

(4.2)

Since $f=\frac{|x|^{2}}{4}$, we have

$$\Delta _{f} x_{\alpha }=\Delta x_{\alpha }- \biggl\langle \nabla \biggl( \frac{|x|^{2}}{4} \biggr), \nabla x_{\alpha }\biggr\rangle =- \frac{1}{2}x_{\alpha }, $$

hence, we infer from the above equality that

$$ \begin{aligned}[b] &\int_{\Omega}\sum_{\alpha =1}^{n} \biggl( \langle \nabla x_{\alpha }, \nabla u_{i}\rangle+\frac{u_{i}\Delta _{f} x_{\alpha }}{2} \biggr)^{2}\,d\mu\\ &\quad= \int_{\Omega}\sum_{\alpha =1}^{n} \biggl(\langle \nabla x_{\alpha }, \nabla u_{i}\rangle^{2}+u_{i} \Delta _{f} x_{\alpha }\langle \nabla x_{\alpha }, \nabla u_{i} \rangle+\frac{u_{i}^{2}(\Delta _{f} x_{\alpha })^{2}}{4} \biggr)\,d\mu\\ &\quad= \int_{\Omega}\Biggl(|\nabla u_{i}|^{2}+\sum _{\alpha =1}^{n} x_{\alpha }u_{i} \frac{\partial u_{i}}{\partial x_{\alpha }}+\sum_{\alpha =1}^{n} \frac{x_{\alpha }^{2}u_{i}^{2}}{16} \Biggr)\,d\mu . \end{aligned} $$

(4.3)

From integration by parts, we have

$$\begin{aligned} &\sum_{\alpha =1}^{n} \int_{\Omega }u_{i} x_{\alpha }\frac{\partial u_{i}}{\partial x_{\alpha }}e^{-\frac{|x|^{2}}{4}}\,d\nu \\ &\quad=\sum_{\alpha =1}^{n} \biggl\{ - \int_{\Omega }u_{i}^{2}e^{-\frac{|x|^{2}}{4}}\,d\nu - \int_{\Omega }u_{i} x_{\alpha }\frac{\partial u_{i}}{\partial x_{\alpha }}e^{-\frac{|x|^{2}}{4}}\,d\nu +\frac{1}{2} \int_{\Omega }u_{i}^{2} x_{\alpha }^{2}e^{-\frac{|x|^{2}}{4}} \,d\nu \biggr\} \\ &\quad=-n-\sum_{\alpha =1}^{n} a \int_{\Omega }u_{i} x_{\alpha }\frac{\partial u_{i}}{\partial x_{\alpha }}e^{-\frac{|x|^{2}}{4}}\,d\nu +\frac{1}{2} \int_{\Omega }u_{i}^{2} |x|^{2}e^{-\frac{|x|^{2}}{4}} \,d\nu , \end{aligned}$$

which implies that

$$\begin{aligned} -2\sum_{\alpha =1}^{n} \int_{\Omega }u_{i} x_{\alpha }\frac{\partial u_{i}}{\partial x_{\alpha }} \,d\mu =n-\frac{1}{2} \int_{\Omega }u_{i}^{2} |x|^{2}\,d \mu . \end{aligned}$$

(4.4)

By a similar computation to (3.12), we have

$$\begin{aligned} \|\nabla u_{i}\|^{2}\leq\frac{\lambda_{i}}{\xi_{1}}. \end{aligned}$$

(4.5)

Using (4.4), we have

$$\begin{aligned} \sum_{\alpha =1}^{n} \int_{\Omega }\biggl(\langle \nabla x_{\alpha }, \nabla u_{i}\rangle-\frac{u_{i}x_{\alpha }}{4} \biggr)^{2}\,d\mu =&\sum _{\alpha =1}^{n} \int_{\Omega }\biggl(\langle \nabla x_{\alpha }, \nabla u_{i}\rangle^{2}-\frac{1}{2}\frac{\partial u_{i}}{\partial x_{\alpha }} u_{i}x_{\alpha }+\frac{u_{i}^{2}x_{\alpha }^{2}}{16} \biggr)\,d\mu \\ =&\frac{n}{4}+ \int_{\Omega }\biggl(|\nabla u_{i}|^{2}- \frac{1}{16}u_{i}^{2} |x|^{2} \biggr)\,d\mu \\ \leq&\frac{n}{4}+\frac{\lambda _{i}}{\xi_{1}}-\frac{1}{16}\min _{x\in\overline{\Omega }} \bigl\{ |x|^{2} \bigr\} . \end{aligned}$$

(4.6)

Taking (4.2) and (4.6) into (4.1), we have

$$\begin{aligned} n\sum_{i=1}^{k}( \lambda_{k+1}-\lambda_{i})^{2} \leq&\sum_{i=1}^{k} \delta( \lambda_{k+1}-\lambda_{i})^{2}n\xi_{2} \\ &{}+\sum_{i=1}^{k}\frac{(\lambda _{k+1}-\lambda_{i})}{\delta} \biggl(\frac{n}{4}+\frac{\lambda_{i}}{\xi_{1}}- \frac{1}{16}\min _{x\in \overline{\Omega }} \bigl\{ |x|^{2} \bigr\} \biggr). \end{aligned}$$

(4.7)

