On the Second Pinching Theorem for Willmore Hypersurfaces in a Sphere

Abstract

In this paper, we consider n-dimensional compact Willmore hypersurface in a unit sphere. We prove a pinching theorem of $\rho ^2$, which is defined as $\rho ^2=S-nH^2$, for n-dimensional compact Willmore hypersurface with constant mean curvature and constant scalar curvature, where H denotes the mean curvature and S the squared norm of the second fundamental form of this hypersurface.

1 Introduction

Studying pinching phenomenon is an important topic in differential geometry. It is well-known that Simons [10] gave a proof of the pinching phenomenon of S for n-dimensional compact minimal hypersurfaces in a unit sphere $\mathbb {S}^{n+1}$, where S denotes the squared norm of the second fundamental form. J. Simons proved if $S\le n$, then $S=0$ or $S=n$. And Chern, do Carmo and Kobayashi [3] and Lawson [5] proved that $S=n$ can be only achieved by Clifford torus independently. So Chern proposed the following conjecture:

Chern’s Conjecture For n-dimensional compact minimal hypersurfaces in $\mathbb {S}^{n+1}$ with constant scalar curvature, $S>n$, then $S\ge 2n$.

In [8, 9], Peng and Terng investigated this conjecture, they proved that $S>n$, then $S\ge n+\frac{1}{12n}$ and they solved this conjecture for $n=3$ (for further classification in the case $n=3$, see [1]). For further improvement see [2, 12,13,14,15,16].

As to the Willmore hypersurface case, in [11], Wang calculated the first variation of Willmore functional and deduced the Willmore equation, later the Willmore equation can be written in the following equivalent form in [4, 7].

Theorem A Let M be an n-dimensional hypersurface in $(n+1)$-dimensional unit sphere $\mathbb {S}^{n+1}$. Then M is a Willmore hypersurface if and only if

$$\begin{aligned}&-\rho ^{n-2}\left( 2HS-nH^3-\sum _{i,j,k}h_{ij}h_{jk}h_{ki}\right) \\&\quad +\,(n-1)\Delta (\rho ^{n-2}H)-\sum _{i,j}(\rho ^{n-2})_{ij}(nH\delta _{ij}-h_{ij})=0. \end{aligned}$$

where $\Delta$ is the Laplacian, $(\cdot )_{ij}$ is the covariant derivative relative to the induced metric and $\rho ^2=S-nH^2$.

In [7], the first author proved the following pinching theorem for Willmore hypersurface in a sphere.

Theorem B [7]. Let M be an n-dimensional, $n\ge 2$, compact Willmore hypersurface in unit sphere $\mathbb {S}^{n+1}$. Then we have

$$\begin{aligned} \int _M\rho ^n(n-\rho ^2)\le 0. \end{aligned}$$

In particular, if

$$\begin{aligned} 0\le \rho ^2\le n, \end{aligned}$$

then either $\rho ^2=0$ and M is totally umbilical, or $\rho ^2=n$ and M is one of the Willmore tori $W_{m,n-m}$.

Here the Willmore tori $W_{m,n-m}$ are defined by (see [4, 7])

$$\begin{aligned} W_{m,n-m}=\mathbb {S}^m\left( \sqrt{\frac{n-m}{n}}\right) \times \mathbb {S}^{n-m}\left( \sqrt{\frac{m}{n}}\right) , \end{aligned}$$

which satisfies $\rho ^2\equiv n$ for any $1\le m\le n-1$.

Comparing with the Chern’s Conjecture for minimal hypersurfaces in a sphere, the following problem is interesting for Willmore hypersurfaces in a sphere.

Conjecture

For n-dimensional compact Willmore hypersurfaces in $\mathbb {S}^{n+1}$ with constant scalar curvature, if $\rho ^2> n$, then $\rho ^2\ge 2n$.

In this paper, we partially solve the conjecture under the assumption of constant mean curvature by proving the following second pinching theorem for Willmore hypersurfaces in a sphere.

Main Theorem Let $M^n$ be an n-dimensional ($n\ge 2$) compact Willmore hypersurface with constant mean curvature H and constant scalar curvature. There exists a positive constant $\gamma (n)$ depending only on n, such that if $|H|\le \gamma (n)$, and $n\le \rho ^2\le n+\frac{2}{3}n$, then $\rho ^2\equiv n$, and M is one of the Willmore tori $W_{m,n-m}$.

Remark 1

Actually $\gamma (n)$ can be written as $\gamma (n)=\frac{C}{\sqrt{n}}$, where C is a constant which does not depend on n. When n is large, we can improve the pinching number $\frac{5}{3}n$ close to 2n.

The paper is organized as follows. In Sect. 2, we give some basic formulas for Willmore hypersurfaces with constant mean curvature. In Sect. 3, we give some lemmas and basic estimates. In Sect. 4, we give the proof of Main Theorem.

2 Preliminary

In this paper we make a convention on the range of indices:

$$\begin{aligned} 1\le A,B,C,\ldots \le n+1; \quad 1\le i,j,k,\ldots \le n. \end{aligned}$$

We always suppose that M is an n-dimensional compact Willmore hypersurface in $\mathbb {S}^{n+1}$ with constant scalar curvature and constant mean curvature, and satisfying the Willmore equation

$$\begin{aligned}&-\rho ^{n-2}\left( 2HS-nH^3-\sum _{i,j,k}h_{ij}h_{jk}h_{ki}\right) \\&\quad +\,(n-1)\Delta (\rho ^{n-2}H)-\sum _{i,j}(\rho ^{n-2})_{ij}(nH\delta _{ij}-h_{ij})=0, \end{aligned}$$

where S is the scalar curvature and H is the mean curvature of M, and $\rho$ defined by $\rho ^2=S-nH^2$, $\Delta$ is the Laplacian, $(\cdot )_{ij}$ is the covariant derivative relative to the induced metric.

For an arbitrary fixed point $x\in M$, we choose an orthonormal local frame field $\{e_A\}$ in $\mathbb {S}^{n+1}$ such that $\{e_i\}$ are tangent to M. Let $\{\omega _A\}$ be the dual frame fields of $\{e_A\}$, and $\{\omega _{AB}\}$ the connection 1-forms of $\mathbb {S}^{n+1}$. Restricting to M, we have

$$\begin{aligned} \omega _{i\text { } n+1}=\sum _jh_{ij}\omega _j, \quad h_{ij}=h_{ji}, \end{aligned}$$

where $h=\sum _{ij}h_{ij}\omega _i\otimes \omega _j$ is the second fundamental form of M. Denote by H and S the mean curvature of M and the squared norm of h, respectively. Thus

$$\begin{aligned} S=\sum _{i,j}h_{ij}^2, \quad H=\frac{1}{n}\sum _ih_{ii}. \end{aligned}$$

Near x we choose $\{e_A\}$ such that

$$\begin{aligned} h_{ij}=\lambda _i\delta _{ij}, \quad S=\sum _i\lambda _i^2. \end{aligned}$$

And we set

$$\begin{aligned}&f_k:=\sum _i\lambda _i^k,\\&h_{ijk}:=\nabla _kh_{ij}, \quad h_{ijkl}:=\nabla _l\nabla _kh_{ij}, \quad h_{ijkls}:=\nabla _s\nabla _l\nabla _kh_{ij}. \end{aligned}$$

Then the Gauss equations, Codazzi equations and the Ricci identities are as follows:

$$\begin{aligned}&R_{ijkl}=\delta _{ik}\delta _{jl}-\delta _{il}\delta _{jk}+h_{ik}h_{jl}-h_{il}h_{jk}, \end{aligned}$$

(1)

$$\begin{aligned}&h_{ijk}=h_{ikj}, \end{aligned}$$

(2)

$$\begin{aligned}&h_{ijkl}-h_{ijlk}=\sum _p(h_{pj}R_{pikl}+h_{ip}R_{pjkl}), \end{aligned}$$

(3)

$$\begin{aligned}&h_{ijkls}-h_{ijksl}=\sum _p(h_{pjk}R_{pils}+h_{ipk}R_{pjls}+h_{ijp}R_{pkls}), \end{aligned}$$

(4)

where $R_{ijkl}$ is the Riemann curvature tensor of M. We also have

$$\begin{aligned} R=n(n-1)+n^2H^2-S, \end{aligned}$$

(5)

where R denote the scalar curvature.

