Uniqueness of Solutions to a Class of Mixed Hessian Quotient Type Equations

Abstract

In this paper, we investigate the uniqueness of uniformly convex solutions to geometric partial differential equations $\frac{\sigma _k(\eta )}{\sigma _l(\eta )}=u^{p-1}$ when $-(k-l)< p-1 <0$. The result implies that the self-similar solutions of the corresponding curvature flows converge to a round point.

1 Introduction

For a smooth positive function u on sphere ${\mathbb {S}}^{n}$, let $\lambda = (\lambda _1,\lambda _2, \ldots ,\lambda _n )$ be the eigenvalue of the matrix $\{u _{ij}+ u \delta _{ij}\}$, and we introduced the Hessian operators $\sigma _k (\eta )$ (see the definition (2.1)) in convex cone $\widetilde{\Gamma }_k$ (see the definition (2.12)), where $\eta = (\eta _1,\eta _2, \ldots ,\eta _n)$ with $\eta _i = \sum _{j \ne i} {\lambda _j }$. In fact, these operators are mixed Hessian operators (see (2.9)–(2.11)). We remark that these Hessian operators appeared in many literatures. Caffarelli–Nirenberg–Spruck [4] introduced these operators by a self-adjoint map, and studied the Dirichlet problems. The “form-type” Calabi–Yau equation which is proposed by Fu–Wang–Wu [13] and solved by Tosatti–Weinkove [39] on Kähler manifolds, is also of this type. See also the Dirichlet problem for the $(n-1)$-plurisubharmonic functions studied by Harvey–Lawson [22]. We last mention that these operators also stemed from the Gauduchon conjecture [17] which was tackled by Székelyhidi–Tosatti–Weinkove [38], and see Guan–Nie [19] for a related work.

In this paper, we consider the mixed Hessian quotient type equation in the following form

$$\begin{aligned} \frac{\sigma _k(\eta )}{\sigma _l(\eta )} = u^{p-1}f(x), \quad x \in {\mathbb {S}}^{n}, \end{aligned}$$

(1.1)

where $f: {\mathbb {S}}^{n} \mapsto \mathbb R$ is a given positive function, $0 \le l < k \le n$ are integers. If u satisfies $\lambda \in \widetilde{\Gamma _k }$ for any $x \in {\mathbb {S}}^{n}$, the Eq. (1.1) is elliptic and u is called $\widetilde{\Gamma _k }$-admissible (see Proposition 2.6). In particular, if u satisfies $\lambda \in \Gamma _n$, u is called uniformly convex.

In fact, the geometric PDE (1.1) corresponds to a class of $L_p$-Minkowski type problem with mixed Hessian. The $L_p$-Minkowski problem introduced by Lutwak [30] is a generalisation of the classical Minkowski problem and has been intensively studied in recent decades, see [1, 2, 7, 11, 24, 29, 31,32,33, 37, 40,41,42] for example. Among many excellent references, we refer the reader to the newly expanded book [36] of Schneider for a comprehensive introduction on the related topics.

Given a Borel measure m on the unit sphere ${\mathbb {S}}^n$, the $L_p$-Minkowski problem investigates the existence of a unique convex body $\mathbb {K}$ in $\mathbb {R}^{n+1}$ such that m is the $L_p$-surface area measure of $\mathbb {K}$, or equivalently

$$\begin{aligned} dm = u^{1-p}d\mu , \end{aligned}$$

(1.2)

where $\mu$ is the ordinary surface area measure of $\mathbb {K}$ and $u: {\mathbb {S}}^n \rightarrow \mathbb {R}$ is the support function of $\mathbb {K}$. Obviously, when $p = 1$, the $L_p$-Minkowski problem reduces to the classical Minkowski problem. We remark that when $p \ne 1$, the $L_p$-Brunn–Minkowski theory is not a translation-invariant theory, and all convex bodies to which this theory is applied must have the origin in their interiors. Throughout this paper, we will always assume that the origin is contained inside the interior of $\mathbb {K}$, in other words, the support function u is positive on ${\mathbb {S}}^n$. When $f = dm/dx$ is a positive continuous function on ${\mathbb {S}}^n$ and the boundary $\partial \mathbb {K}$ is in a smooth category, for example $C^4$ smooth, (1.2) can be described by the following Monge–Amp$\grave{e}$re type equation:

$$\begin{aligned} det(u_{ij}+u\delta _{ij})=u^{p-1}f, \quad \text {on}\; {\mathbb {S}}^{n}. \end{aligned}$$

(1.3)

When $p \ge 1$, the existence and uniqueness of solutions are well understood.