Taking

$$\begin{aligned} \delta= \biggl\{ \frac{\sum_{i=1}^{k}(\lambda_{k+1}-\lambda_{i}) \{\frac{n}{4}+\frac {\lambda_{i}}{\xi_{1}}-\frac{1}{16}\min_{x\in\overline{\Omega }}\{|x|^{2}\} \}}{\sum_{i=1}^{k}(\lambda_{k+1}-\lambda_{i})^{2} n\xi_{2}} \biggr\} ^{\frac{1}{2}}, \end{aligned}$$

we obtain (1.7). This finishes the proof of Theorem 1.5. □

Proof of Theorem 1.7

Let $a=0, b=1,V\equiv0, \rho \equiv1$ in (2.1), Set $\overline{x}=(x,t)\in \Omega$, where $x\in\Sigma, t\in \mathbb {R}$. By a direct computation, we know that $|\nabla t|=1, \Delta_{f} t=-\kappa t$, then taking $h=t$, we have

$$\begin{aligned} \sum_{i=1}^{k}( \lambda_{k+1}-\lambda_{i})^{2} \leq&\sum_{i=1}^{k} \delta( \lambda_{k+1}-\lambda_{i})^{2} \int_{\Omega}t u_{i} p_{ i}\,d\mu \\ &{}+\sum_{i=1}^{k}\frac{\lambda_{k+1}-\lambda_{i}}{\delta} \int_{\Omega}\biggl(\langle \nabla t, \nabla u_{i}\rangle- \frac{\kappa tu_{i} }{2} \biggr)^{2}\,d\mu , \end{aligned}$$

(4.8)

where $p_{ i}=-2\langle \nabla t, A\nabla u_{i}\rangle+u_{i} \mathfrak{L}_{f}t$.

In the following, let us estimate the right side of (4.8), first of all, by a similar computation to (3.11), we infer from $A\leq\xi_{2} I$ that

$$\begin{aligned} \int_{\Omega}t u_{i}p_{ i}\,d\mu = \int_{\Omega}t u_{i} \bigl(-2\langle \nabla t, A\nabla u_{i}\rangle+u_{i} \mathfrak{L}_{f}t \bigr)\,d\mu = \int_{\Omega}u_{i}^{2}\langle \nabla t,A \nabla t \rangle\,d\mu \leq \xi_{2}. \end{aligned}$$

(4.9)

By a direct computation,

$$\begin{aligned} \int_{\Omega}t u_{i}\langle \nabla t,\nabla u_{i} \rangle\,d\mu =- \int_{\Omega}u_{i}^{2}|\nabla t|^{2}\,d \mu - \int_{\Omega}t u_{i} \langle \nabla t,\nabla u_{i} \rangle\,d\mu - \int_{\Omega}tu_{i}^{2} \Delta_{f} t\,d\mu , \end{aligned}$$

which implies that

$$\begin{aligned} 2 \int_{\Omega}t u_{i}\langle \nabla t,\nabla u_{i} \rangle \,d\mu =-1+\kappa \int_{\Omega}t^{2}u_{i}^{2}\,d\mu . \end{aligned}$$

(4.10)

From the above equality, we have

$$\begin{aligned} 2 \int_{\Omega}\biggl(\langle \nabla t, \nabla u_{i}\rangle- \frac{\kappa tu_{i} }{2} \biggr)^{2}\,d\mu =& \int_{\Omega}\biggl(\langle \nabla t, \nabla u_{i} \rangle^{2}-\kappa tu_{i}\langle \nabla t, \nabla u_{i} \rangle+\frac{\kappa^{2}}{4}t^{2}u_{i}^{2} \biggr)\,d \mu \\ =& \int_{\Omega}\biggl(\langle \nabla t, \nabla u_{i} \rangle^{2}+1- \frac{\kappa^{2}}{4}t^{2}u_{i}^{2} \biggr)\,d\mu \\ \leq&1+\|\nabla u_{i}\|^{2}-\frac{\kappa^{2}}{4}\min _{(x,t)\in\overline{\Omega }} \bigl\{ |t|^{2} \bigr\} \\ \leq&1+\frac{\lambda_{i}}{\xi_{1}}-\frac{\kappa^{2}}{4} \min_{(x,t)\in\overline{\Omega }} \bigl\{ |t|^{2} \bigr\} . \end{aligned}$$

(4.11)

Taking (4.9) and (4.11) into (4.8), we have

$$\begin{aligned} \sum_{i=1}^{k}( \lambda_{k+1}-\lambda_{i})^{2} \leq&\sum_{i=1}^{k} \delta( \lambda_{k+1}-\lambda_{i})^{2}\xi_{2} \\ &{}+\sum_{i=1}^{k}\frac{\lambda_{k+1}-\lambda_{i}}{\delta} \biggl(1+\frac{\lambda_{i}}{\xi_{1}}-\frac{\kappa^{2}}{4} \min_{(x,t)\in\overline{\Omega }} \bigl\{ |t|^{2} \bigr\} \biggr). \end{aligned}$$

(4.12)

Taking

$$\begin{aligned} \delta= \biggl\{ \frac{\sum_{i=1}^{k}(\lambda_{k+1}-\lambda_{i}) \{1+\frac{\lambda_{i}}{\xi _{1}}-\frac{\kappa^{2}}{4}\min_{(x,t)\in\overline{\Omega }}\{|t|^{2}\} \}}{ \sum_{i=1}^{k}(\lambda_{k+1}-\lambda_{i})^{2} \xi_{2}} \biggr\} ^{\frac{1}{2}}, \end{aligned}$$

we obtain (1.9). This finishes the proof of Theorem 1.7. □