We need the following useful Ricci identities

$$\begin{aligned} h_{ijij}-h_{jiji}=(\lambda _i-\lambda _j)(1+\lambda _i\lambda _j), \end{aligned}$$

(6)

which follows from (1), (3).

By use of Gauss equations, Codazzi equations and Ricci identities, we have (see [6, 12, 13])

$$\begin{aligned}&0=\frac{1}{2}\Delta S=S(n-S)-n^2H^2+nHf_3+|\nabla h|^2, \end{aligned}$$

(7)

$$\begin{aligned}&\frac{1}{3}\Delta f_3=(n-S)f_3+nH(f_4-S)+2C, \end{aligned}$$

(8)

$$\begin{aligned}&\frac{1}{4}\Delta f_4=(n-S)f_4+nH(f_5-f_3)+2A+B, \end{aligned}$$

(9)

$$\begin{aligned}&\frac{1}{2}\Delta (|\nabla h|^2)=(2n+3-S)|\nabla h|^2+6B-3A+3nHC+|\nabla ^2h|^2, \end{aligned}$$

(10)

where

$$\begin{aligned} A:=\sum _{i,j,k}\lambda _i^2h_{ijk}^2, \quad B:=\sum _{i,j,k}\lambda _i\lambda _jh_{ijk}^2, \quad C:=\sum _{i,j,k}\lambda _ih_{ijk}^2. \end{aligned}$$

(11)

Since H and S are constant, Willmore equation can be simplified as

$$\begin{aligned} 2HS-nH^3-f_3=0. \end{aligned}$$

(12)

From (12), $f_3$ is also a constant, and we can get following equation

$$\begin{aligned} A-2B=Sf_4-f_3^2+nHf_3-S^2, \end{aligned}$$

(13)

for completeness we give a proof of (13) by use of (6):

$$\begin{aligned} 0=&\frac{1}{3}\sum _{i,j}h_{ij}(f_3)_{ij}\\ =&\sum _{k}\lambda _{k}(\sum _ih_{iikk}\lambda _i^2+2\sum _{i,j}h_{ijk}^2\lambda _i)\\ =&\sum _{i,k}h_{iikk}\lambda _i\lambda _k^2+2B\\ =&\sum _{i,k}(h_{kkii}+(\lambda _i-\lambda _k)(1+\lambda _i\lambda _k))\lambda _k\lambda _i^2+2B\\ =&\sum _{i}\left( \frac{S_{ii}}{2}-\sum _{jk}h_{ijk}^2\right) \lambda _i^2+\sum _{i,k} (\lambda _i-\lambda _k)(1+\lambda _i\lambda _k)\lambda _k\lambda _i^2+2B\\ =&-A+Sf_4-f_3^2+nHf_3-S^2+2B. \end{aligned}$$

Since M is a Willmore hypersurface, we use some notations to simplify calculation, define

$$\begin{aligned} \bar{h}_{ij}=h_{ij}-H\delta _{ij}, \quad \bar{\lambda }_i=\lambda _i-H, \end{aligned}$$

so $\bar{h}_{ijk}=h_{ijk}$ and $\rho ^2=|\bar{h}|^2$, and we define $\bar{f_k}=\sum _i\bar{\lambda }_i^k$, by a direct computation

$$\begin{aligned} f_3=&\sum _i(\bar{\lambda }_i+H)^3=\bar{f_3}+3H\rho ^2+nH^3,\\ f_4=&\sum _i(\bar{\lambda }_i+H)^4=\bar{f_4}+4H\bar{f_3}+6H^2\rho ^2+nH^4,\\ f_5=&\sum _i(\bar{\lambda }_i+H)^5=\bar{f_5}+5H\bar{f_4}+10H^2\bar{f_3}+10H^3\rho ^2+nH^5. \end{aligned}$$

So Willmore equation (12) becomes

$$\begin{aligned} \bar{f_3}=-H\rho ^2, \end{aligned}$$

(14)

and using (14) we have

$$\begin{aligned} f_3=&2H\rho ^2+nH^3, \end{aligned}$$

(15)

$$\begin{aligned} f_4=&\bar{f_4}+2H^2\rho ^2+nH^4, \end{aligned}$$

(16)

$$\begin{aligned} f_5=&\bar{f_5}+5H\bar{f_4}+nH^5, \end{aligned}$$

(17)

so (7) can be written as

$$\begin{aligned} |\nabla \bar{h}|^2=(\rho ^2-n)\rho ^2. \end{aligned}$$

(18)

We also define

$$\begin{aligned} \bar{A}:=\sum _{i,j,k}\bar{\lambda }_i^2\bar{h}_{ijk}^2, \quad \bar{B}:=\sum _{i,j,k}\bar{\lambda }_i\bar{\lambda }_j\bar{h}_{ijk}^2, \quad \bar{C}:=\sum _{i,j,k}\bar{\lambda }_i\bar{h}_{ijk}^2, \end{aligned}$$

(19)

using (11), (18), (19) we have

$$\begin{aligned} A=&\bar{A}+2H\bar{C}+H^2(\rho ^2-n)\rho ^2, \end{aligned}$$

(20)

$$\begin{aligned} B=&\bar{B}+2H\bar{C}+H^2(\rho ^2-n)\rho ^2, \end{aligned}$$

(21)

$$\begin{aligned} C=&\bar{C}+H(\rho ^2-n)\rho ^2, \end{aligned}$$

(22)

and through a direct computation using (15)–(18), (20)–(22), we can write (8)–(10), (13) as

$$\begin{aligned}&2\bar{C}+nH\bar{f_4}-nH^3\rho ^2-nH\rho ^2=0, \end{aligned}$$

(23)

$$\begin{aligned}&\frac{1}{4}\Delta f_4=-(\rho ^2-n)\bar{f_4}+2\bar{A}+\bar{B}+nH\left( \bar{f_5}+H\bar{f_4}+\frac{1}{n}H\rho ^4\right) , \end{aligned}$$

(24)

$$\begin{aligned}&|\nabla ^2\bar{h}|^2=3(\bar{A}-2\bar{B})-(6+3n)H\bar{C}-(3+2n)H^2\rho ^2(\rho ^2-n)\nonumber \\&\qquad \qquad +\,(\rho ^2-2n)(\rho ^2-n)\rho ^2-3\rho ^2(\rho ^2-n), \end{aligned}$$

(25)

$$\begin{aligned}&\bar{A}-2\bar{B}=\rho ^2(\bar{f_4}-H^2\rho ^2-\rho ^2 ), \end{aligned}$$

(26)

where in the calculations of (26) we used Eq. (23).