Recall that the tool used to establish uniqueness in the classical Minkowski problem is the Brunn–Minkowski inequality (among several equivalent forms in Gardner [15]), which was generalized to the $L_p$-Brunn–Minkowski inequality in [30]. $L_p$-Brunn-Minkowski inequalities are among the most important results in $L_p$-Brunn–Minkowski theory, which were intensively investigated since 1990s. They are essentially the isoperimetric type inequalities, and are useful to show the uniqueness of solutions to Minkowski type problems, see for instance [12, 16, 26, 28, 34, 35, 43]. However, the $L_p$-Brunn–Minkowski inequality does not hold when $p < 1$. So the lack of such an important ingredient causes the uniqueness to be a very difficult and challenging problem for the case of $p < 1$. Indeed it was shown in [25] that the uniqueness fails when $p < 0$ even restricted to smooth origin-symmetric convex bodies. Hence, to study the uniqueness of the $L_p$-Minkowski problem for $p < 1$, one needs to impose more conditions on the convex body $\mathbb {K}$ or on the function f. Recently, Brendle–Choi–Daskaspoulos’s work [3] implies the uniqueness holds true for $1> p > -1 -n$ and $f \equiv 1$, and Chen–Huang–Li–Liu [8] prove the uniqueness for p close to 1 and even positive function f.

The associated $L_p$-Christoffel–Minkowski problem in the $L_p$-Brunn–Minkowski theory can be described by the following k-Hessian type equation on ${\mathbb {S}}^{n}$ in smooth case:

$$\begin{aligned} \sigma _k(u_{ij}+u\delta _{ij})=u^{p-1}f, \quad \text {on}\; {\mathbb {S}}^{n}. \end{aligned}$$

(1.4)

The $L_p$-Christoffel–Minkowski problem is difficult to deal with, since the admissible solution to Eq. (1.4) is not necessary a geometric solution to $L_p$-Christoffel–Minkowski problem if $k<n$. So, one needs to deal with the convexity of the solutions of (1.4). Under a sufficient condition on the prescribed function, Guan–Ma [18] proved the existence of a unique convex solution. The key tool to handle the convexity is the constant rank theorem for fully nonlinear partial differential equations. Later, the equation (1.4) has been studied by Hu–Ma–Shen [23] for $p \ge k +1$ and Guan–Xia [20] for $1< p < k+1$ and even prescribed data, by using the constant rank theorem. But for $p < 1$, since the lack of $L_p$ Brunn–Minkowski inequality and constant rank theorem, the existence and uniqueness are still open. Recently, Chen [5] make some progresses on the uniqueness for $1-k \le p < 1$ and $f \equiv 1$.

For the $L_p$-Minkowski type problem with mixed Hessian, Chen–Xu [6] obtained the existence and uniqueness of $\widetilde{\Gamma _k }$-admissible solutions and uniformly convex solutions of (1.1) for $p \ge k-l +1$, and established a full rank theorem of (1.1) for $p \ge 1$. When $p < 1$, the existence and uniqueness of solutions are difficult problems, due to the lack of a priori estimates and full rank Theorem. In this paper, we consider the case $p<1$ and $f\equiv 1$, we obtain the uniqueness of uniformly convex solutions as follows.

Theorem 1.1

For $0\le l< k \le n$, and $-( k-l)< p-1 <0$, the equation

$$\begin{aligned} \frac{\sigma _k(\eta )}{\sigma _l(\eta )} = u^{p-1}, \quad x \in {\mathbb {S}}^{n}, \end{aligned}$$

(1.5)

has a unique, positive, uniformly convex solution $u \equiv \big [ \frac{C_n^l (n-1)^l}{C_n^k (n-1)^k} \big ]^{\frac{1}{k-l-(p-1)}}$.

Remark 1.2

By a similar proof, we can generalize Theorem 1.1 to a class fully nonlinear Hessian equation. That is for the geometric PDE

$$\begin{aligned} F(u _{ij}+ u \delta _{ij}) = u^{\alpha }, \quad x \in {\mathbb {S}}^{n}, \end{aligned}$$

(1.6)

where $F(u _{ij}+ u \delta _{ij}) = f (\lambda )$, and $\lambda =(\lambda _1, \lambda _2, \ldots , \lambda _n)$ are the eigenvalues of $\{u _{ij}+ u \delta _{ij}\}$, and $f (\lambda )$ is smooth, symmetric, positive, homogeneous of degree 1, strictly increasing and concave with respect to $\lambda$ in $\Gamma _n$. Then (1.6) has a unique, positive, uniformly convex solution $u = constant$ for $-1< \alpha <0$.

Remark 1.3

Remark 1.2 is equivalent to Theorem 1.6 in Gao–Li–Wang [14]. In fact, the “inverse concave" condition in [14] is equivalent to that $f (\lambda )$ is concave with respect to $\lambda$, since the equation in [14] is about principal curvatures $\kappa$ (that is the eigenvalue of $\{u _{ij}+ u \delta _{ij} \} ^{-1}$) and $f (\lambda )$ in this paper is about principal radii $\lambda$ (that is the eigenvalue of $\{u _{ij}+ u \delta _{ij}\}$).

In fact, Brendle–Choi–Daskapoulous [3] show the self-similar solution of $\alpha$-Gauss curvature flow, i.e., an embedded, strictly convex hypersurface $\Sigma$ in $\mathbb {R}^{n+1}$ given by $X: {\mathbb {S}}^n \rightarrow \mathbb {R}^{n+1}$ satisfying the equation,

$$\begin{aligned} K^\alpha = <X,\nu >, \end{aligned}$$

(1.7)

is a sphere when $\alpha > \frac{1}{n+2}$, where K and $\nu$ are the Gauss curvature and outward unit normal vector of $\Sigma$ respectively. They apply maximum principle with the following two important auxiliary quantities which are introduced in [3, 10]:

$$\begin{aligned} W(x) = K^\alpha \lambda _1^{-1}(h_{ij}) - \frac{n \alpha - 1}{2n\alpha }|X|^2 = u \lambda _1(b) - \frac{n\alpha - 1}{2n\alpha }\big (u^2 + |Du|^2\big ), \end{aligned}$$

(1.8)

and

$$\begin{aligned} Z(x) = K^\alpha tr(b) - \frac{n\alpha -1}{2\alpha }|X|^2 = u tr(b) - \frac{n\alpha - 1}{2\alpha }\big (u^2 + |Du|^2\big ), \end{aligned}$$

(1.9)

where b is the inverse matrix of the second fundamental form $h_{ij}$ of $\Sigma$, $\lambda _1(b)$ is the biggest eigenvalues of b, $\lambda _1^{-1} (h_{ij}) =\lambda _1(b)$, and $u: {\mathbb {S}}^n \rightarrow \mathbb {R}$ is the support function of $\Sigma$. However, to deal with (1.6) Gao–Li–Wang [14] use the following auxiliary quantities:

$$\begin{aligned} W(x) = F^{1 \over \alpha } \lambda _1^{-1}(h_{ij}) - \frac{ \alpha +1}{2}|X|^2 = u \lambda _1(b) - \frac{\alpha + 1}{2}\big (u^2 + |Du|^2\big ), \end{aligned}$$

(1.10)

and

$$\begin{aligned} Z(x) = F^{1 \over \alpha } \frac{|b|^2}{tr(b)} - \frac{\alpha +1}{2}|X|^2 = u \frac{|b|^2}{tr(b)} - \frac{\alpha + 1}{2}\big (u^2 + |Du|^2\big ). \end{aligned}$$

(1.11)

In view of [3, 14], we consider the auxiliary functions as follows:

$$\begin{aligned} W(x) = u\lambda _{\max } (b ) - \beta \big (u^2 + |\nabla u|^2 \big ), \end{aligned}$$

(1.12)

and

$$\begin{aligned} Z(x) = u g(b) - \beta \big (u^2 + |\nabla u|^2 \big ), \end{aligned}$$

(1.13)

with

$$\begin{aligned} g(b) =: \frac{{\sum \limits _{i = 1}^n {\lambda _i ^2 } }}{{\sigma _1 (\lambda )}} = \sigma _1 (\lambda ) - 2\frac{{\sigma _2 (\lambda )}}{{\sigma _1 (\lambda )}}, \end{aligned}$$

(1.14)

where $\lambda = (\lambda _1,\lambda _2, \ldots ,\lambda _n )$ be the eigenvalues of b and $\lambda _{\max } (b )$ is the maximum eigenvalue, and $\beta = \frac{k-l+p-1}{2(k-l)} \in (0, \frac{1}{2})$.