Remark

Combining (23), (26), we can see

$$\begin{aligned} 2\bar{C}=-\frac{n}{\rho ^2}H(\bar{A}-2\bar{B}). \end{aligned}$$

(27)

3 Some Lemmas and Basic Estimates

In this part we give an estimate of $|\nabla ^2\bar{h}|^2$, the idea is similar to Yang and Cheng [15] and Cheng et al. [2], we define

$$\begin{aligned} u_{ijkl}:=\frac{1}{4}(\bar{h}_{ijkl}+\bar{h}_{lijk}+\bar{h}_{klij}+\bar{h}_{jkli}), \end{aligned}$$

a direct computation give

$$\begin{aligned} 16\sum _{i,j,k,l}u_{ijkl}^2=&\sum _{i,j,k,l}(\bar{h}_{ijkl}+\bar{h}_{lijk}+\bar{h}_{klij}+\bar{h}_{jkli})^2\\ =&\sum _{i,j,k,l}(h_{ijkl}+h_{lijk}+h_{klij}+h_{jkli})^2\\ =&\sum _{i,j,k,l}(4h_{ijkl}^2+12h_{ijkl}h_{ijlk})\\ =&4\sum _{i,j,k,l}h_{ijkl}^2+12\sum _{i,j,k,l}h_{ijkl}(h_{ijkl}-\lambda _jR_{jikl}-\lambda _iR_{ijkl})\\ =&16\sum _{i,j,k,l}h_{ijkl}^2-12\sum _{i,j,k,l}(\lambda _i-\lambda _j)h_{ijkl}R_{ijkl}\\ =&16\sum _{i,j,k,l} h_{ijkl}^2\\&-12\sum _{i,j,k,l}(\lambda _i-\lambda _j)h_{ijkl} (\delta _{ik}\delta _{jl}-\delta _{il}\delta _{jk}+h_{ik}h_{jl}-h_{il}h_{jk})\\ =&16\sum _{i,j,k,l}h_{ijkl}^2-12\sum _{i,j}(\lambda _i-\lambda _j)(1+\lambda _i\lambda _j)(h_{ijij}-h_{jiji})\\ =&16\sum _{i,j,k,l}h_{ijkl}^2-12\sum _{i,j}(\lambda _i-\lambda _j)^2(1+\lambda _i\lambda _j)^2\\ =&16\sum _{i,j,k,l}h_{ijkl}^2-24(Sf_4-f_3^2-2S^2+nS-n^2H^2+2nHf_3), \end{aligned}$$

that is

$$\begin{aligned} \sum _{i,j,k,l}\bar{h}_{ijkl}^2=\sum _{i,j,k,l}u_{ijkl}^2+ \frac{3}{2}(Sf_4-f_3^2-2S^2+nS-n^2H^2+2nHf_3). \end{aligned}$$

(28)

We define $t:=\frac{\rho ^2-n}{\rho ^2}$, using our convention and (15), (16), (26), then (28) can be rewritten as

$$\begin{aligned} \sum _{i,j,k,l}\bar{h}_{ijkl}^2= & {} \sum _{i,j,k,l}u_{ijkl}^2+ \frac{3}{2}(\bar{A}-2\bar{B}-2H\bar{C} -t\rho ^4-tH^2\rho ^4). \end{aligned}$$

(29)

So combining (25) and (29), we get

$$\begin{aligned} (\rho ^2-2n)(\rho ^2-n)\rho ^2= & {} \sum _{i,j,k,l}u_{ijkl}^2- \frac{3}{2}(\bar{A}-2\bar{B})+\frac{3}{2}t\rho ^4\nonumber \\{} & {} +\,3(n+1)H\bar{C}+\left( \frac{3}{2}+2n\right) tH^2\rho ^4. \end{aligned}$$

(30)

Next we estimate the term $\sum _{i,j,k,l}u_{ijkl}^2$.

Since $\rho ^2$, $|\nabla \bar{h}|^2$and $\bar{f_3}$ are constants, derivative of these terms give

$$\begin{aligned}{} & {} \sum _{i,j}(\bar{h}_{ij}\bar{h}_{ijkl}+\bar{h}_{ijk} \bar{h}_{ijl})=0,\quad \forall k,l,\nonumber \\{} & {} \sum _{i,j,k}\bar{h}_{ijk}\bar{h}_{ijkl}=0,\quad \forall ,l,\nonumber \\{} & {} \sum _{i}\bar{\lambda }_i^2\bar{h}_{iik}=0,\quad \forall k. \end{aligned}$$

(31)

For convenience, defining

$$\begin{aligned} \bar{f}:&=\bar{f_4}-H^2\rho ^2-\frac{\rho ^4}{n}, \end{aligned}$$

(32)

$$\begin{aligned} a_{ij}:&=\sum _{p}\bar{h}_{ip}\bar{h}_{pj}+H\bar{h}_{ij}-\frac{\rho ^2}{n}\delta _{ij}, \end{aligned}$$

(33)

we have

$$\begin{aligned} \sum _{i,j}a_{ij}^2=\bar{f},\quad \sum _{i,j}a_{ij}\bar{h}_{ij}=0,\quad \sum _{i,j}a_{ij}\delta _{ij}=0, \end{aligned}$$

(34)

by using (14).

For any $\alpha ,\beta , \gamma \in \mathbb {R}$,

$$\begin{aligned} \sum _{i,j,k,l}[u_{ijkl}+\alpha (a_{ij}\bar{h}_{kl}+\bar{h}_{ij} a_{kl})+\beta \bar{h}_{ij}\bar{h}_{kl}+\gamma (\bar{h}_{ij} \delta _{kl}+\delta _{ij}\bar{h}_{kl})]^2\ge 0, \end{aligned}$$

(35)

we have

$$\begin{aligned}{} & {} \sum _{i,j,k,l}u_{ijkl}\bar{h}_{ij}\delta _{kl}=-\frac{t}{2} \rho ^4,\quad \sum _{i,j,k,l}u_{ijkl}\bar{h}_{ij}\bar{h}_{kl}=-\bar{C},\nonumber \\{} & {} \sum _{i,j,k,l}u_{ijkl}a_{ij}\bar{h}_{kl}=-\bar{B}- \frac{1}{2}\bar{A}-H\bar{C}+\frac{t}{2(1-t)}\rho ^4, \end{aligned}$$

(36)

here the first two equation is easy to check, for completeness we compute the third equation and we follow the Einstein summation convention,