The rest of the paper is organized as follows. In Sect. 2, we give the definitions and some basic properties of elementary symmetric functions, and introduce Hessian operators $\sigma _k (\eta )$ and further give some properties. At last, we prove Theorem 1.1 in Sect. 3.

2 Preliminaries

In this section, we recall the definition and some basic properties of elementary symmetric functions, which could be found in [27].

Definition 2.1

For any $k = 1, 2,\ldots , n,$ we set

$$\begin{aligned} \sigma _k(\lambda ) = \sum _{1 \le i_1< i_2<\cdots <i_k\le n}\lambda _{i_1}\lambda _{i_2}\cdots \lambda _{i_k}, \qquad \text{ for } \text{ any } \quad \lambda =(\lambda _1,\dots ,\lambda _n)\in {\mathbb R}^n. \end{aligned}$$

(2.1)

We also set $\sigma _0=1$ and $\sigma _k =0$ for $k>n$.

We denote by $\sigma _k (\lambda \left| i \right. )$ the symmetric function with $\lambda _i = 0$ and $\sigma _k (\lambda \left| ij \right. )$ the symmetric function with $\lambda _i =\lambda _j = 0$.

We need the following standard formulas of elementary symmetric functions.

Proposition 2.2

Let $\lambda =(\lambda _1,\dots ,\lambda _n)\in \mathbb {R}^n$ and $k =0, 1, \dots , n$, then

$$\begin{aligned}&\sigma _k(\lambda )=\sigma _k(\lambda |i)+\lambda _i\sigma _{k-1}(\lambda |i), \quad \forall \,1\le i\le n,\\&\sum _{i=1}^n \lambda _i\sigma _{k-1}(\lambda |i)=k\sigma _{k}(\lambda ),\\&\sum _{i=1}^n \sigma _{k}(\lambda |i)=(n-k)\sigma _{k}(\lambda ). \end{aligned}$$

Proposition 2.3

Let $W= \{W_{ij}\}$ is an $n \times n$ symmetric matrix and $\lambda (W)= (\lambda _1,\lambda _2, \dots ,\lambda _{n})$ are the eigenvalues of the symmetric matrix W. Suppose that $W= \{W_{ij}\}$ is diagonal and $\lambda _i= W_{ii}$, then we have

$$\begin{aligned}&\frac{{\partial \lambda _i }}{{\partial W_{ii} }} = 1,\quad \frac{{\partial \lambda _k }}{{\partial W_{ij} }} = 0,~~ \text {otherwise} , \end{aligned}$$

(2.2)

$$\begin{aligned}&\frac{{\partial ^2 \lambda _i }}{{\partial W_{ij} \partial W_{ji} }} = \frac{1}{{\lambda _i - \lambda _j }}, \quad i \ne j \text { and } \lambda _i \ne \lambda _j,\end{aligned}$$

(2.3)

$$\begin{aligned}&\frac{{\partial ^2 \lambda _i }}{{\partial W_{kl} \partial W_{pq} }} = 0, \text {otherwise}. \end{aligned}$$

(2.4)

Definition 2.1 can be extended to symmetric matrices by letting $\sigma _k(W) = \sigma _k(\lambda (W))$, where $\lambda (W)= (\lambda _1(W),\lambda _2 (W), \ldots ,\lambda _{n}(W))$ are the eigenvalues of the symmetric matrix W.

Recall that the Gårding’s cone is defined as

$$\begin{aligned} \Gamma _k = \{ \lambda \in \mathbb {R}^n:\sigma _i (\lambda ) > 0,\forall 1 \le i \le k\}, \end{aligned}$$

(2.5)

and the following properties are well known.