$$\begin{aligned}&\sum _{i,j,k,l}u_{ijkl}a_{ij}\bar{h}_{kl}\\&\quad =\frac{1}{4}(\bar{h}_{ijkl}+\bar{h}_{lijk}+\bar{h}_{klij}+\bar{h}_{jkli})(\bar{h}_{ip}\bar{h}_{pj}+H\bar{h}_{ij}-\frac{\rho ^2}{n}\delta _{ij})\bar{h}_{kl}\\&\quad =\frac{1}{4}(\bar{h}_{ijkl}+\bar{h}_{lijk}+\bar{h}_{klij}+\bar{h}_{jkli})\bar{h}_{ip}\bar{h}_{pj}\bar{h}_{kl}-H\bar{C}+\frac{t}{2(1-t)}\rho ^4\\&\quad =\frac{1}{2}\bar{h}_{ijkl}\bar{h}_{ip}\bar{h}_{pj}\bar{h}_{kl}+\frac{1}{2}\bar{h}_{klij}\bar{h}_{ip}\bar{h}_{pj}\bar{h}_{kl}-H\bar{C}+\frac{t}{2(1-t)}\rho ^4\\&\quad =\frac{1}{2}(\bar{h}_{ijkl}-\bar{h}_{klij})\bar{h}_{ip}\bar{h}_{pj}\bar{h}_{kl}+\bar{h}_{klij}\bar{h}_{ip}\bar{h}_{pj}\bar{h}_{kl}-H\bar{C}+\frac{t}{2(1-t)}\rho ^4\\&\quad =\frac{1}{2}(\bar{h}_{ikik}-\bar{h}_{kiki})\bar{\lambda }_i^2\bar{\lambda }_k-\bar{A}-H\bar{C}+\frac{t}{2(1-t)}\rho ^4\\&\quad =\frac{1}{2}(\lambda _i-\lambda _k)(1+\lambda _i\lambda _k)\bar{\lambda }_i^2\bar{\lambda }_k-\bar{A}-H\bar{C}+\frac{t}{2(1-t)}\rho ^4\\&\quad =\frac{1}{2}(\bar{\lambda }_i-\bar{\lambda }_k)(1+(\bar{\lambda }_i+H)(\bar{\lambda }_k+H))\bar{\lambda }_i^2\bar{\lambda }_k-\bar{A}-H\bar{C}+\frac{t}{2(1-t)}\rho ^4\\&\quad =\frac{1}{2}(\rho ^2\bar{f_4}-H^2\rho ^4-\rho ^4)-\bar{A}-H\bar{C}+\frac{t}{2(1-t)}\rho ^4\\&\quad =-\bar{B}-\frac{1}{2}\bar{A}-H\bar{C}+\frac{t}{2(1-t)}\rho ^4, \end{aligned}$$

where the second and fifth equalities used (31), the last equality followed from (26).

By taking $\beta =\frac{\bar{C}}{\rho ^4}$ and $\gamma =\frac{t}{2(1-t)}$, we obtain, from (34) to (36),

$$\begin{aligned} \sum _{i,j,k,l}u_{ijkl}^2\ge 2\alpha \left( 2\bar{B}+\bar{A}+2H \bar{C}-\frac{t}{1-t}\rho ^4\right) -2\alpha ^2\rho ^2\bar{f}+ \frac{\bar{C}^2}{\rho ^4}+\frac{t^2}{2(1-t)}\rho ^4.\nonumber \\ \end{aligned}$$

(37)

So putting (27) and (37) into (30), and noting that $n=(1-t)\rho ^2$ we get

$$\begin{aligned} t(2t-1)\rho ^6\ge & {} \left[ 2\alpha -2\alpha ^2-\frac{3}{2}-\frac{3}{2}(1-t)^2H^2 \rho ^2-\left( 2\alpha +\frac{3}{2}\right) (1-t)H^2\right] \bar{A}\nonumber \\{} & {} +\,[4\alpha +4\alpha ^2+3+3(1-t)^2H^2\rho ^2+(4\alpha +3)(1-t)H^2]\bar{B}\nonumber \\{} & {} +\,\frac{t}{1-t}\left( 2\alpha ^2-2\alpha +\frac{3}{2}-t\right) \rho ^4+\frac{3}{2}tH^2\rho ^4\nonumber \\{} & {} +2t(1-t)H^2\rho ^6+\,\frac{\bar{C}^2}{\rho ^4}. \end{aligned}$$

(38)

Finally, we need the following algebraic lemma.

Lemma 1

For $\{\lambda _i\}$, $i=1,\ldots ,n$, and $\sum _{i=1}^n\lambda _i^2=\rho ^2$, $\rho >0$ then $-\rho ^5\le \sum _{i=1}^n\lambda _i^5\le \rho ^5$.

Proof

Using Lagrange multiplier method, let

$$\begin{aligned} f=\sum _{i=1}^n\lambda _i^5+b\left( \sum _{i=1}^n\lambda _i^2-\rho ^2\right) , \end{aligned}$$

then the extreme points are determined by

$$\begin{aligned}&\frac{\partial }{\partial \lambda _i}f=5\lambda _i^4+2b\lambda _i=5 \lambda _i\left( \lambda _i^3+\frac{2b}{5}\right) =0,\\&\frac{\partial }{\partial \alpha }f=\sum _{n=1}^n\lambda _i^2-\rho ^2=0, \end{aligned}$$

then $\lambda _i$ can only be 0 or $-\left( \frac{2b}{5}\right) ^\frac{1}{3}$, so suppose there are $m(1\le m\le n)$ terms $-\left( \frac{2b}{5}\right) ^\frac{1}{3}$ in $\{\lambda _i\}$, then

$$\begin{aligned} m\left( \frac{2b}{5}\right) ^\frac{2}{3}=\rho ^2, \end{aligned}$$

thus

$$\begin{aligned} \left| \sum _{i=1}^n\lambda _i^5\right| =\left| m\left( \frac{2b}{5}\right) ^\frac{5}{3}\right| =\left| \frac{\rho ^5}{m^{\frac{3}{2}}}\right| \le \rho ^5. \end{aligned}$$

$\square$

We also need following estimates of which proofs can also be found in [14]. For convenience, we give their proofs here.

Lemma 2

At any point $x\in M$, if $\bar{\lambda }_1=max_i\bar{\lambda }_i$ and $\bar{\lambda }_2=min_i\bar{\lambda }_i$, then

$$\begin{aligned} \rho ^2\bar{f_4}-\bar{f_3}^2-2\rho ^4+n\rho ^2\ge \frac{(\bar{\lambda }_1- \bar{\lambda }_2)^2}{\bar{\lambda }_1^2+\bar{\lambda }_2^2}(1+\bar{\lambda }_1\bar{\lambda }_2)^2\rho ^2. \end{aligned}$$

(39)

Proof

From

$$\begin{aligned} \rho ^2\bar{f_4}-\bar{f_3}^2-2\rho ^4+n\rho ^2=\frac{1}{\rho ^2} \sum _{i}(\bar{\lambda }_i^2\rho ^2-\bar{f_3}\bar{\lambda }_i-\rho ^2)^2, \end{aligned}$$

we see

$$\begin{aligned}&\rho ^2\bar{f_4}-\bar{f_3}^2-2\rho ^4+n\rho ^2\\&\quad \ge \frac{1}{\rho ^2}(\bar{\lambda }_1^2\rho ^2-\bar{f_3}\bar{\lambda }_1-\rho ^2)^2 +\frac{1}{\rho ^2}(\bar{\lambda }_2^2\rho ^2-\bar{f_3}\bar{\lambda }_2-\rho ^2)^2\\&\quad =\frac{1}{\rho ^2}\{[(\bar{\lambda }_1^2-1)^2+(\bar{\lambda }_2^2-1)^2] \rho ^4+\bar{f_3}^2(\bar{\lambda }_1^2+\bar{\lambda }_2^2)\\&\qquad -\,2[(\bar{\lambda }_1^2-1)\bar{\lambda }_1+(\bar{\lambda }_2^2-1) \bar{\lambda }_2]\bar{f_3}\rho ^2\}. \end{aligned}$$