Proposition 2.4

(Generalized Newton–MacLaurin inequality) For $\lambda \in \Gamma _k$ and $k > l \ge 0$, $r > s \ge 0$, $k \ge r$, $l \ge s$, we have

$$\begin{aligned} \Bigg [\frac{{\sigma _k (\lambda )}/{C_n^k }}{{\sigma _l (\lambda )}/{C_n^l }}\Bigg ]^{\frac{1}{k-l}} \le \Bigg [\frac{{\sigma _r (\lambda )}/{C_n^r }}{{\sigma _s (\lambda )}/{C_n^s }}\Bigg ]^{\frac{1}{r-s}}, \end{aligned}$$

(2.6)

and the equality holds if and only if $\lambda _1 = \lambda _2 = \cdots =\lambda _n >0$.

Proposition 2.5

• (1) $\Gamma _k$ are convex cones, and $\Gamma _1 \supset \Gamma _2 \supset \cdots \supset \Gamma _n$.

• (2) If $\lambda =(\lambda _1,\ldots ,\lambda _n) \in \Gamma _k$ with $\lambda _1 \ge \lambda _2 \ge \cdots \ge \lambda _n$, then

  $$\begin{aligned} \sigma _{k-1} (\lambda |n) \ge \sigma _{k-1} (\lambda |n-1) \ge \cdots \ge \sigma _{k-1} (\lambda |1) >0. \end{aligned}$$

  (2.7)

• (3) If $\lambda =(\lambda _1,\ldots ,\lambda _n) \in \Gamma _k$, then $\sigma _k(\lambda )^{\frac{1}{k}}$ and $\Big [\frac{\sigma _k(\lambda )}{\sigma _l(\lambda )}\Big ]^{\frac{1}{k-l}}$ ($0 \le l <k \le n$) are concave with respect to $\lambda$. Equivalently, for any $(\xi _1, \ldots , \xi _n)$,

  $$\begin{aligned} \sum \limits _{i,j=1}^n\frac{\partial ^2 \Big [\frac{\sigma _k(\lambda )}{\sigma _l(\lambda )}\Big ]}{\partial \lambda _i \partial \lambda _j} \xi _i \xi _j \le&(1- \frac{1}{k-l}) \frac{[\sum \limits _{i=1}^n \frac{\partial \Big [\frac{\sigma _k(\lambda )}{\sigma _l(\lambda )}\Big ]}{\partial \lambda _i} \xi _i ]^2}{\frac{\sigma _k(\lambda )}{\sigma _l(\lambda )}}. \end{aligned}$$

  (2.8)

In the following, for $\lambda = (\lambda _1,\lambda _2, \ldots ,\lambda _n )$, we assume $\eta = (\eta _1,\eta _2, \ldots ,\eta _n)$ with $\eta _i = \sum _{p \ne i} {\lambda _p }$. We will give some important properties of $\sigma _k(\eta )$ referring to Chen–Dong–Han [9] or Chen–Xu [6]. Firstly, we point out that $\sigma _k(\eta )$ are really mixed Hessian operators since direct computations yield

$$\begin{aligned}&\sigma _\mathrm{{1}} (\eta ) = \sum \limits _{i = 1}^n {\sigma _\mathrm{{1}} (\lambda |i)} = (n - 1)\sigma _\mathrm{{1}} (\lambda ), \end{aligned}$$

(2.9)

$$\begin{aligned}&\sigma _2 (\eta ) =\frac{{(n - 1)(n - 2)}}{2}\sigma _1 (\lambda )^2 + \sigma _\mathrm{{2}} (\lambda ), \end{aligned}$$

(2.10)

and

$$\begin{aligned} \sigma _3 (\eta ) = \frac{{(n - 1)(n - 2)(n-3)}}{6}\sigma _1 (\lambda )^3 + (n-2) \sigma _1 (\lambda )\sigma _\mathrm{{2}} (\lambda ) - \sigma _3 (\lambda ). \end{aligned}$$

(2.11)

As in Chen–Dong–Han [9], we define the cone

$$\begin{aligned} \widetilde{\Gamma _k } = \{ \lambda = (\lambda _1 ,\lambda _2 , \ldots ,\lambda _n ): \sigma _i (\eta ) > 0,1 \le i \le k\}, \end{aligned}$$

(2.12)

and we have the following properties.