Since $\bar{f_3}^2+a\bar{f_3}\ge -\frac{a^2}{4}$, we deduce

$$\begin{aligned}&\rho ^2\bar{f_4}-\bar{f_3}^2-2\rho ^4+n\rho ^2\\&\quad \ge [(\bar{\lambda }_1^2-1)^2+(\bar{\lambda }_2^2-1)^2]\rho ^2- \frac{[(\bar{\lambda }_1^2-1)\bar{\lambda }_1+(\bar{\lambda }_2^2-1) \bar{\lambda }_2]^2\rho ^2}{\bar{\lambda }_1^2+\bar{\lambda }_2^2}\\&\quad =\frac{(\bar{\lambda }_1-\bar{\lambda }_2)^2}{\bar{\lambda }_1^2+ \bar{\lambda }_2^2}(1+\bar{\lambda }_1\bar{\lambda }_2)^2\rho ^2. \end{aligned}$$

$\square$

Lemma 3

At any point $x\in M$, we have

$$\begin{aligned} \bar{A}-\bar{B}\le \frac{1}{3}(\bar{\lambda }_1-\bar{\lambda }_2)^2t\rho ^4. \end{aligned}$$

(40)

Proof

From the definition we get

$$\begin{aligned} \bar{A}-\bar{B}&=\sum _{i,j,k}(\bar{\lambda }_i^2-\bar{\lambda }_i\bar{\lambda }_j)\bar{h}_{ijk}^2\\&=\frac{1}{3}\sum _{i,j,k}(\bar{\lambda }_i^2+\bar{\lambda }_j^2+\bar{\lambda }_k^2-\bar{\lambda }_i\bar{\lambda }_j- \bar{\lambda }_j\bar{\lambda }_k-\bar{\lambda }_k\bar{\lambda }_i)\bar{h}_{ijk}^2\\&=\frac{1}{3}\sum _{i\ne j\ne k\ne i}(\bar{\lambda }_i^2+\bar{\lambda }_j^2+ \bar{\lambda }_k^2-\bar{\lambda }_i\bar{\lambda }_j-\bar{\lambda }_j\bar{\lambda }_k-\bar{\lambda }_k\bar{\lambda }_i)\bar{h}_{ijk}^2\\&\quad +\,\sum _{i,j}(\bar{\lambda }_i-\bar{\lambda }_j)^2\bar{h}_{iij}^2. \end{aligned}$$

Without loss of generality, letting $\bar{\lambda }_i\le \bar{\lambda }_j\le \bar{\lambda }_k$, consider $g=\bar{\lambda }_i^2+\bar{\lambda }_j^2+\bar{\lambda }_k^2-\bar{\lambda }_i\bar{\lambda }_j-\bar{\lambda }_j\bar{\lambda }_k-\bar{\lambda }_k\bar{\lambda }_i$ as a function of $\bar{\lambda }_j$. Since it is a convex function, it takes its maximum on the boundary point $\bar{\lambda }_j=\bar{\lambda }_i$ or $\bar{\lambda }_j=\bar{\lambda }_k$. But since

$$\begin{aligned}&g|_{\bar{\lambda }_j=\bar{\lambda }_i}=(\bar{\lambda }_i-\bar{\lambda }_k)^2\le (\bar{\lambda }_1-\bar{\lambda }_2)^2,\\&g|_{\bar{\lambda }_j=\bar{\lambda }_k}=(\bar{\lambda }_i-\bar{\lambda }_k)^2\le (\bar{\lambda }_1-\bar{\lambda }_2)^2, \end{aligned}$$

we have

$$\begin{aligned} \bar{A}-\bar{B}\le&\frac{1}{3}\left[ \sum _{i\ne j}3(\bar{\lambda }_1-\bar{\lambda }_2)^2\bar{h}_{iij}^2+ \sum _{i\ne j\ne k\ne i}(\bar{\lambda }_1-\bar{\lambda }_2)^2\bar{h}_{ijk}^2\right] \\ =&\frac{1}{3}(\bar{\lambda }_1-\bar{\lambda }_2)^2\left( \sum _{i,j,k}\bar{h}_{ijk}^2- \sum _{i}\bar{h}_{iii}^2\right) \\ \le&\frac{1}{3}(\bar{\lambda }_1-\bar{\lambda }_2)^2|\nabla \bar{h}|^2= \frac{1}{3}(\bar{\lambda }_1-\bar{\lambda }_2)^2t\rho ^4. \end{aligned}$$

$\square$

4 Proof of Main Theorem

When $n=2,3,$ M is isoparametric, so we only need to consider $n\ge 4$.In this part, we first get a small pinching result of t, then use it to prove our main theorem.

Step 1: $t\ge \frac{1}{22}$.

Combining (14), (26) and (39), and noticing that $\bar{\lambda }_1\bar{\lambda }_2\le 0$ we get

$$\begin{aligned} \bar{A}-2\bar{B}&\ge \frac{(\bar{\lambda }_1-\bar{\lambda }_2)^2}{\bar{\lambda }_1^2+\bar{\lambda }_2^2}(1+\bar{\lambda }_1 \bar{\lambda }_2)^2\rho ^2+t\rho ^4\\&\ge (1+\bar{\lambda }_1\bar{\lambda }_2)^2\rho ^2+t\rho ^4, \end{aligned}$$

this combining with (40) yields: for any $\epsilon >0$, we have

$$\begin{aligned} \bar{A}-\bar{B}-\epsilon (\bar{A}-2\bar{B})\le & {} \frac{1}{3}(\bar{\lambda }_1- \bar{\lambda _2})^2t\rho ^4-\epsilon [(1+\bar{\lambda }_1\bar{\lambda }_2)^2\rho ^2+t\rho ^4]\nonumber \\= & {} \frac{1}{3}(\bar{\lambda }_1^2+\bar{\lambda }_2^2)t\rho ^4-\epsilon (1+\bar{\lambda }_1\bar{\lambda }_2)^2\rho ^2-\frac{2}{3} (1+\bar{\lambda }_1\bar{\lambda }_2)t\rho ^4\nonumber \\{} & {} +\,\left( \frac{2}{3}-\epsilon \right) t\rho ^4\nonumber \\\le & {} \frac{1}{3}t\rho ^6+\frac{1}{9\epsilon }t^2\rho ^6+\left( \frac{2}{3}-\epsilon \right) t\rho ^4. \end{aligned}$$

(41)

Using (27) and (30), we can see

$$\begin{aligned} t(2t-1)\rho ^6\ge -\frac{3}{2}\left( 1+\frac{n(n+1)H^2}{\rho ^2}\right) (\bar{A}-2\bar{B})+\frac{3}{2}t\rho ^4+\left( \frac{3}{2}+2n\right) tH^2\rho ^4,\nonumber \\ \end{aligned}$$

(42)

where we only used $\sum _{i,j,k,l}u_{ijkl}^2\ge 0$. Noticing that

$$\begin{aligned} \bar{A}+2\bar{B}=\frac{1}{3}\sum _{i,j,k}(\bar{\lambda }_i+ \bar{\lambda }_j+\bar{\lambda }_k)^2h_{ijk}^2\ge 0. \end{aligned}$$

(43)

Then for any $r>0$

$$\begin{aligned}{} & {} -\frac{3}{2}\left( 1+\frac{n(n+1)H^2}{\rho ^2}\right) (\bar{A}-2\bar{B})\nonumber \\{} & {} \quad =r(\bar{A}-2\bar{B})-\left( \frac{3}{2}+\frac{3n(n+1)H^2}{2\rho ^2}+r\right) (\bar{A}-2\bar{B})\nonumber \\{} & {} \quad =r(\bar{A}-2\bar{B})-\left( \frac{3}{2}+\frac{3n(n+1)H^2}{2\rho ^2}+r\right) \nonumber \\{} & {} \qquad \left[ \frac{4}{3}(\bar{A}-\bar{B})-\frac{1}{3}(\bar{A}+2\bar{B})\right] \nonumber \\{} & {} \quad \ge r(\bar{A}-2\bar{B})-\left( 2+\frac{2n(n+1)H^2}{\rho ^2}+\frac{4}{3}r\right) (\bar{A}-\bar{B})\nonumber \\{} & {} \quad =-\left( 2+\frac{2n(n+1)H^2}{\rho ^2}+\frac{4}{3}r\right) \nonumber \\{} & {} \qquad \left[ \bar{A}-\bar{B}-\frac{r}{2+\frac{2n(n+1)H^2}{\rho ^2}+\frac{4}{3}r}(\bar{A}-2\bar{B})\right] \nonumber \\{} & {} \quad \ge -\left( 2+\frac{2n(n+1)H^2}{\rho ^2}+\frac{4}{3}r\right) \left[ \frac{1}{3}t\rho ^6+\frac{2+\frac{2n(n+1)H^2}{\rho ^2}+\frac{4}{3}r}{9r}t^2\rho ^6\right. \nonumber \\{} & {} \qquad \left. +\,\left( \frac{2}{3}-\frac{r}{2+\frac{2n(n+1)H^2}{\rho ^2}+\frac{4}{3}r}\right) t\rho ^4\right] \nonumber \\{} & {} \quad =-t\rho ^6\left[ \frac{2}{3}+\frac{2n(n+1)H^2}{3\rho ^2}+\frac{16t}{27}\left( 1+\frac{n(n+1)H^2}{\rho ^2}\right) \right. \nonumber \\{} & {} \qquad \left. +\,\left( \frac{4}{9}+\frac{16t}{81}\right) r+\frac{4\left( 1+\frac{n(n+1)H^2}{\rho ^2}\right) ^2}{9r}t\right] \nonumber \\{} & {} \qquad -\,\left( 2+\frac{2n(n+1)H^2}{\rho ^2}+\frac{4}{3}r\right) \nonumber \\{} & {} \qquad \left( \frac{2}{3}-\frac{r}{2+\frac{2n(n+1)H^2}{\rho ^2}+\frac{4}{3}r}\right) t\rho ^4 \end{aligned}$$

(44)

where in the last inequality we used (41) by taking $\epsilon =\frac{r}{2+\frac{2n(n+1)H^2}{\rho ^2}+\frac{4}{3}r}$.

Putting (44) into (42) and let $r=\frac{1+\frac{n(n+1)H^2}{\rho ^2}}{\sqrt{\frac{1}{t}+\frac{4}{9}}}$ such that the coefficient of $t\rho ^6$ achieve its maximum value, we get

$$\begin{aligned} t(2t-1)\rho ^6\ge & {} -t\rho ^6\left( 1+\frac{n(n+1)H^2}{\rho ^2}\right) \left( \frac{2}{3}+\frac{16}{27}t+\frac{8}{9}\sqrt{\left( 1+\frac{4}{9}t\right) t}\right) \nonumber \\{} & {} -\,\left( 1+\frac{n(n+1)H^2}{\rho ^2}\right) \left( 2+\frac{4}{\sqrt{4+\frac{9}{t}}}\right) \left( \frac{2}{3}-\frac{3}{4+6\sqrt{\frac{1}{t}+\frac{4}{9}}}\right) t\rho ^4\nonumber \\{} & {} +\,\frac{3}{2}t\rho ^4+\left( \frac{3}{2}+2n\right) tH^2\rho ^4, \end{aligned}$$

(45)

by dividing $t\rho ^6$, and notice that $\rho ^2=\frac{n}{1-t}$, we have

$$\begin{aligned} 2t-1\ge & {} -\left( \frac{2}{3}+\frac{8}{9}\sqrt{\left( 1+\frac{4}{9}t\right) t}+\frac{16}{27}t\right) \nonumber \\{} & {} -\,\left( 2+\frac{4}{\sqrt{4+\frac{9}{t}}}\right) \left( \frac{2}{3}-\frac{3}{4+6\sqrt{\frac{1}{t}+\frac{4}{9}}}\right) \frac{1-t}{n}+\frac{3}{2}\frac{1-t}{n}+F_1,\nonumber \\ \end{aligned}$$

(46)

where

$$\begin{aligned} \begin{aligned} F_1&:=-(1-t)(n+1)H^2\left( \frac{2}{3}+\frac{8}{9} \sqrt{\left( 1+\frac{4}{9}t\right) t}+\frac{16}{27}t\right) \\&\quad -\,(1-t)(n+1)H^2\left( 2+\frac{4}{\sqrt{4+\frac{9}{t}}}\right) \left( \frac{2}{3}- \frac{3}{4+6\sqrt{\frac{1}{t}+\frac{4}{9}}}\right) \frac{1-t}{n}\\&\quad +\,\frac{3}{2}H^2\frac{1-t}{n}+2H^2(1-t). \end{aligned} \end{aligned}$$

So $F_1=H^2C_1(n,t)$, since t is bounded, then $\left| \frac{C_1(n,t)}{n}\right|$ is bounded by some constant $C_2>0$ which does not depend on n.

So if $t<\frac{1}{22}$,

$$\begin{aligned} \frac{70}{27}t\ge & {} \frac{1}{3}-\frac{8}{9}\sqrt{\left( 1+\frac{4}{9}t\right) t}-\,\left( 2+\frac{4}{\sqrt{4+\frac{9}{t}}}\right) \nonumber \\{} & {} \left( \frac{2}{3}- \frac{3}{4+\sqrt{\frac{1}{t}+\frac{4}{9}}}\right) \frac{1-t}{n}+\,\frac{3}{2}\frac{1-t}{n}+F_1\nonumber \\\ge & {} \frac{1}{3}-\frac{8}{9}\sqrt{\left( 1+\frac{4}{9}t\right) t}- \left( 2+\frac{4}{\sqrt{4+\frac{9}{t}}}\right) \frac{2}{3}\frac{1-t}{n}+\frac{3}{2}\frac{1-t}{n}+F_1\nonumber \\\ge & {} \frac{1}{3}-\frac{8}{9}\sqrt{\left( 1+\frac{4}{9}t\right) t}- \left( 2+\frac{4}{\sqrt{4+\frac{9}{t}}}\right) \frac{2}{3n}+\frac{3}{2}\frac{1-t}{n}+F_1\nonumber \\\ge & {} \frac{1}{3}-\frac{4}{297}\sqrt{202}-\left( \frac{4}{3}+ \frac{8}{3\sqrt{202}}\right) \frac{1}{n}+\frac{63}{44}\frac{1}{n}+F_1\nonumber \\\ge & {} 0.11963+F_1, \end{aligned}$$

(47)

the last inequality followed from $n\ge 4$, so

$$\begin{aligned} t\ge \frac{1}{22}+0.0006+\frac{27}{70}F_1. \end{aligned}$$

(48)

If we take $H^2\le \frac{21}{13500nC_2}$, we have $t\ge \frac{1}{22}$.

Step 2: $t>\frac{2}{5}$.

We assume $t\le \frac{2}{5}$ to deduce a contradiction.

Because M is compact, there exists a point p such that $\Delta \bar{f_4}=0$ holds at p, so by taking (26) into (24) at p, we have

$$\begin{aligned}&[2-t+(1-t)H^2]\bar{A}+[2t+1-2(1-t)H^2]\bar{B}\nonumber \\&\quad -\,t\rho ^4+(1-t)H\rho ^2(\bar{f_5}+H^3\rho ^2+2H\rho ^2)=0. \end{aligned}$$

(49)

Through this section all the computation is done at p, for convenience we let

$$\begin{aligned} g:=(1-t)H\rho ^2(\bar{f_5}+H^3\rho ^2+2H\rho ^2), \end{aligned}$$

(50)

so (49) implies

$$\begin{aligned} \bar{A}=-\frac{2t+1-2(1-t)H^2}{2-t+(1-t)H^2}\bar{B}+\frac{t\rho ^4-g}{2-t+(1-t)H^2}, \end{aligned}$$

(51)

and combining (43) and (51), we have

$$\begin{aligned} \bar{B}\ge -\frac{t\rho ^4-g}{3-4t+4(1-t)H^2}. \end{aligned}$$

(52)

Using (51), then (38) turns into

$$\begin{aligned} t(2t-1)\rho ^6\ge & {} [2\alpha -2\alpha ^2-\frac{3}{2}-\frac{3}{2}(1-t)^2H^2\rho ^2- (2\alpha +\frac{3}{2})(1-t)H^2]\nonumber \\{} & {} \times \left( -\frac{2t+1-2(1-t)H^2}{2-t+(1-t)H^2} \bar{B}+\frac{t\rho ^4-g}{2-t+(1-t)H^2}\right) \nonumber \\{} & {} +\,[4\alpha +4\alpha ^2+3+3(1-t)^2H^2\rho ^2+(4\alpha +3)(1-t)H^2]\bar{B}\nonumber \\{} & {} +\,\frac{t}{1-t}\left( 2\alpha ^2-2\alpha +\frac{3}{2}-t\right) \rho ^4\nonumber \\{} & {} +\frac{3}{2}tH^2\rho ^4+2t(1-t)H^2\rho ^6+\,\frac{\bar{C}^2}{\rho ^4}, \end{aligned}$$

(53)

let

$$\begin{aligned} (I)&:=-\left[ 2\alpha -2\alpha ^2-\frac{3}{2}- \frac{3}{2}(1-t)^2H^2\rho ^2-\left( 2\alpha +\frac{3}{2}\right) (1-t)H^2\right] \\&\quad \times \,\frac{2t+1-2(1-t)H^2}{2-t+(1-t)H^2}\bar{B}+[4\alpha +4\alpha ^2+3+3(1-t)^2H^2\rho ^2\\&\quad +\,(4\alpha +3)(1-t)H^2]\bar{B},\\ (II)&:=[2\alpha -2\alpha ^2-\frac{3}{2}- \frac{3}{2}(1-t)^2H^2\rho ^2-\left( 2\alpha +\frac{3}{2}\right) (1-t)H^2]\\&\quad \times \frac{t\rho ^4-g}{2-t+(1-t)H^2}+\frac{t}{1-t}(2\alpha ^2-2\alpha )\rho ^4\\&=\frac{1}{(1-t)(2-t+(1-t)H^2)}\Bigg \{\Bigg [(2+2(1-t)H^2)\alpha ^2\\&\quad -\,(2+2(2-t)(1-t)H^2)\alpha -\frac{3}{2}(1-t)-\frac{3}{2}(1-t)^3H^2\rho ^2\\&\quad -\frac{3}{2}(1-t)^2H^2\Bigg ]t\rho ^4-(1-t)\left[ 2\alpha -2\alpha ^2-\frac{3}{2}- \frac{3}{2}(1-t)^2H^2\rho ^2\right. \\&\quad \left. -\,\left( 2\alpha +\frac{3}{2}\right) (1-t)H^2\right] g\\&=\frac{1}{(1-t)(2-t+(1-t)H^2)}\left\{ \left[ 2\alpha ^2-2\alpha -\frac{3}{2}(1-t)+l_2(H)\right] t\rho ^4+l_3(H)\right\} , \end{aligned}$$

in convenience of computation, here we denote

$$\begin{aligned} l_2(H)&:=2(1-t)H^2\alpha ^2-2(2-t)(1-t)H^2\alpha -\frac{3}{2}(1-t)^3H^2\rho ^2\\&\quad -\,\frac{3}{2}(1-t)^2H^2,\\ l_3(H)&:=(1-t)\left[ 2\alpha -2\alpha ^2-\frac{3}{2}-\frac{3}{2}(1-t)^2H^2\rho ^2\right. \\&\quad \left. -\,\left( 2\alpha +\frac{3}{2}\right) (1-t)H^2\right] g, \end{aligned}$$

through a direct computation and using (52), we obtain

$$\begin{aligned} (I)&=\frac{\bar{B}}{2-t+(1-t)H^2}\\&\quad \left[ 10\alpha ^2+(6-8t+18(1-t)H^2)\alpha +\frac{15}{2}+\frac{15}{2}(1-t)^2H^2\rho ^2+\frac{15}{2}(1-t)H^2\right] ,\\&\ge -\frac{t\rho ^4-g}{(2-t+(1-t)H^2)(3-4t+4(1-t)H^2)}\\&\quad \left[ 10\alpha ^2+(6-8t+18(1-t)H^2)\alpha +\frac{15}{2}+\frac{15}{2}(1-t)^2H^2\rho ^2+\frac{15}{2}(1-t)H^2\right] \\&=-\frac{t\rho ^4-g}{(2-t+(1-t)H^2)(3-4t+4(1-t)H^2)}\\&\quad \left[ 10\alpha ^2+(6-8t)\alpha +\frac{15}{2}+l_1(H)\right] , \end{aligned}$$

where the inequality used the coefficient of $\bar{B}$ is always positive for any $\alpha$ since $0<t<\frac{2}{5}$ and $H^2\le \frac{21}{13500nC_2}$ small and

$$\begin{aligned} l_1(H):=&18(1-t)H^2\alpha +\frac{15}{2}(1-t)^2H^2\rho ^2+\frac{15}{2}(1-t)H^2, \end{aligned}$$

so

$$\begin{aligned} (I)+(II)&\ge \frac{1}{(2-t+(1-t)H^2)(3-4t+4(1-t)H^2)(1-t)}\\&\quad \Bigg \{-(1-t)\left[ 10\alpha ^2+(6-8t)\alpha +\frac{15}{2}+l_1(H)\right] (t\rho ^4-g)\\&\quad +\,(3-4t+4(1-t)H^2)\left[ (2\alpha ^2-2\alpha -\frac{3}{2}(1-t)+l_2(H))t\rho ^4 +l_3(H)\right] \Bigg \}\\&=\frac{1}{(2-t+(1-t)H^2)(3-4t+4(1-t)H^2)(1-t)}\\&\quad \Bigg \{-(1-t)\left[ 10\alpha ^2+(6-8t)\alpha +\frac{15}{2}\right] t\rho ^4+(3-4t)\\&\quad \left( 2\alpha ^2-2\alpha -\frac{3}{2}(1-t)\right) t\rho ^4+G_1\Bigg \}\\&=\frac{1}{(2-t+(1-t)H^2)(3-4t+4(1-t)H^2)(1-t)}\\&\quad \{-2(2-t)[\alpha ^2+(3-4t)\alpha +3(1-t)]t\rho ^4+G_1\}, \end{aligned}$$

where

$$\begin{aligned} G_1&:=-(1-t)l_1(H)t\rho ^4+(1-t)\left[ 10\alpha ^2+(6-8t)\alpha +\frac{15}{2}+l_1(H)\right] g\\&\quad +\left[ (3\!-\!4t\!+\!4(1\!-\!t)H^2)l_2(H)\!+\!4(1-t)H^2\!\left( 2\alpha ^2\!-\!2\alpha \!-\!\frac{3}{2}(1-t)\right) \right] t\rho ^4\\&\quad +\,(3-4t+4(1-t)H^2)l_3(H), \end{aligned}$$

then

$$\begin{aligned} (I)+(II)\ge & {} \frac{1}{(2-t+(1-t)H^2)(3-4t+4(1-t)H^2)(1-t)}\nonumber \\{} & {} \quad \left\{ -2(2-t)[\alpha ^2+(3-4t)\alpha +3(1-t)]t\rho ^4+G_1\right\} \nonumber \\= & {} -\frac{2}{(3-4t)(1-t)}[\alpha ^2+(3-4t)\alpha +3(1-t)]t\rho ^4+G_2, \end{aligned}$$

(54)

and denote

$$\begin{aligned} G_2&:=\left( \frac{1}{(2-t)(3-4t)}-\frac{1}{(2-t+(1-t)H^2)(3-4t+4(1-t)H^2)}\right) \frac{2(2-t)}{1-t}\\&\quad [\alpha ^2+(3-4t)\alpha +3(1-t)]t\rho ^4\\&\quad +\,\frac{G_1}{(2-t+(1-t)H^2)(3-4t+4(1-t)H^2)(1-t)}. \end{aligned}$$

So combining (54), putting $\alpha =\frac{4t-3}{2}$, (53) transforms into

$$\begin{aligned} t(2t-1)\rho ^6\ge & {} (I)+(II)+\frac{t}{1-t} \left( \frac{3}{2}-t\right) \rho ^4+\frac{3}{2}tH^2\rho ^4+2t(1-t)H^2\rho ^6+\frac{\bar{C}^2}{\rho ^4}\nonumber \\\ge & {} t\rho ^4(3-\frac{6}{3-4t})+G_3, \end{aligned}$$

(55)

and

$$\begin{aligned} G_3:=G_2+\frac{3}{2}tH^2\rho ^4+2t(1-t)H^2\rho ^6, \end{aligned}$$

to investigate the order of H compared with n, we need to estimate $|\frac{G_3}{t\rho ^6}|$.

Since $t\ge \frac{1}{22}$ and let $H^2n=\delta ^2\le \frac{21}{13500C_2}$, then $|H|\rho =\frac{\delta }{\sqrt{1-t}}$

$$\begin{aligned} |g|= & {} (1-t)|H|\rho ^2|(\bar{f_5}+H^3\rho ^2+2H\rho ^2)|\nonumber \\\le & {} (1-t)|H|\rho ^2(\rho ^5+|H|^3\rho ^2+2|H|\rho ^2)\nonumber \\= & {} \sqrt{1-t}\rho ^6\delta +\frac{\delta ^4}{1-t}+2\rho ^2\delta ^2, \end{aligned}$$

(56)

the first inequality derived from Lemma 1.

Thus

$$\begin{aligned} |l_1(H)|&=|18(1-t)H^2\alpha +\frac{15}{2}(1-t)^2H^2\rho ^2+\frac{15}{2}(1-t)H^2|\\&\le |18(1-t)H^2\alpha |+\left| \frac{15}{2}(1-t)^2H^2\rho ^2\right| +\left| \frac{15}{2}(1-t)H^2\right| \\&=9(1-t)(3-4t)\frac{\delta ^2}{n}+\frac{15}{2}(1-t)\delta ^2+\frac{15}{2}(1-t)\frac{\delta ^2}{n}\\&\le C_3\delta ^2,\\ |l_2(H)|&=|2(1-t)H^2\alpha ^2-2(2-t)(1-t)H^2\alpha -\frac{3}{2}(1-t)^3H^2\rho ^2\\&\quad -\,\frac{3}{2}(1-t)^2H^2|\\&\le \frac{(3-4t)^2(1-t)\delta ^2}{2n}+(2-t)(1-t)(3-4t)\frac{\delta ^2}{n}\\&\quad +\,\frac{3}{2}(1-t)^2\delta ^2+\frac{3}{2}(1-t)^2\frac{\delta ^2}{n}\\&\le C_4\delta ^2,\\ |\frac{l_3(H)}{t\rho ^6}|&=(1-t)|2\alpha -2\alpha ^2-\frac{3}{2}-\frac{3}{2}(1-t)^2H^2\rho ^2\\&\quad -\,\left( 2\alpha +\frac{3}{2}\right) (1-t)H^2|\left| \frac{g}{t\rho ^6}\right| \\&\le (1-t)\left[ 3-4t+\frac{(3-4t)^2}{2}+\frac{3}{2}+\frac{3}{2}(1-t)\delta ^2\right. \\&\left. \quad +\,\left| 4t-\frac{3}{2}\right| (1-t)\frac{\delta ^2}{n}\right] \left( \frac{\sqrt{1-t}}{t}\delta + \frac{(1-t)^2\delta ^4}{tn^3}+2\frac{\delta ^2(1-t)^2}{tn^2}\right) \\&\le C_5\delta , \end{aligned}$$

here $C_3$, $C_4$, $C_5$ are constants which is independent of n because of $\frac{1}{22}\le t\le \frac{2}{5}$ and $\delta ^2\le \frac{21}{13500C_2}$, this leads to

$$\begin{aligned} \left| \frac{G_1}{t\rho ^6}\right| \le C_6\delta , \end{aligned}$$

also $C_6$ is independent of n. And notice that

$$\begin{aligned}&\frac{1}{(2-t)(3-4t)}-\frac{1}{(2-t+(1-t)H^2)(3-4t+4(1-t)H^2)}\\&\quad =\frac{5(1-t)^2\frac{\delta ^2}{n}+4(1-t)^2 \frac{\delta ^4}{n^2}}{(2-t)(3-4t)(2-t+(1-t)H^2)(3-4t+4(1-t)H^2)}\\&\quad \le C_7\frac{\delta ^2}{n}, \end{aligned}$$

then

$$\begin{aligned} \left| \frac{G_2}{t\rho ^6}\right| \le&C_8\delta \\ \left| \frac{G_3}{t\rho ^6}\right| \le&\left| \frac{G_2}{t\rho ^6}\right| +\frac{3}{2} t(1-t)\frac{\delta ^2}{n^2}+2(1-t)\frac{\delta ^2}{n}\\ \le&C_9\delta \end{aligned}$$

and $C_7$, $C_8$, $C_9$ are constants independent of n, we suppose $t\le \frac{2}{5}$, since $n\ge 4$, (55) implies

$$\begin{aligned} 2t-1\ge \frac{3-12t}{3-4t}\frac{1-t}{n}+\frac{G_3}{t\rho ^6}. \end{aligned}$$

(57)

Because $\frac{(3-12t)(1-t)}{3-4t}$ decreases on $[\frac{1}{22},\frac{2}{5}]$, we get

$$\begin{aligned} t\ge \frac{1}{2}-\frac{27}{35n}+\frac{G_3}{2t\rho ^6} \ge \frac{2}{5}+\frac{1}{280}+\frac{G_3}{2t\rho ^6}, \end{aligned}$$

(58)

if we take $\delta <\frac{1}{140C_9}$ such that $\frac{1}{280}+\frac{G_3}{2t\rho ^6}>0$, then $t>\frac{2}{5}$, which is a contradiction.

So

$$\begin{aligned} \rho ^2=\frac{n}{1-t}> n+\frac{2}{3}n, \end{aligned}$$

(59)

and this completes the proof of the main theorem.