Proposition 2.6

• (1) $\widetilde{\Gamma _k }$ are convex cones, and

  $$\begin{aligned} \Gamma _1 = \widetilde{\Gamma _1} \supset \widetilde{\Gamma _2 } \supset \cdots \supset \widetilde{\Gamma _n} \supset \Gamma _2 \supset \cdots \supset \Gamma _n. \end{aligned}$$

  (2.13)

• (2) If $\lambda =(\lambda _1,\dots ,\lambda _n) \in \widetilde{\Gamma _k}$, then

  $$\begin{aligned} \frac{\partial \Big [\frac{\sigma _k(\eta )}{\sigma _l(\eta )} \Big ] }{\partial \lambda _i}&= \sum _{p \ne i} \frac{\sigma _{k-1} (\eta |p) \sigma _l(\eta )- \sigma _k(\eta )\sigma _{l-1}(\eta |p)}{\sigma _l(\eta )^2} \nonumber \\&\ge \frac{n(k-l)}{k(n-l)} \sum _{p \ne i}\frac{\sigma _{k-1} (\eta |p) \sigma _l(\eta |p)}{\sigma _l(\eta )^2} > 0, \end{aligned}$$

  (2.14)

  for $i=1, 2, \dots , n$ and $0 \le l <k \le n$.

• (3) If $\lambda =(\lambda _1,\dots ,\lambda _n) \in \widetilde{\Gamma _k}$, then $\Big [\frac{\sigma _k(\eta )}{\sigma _l(\eta )} \Big ]^{\frac{1}{k-l}}$ ($0 \le l <k \le n$) are concave with respect to $\lambda$. Equivalently, for any $(\xi _1, \dots , \xi _n)$

  $$\begin{aligned} \sum \limits _{i,j=1}^n\frac{\partial ^2 \Big [\frac{\sigma _k(\eta )}{\sigma _l(\eta )} \Big ]}{\partial \lambda _i \partial \lambda _j} \xi _i \xi _j \le&(1- \frac{1}{k-l}) \frac{[\sum \limits _{i=1}^n \frac{\partial \Big [\frac{\sigma _k(\eta )}{\sigma _l(\eta )} \Big ]}{\partial \lambda _i} \xi _i ]^2}{\frac{\sigma _k(\eta )}{\sigma _l(\eta )}}. \end{aligned}$$

  (2.15)

Proof

First, we prove $\widetilde{\Gamma _k }$ are convex cones. For any $\lambda \in \widetilde{\Gamma _k }$ and any $\widetilde{\lambda } \in \widetilde{\Gamma _k }$, we have

$$\begin{aligned} \sigma _i (\eta )> 0, \sigma _i (\widetilde{\eta }) >0, \end{aligned}$$

for $i=1, 2, \dots , k$. Hence $\eta \in \Gamma _k$ and $\widetilde{\eta } \in \Gamma _k$. Since $\Gamma _k$ is convex (see Proposition 2.5), we know $\frac{\eta +\widetilde{\eta }}{2} \in \Gamma _k$, that is

$$\begin{aligned} \sigma _i \big (\frac{\eta +\widetilde{\eta }}{2}\big ) > 0, \end{aligned}$$

for $i=1, 2, \dots , k$. Hence $\frac{\lambda +\widetilde{\lambda }}{2}\in \widetilde{\Gamma _k }$. This means $\widetilde{\Gamma _k }$ is convex.

Now we prove (2.13). From (2.9), we know $\Gamma _1 = \widetilde{\Gamma _1}$. Given any $\lambda \in \Gamma _2$, by (2.7) we know

$$\begin{aligned} \eta _i = \sigma _1(\lambda |i) >0, \end{aligned}$$

which means $\Gamma _2 \subset \widetilde{\Gamma _n}$. Hence (2.13) holds.

Second, for $\lambda =(\lambda _1,\dots ,\lambda _n) \in \widetilde{\Gamma _k}$, then $\eta \in \Gamma _k$ and we can directly get $\sigma _{k-1} (\eta |p) >0$ and $\sigma _{l} (\eta |p) >0$. Combining these facts with Propositions 2.2 and 2.4, we obtain

$$\begin{aligned} \frac{\partial \Big [\frac{\sigma _k(\eta )}{\sigma _l(\eta )} \Big ]}{\partial \lambda _i}&= \sum _{p}\frac{\partial \Big [\frac{\sigma _k(\eta )}{\sigma _l(\eta )} \Big ]}{\partial \eta _p} \frac{\partial \eta _p}{\partial \lambda _i} = \sum _{p \ne i} \frac{\sigma _{k-1} (\eta |p) \sigma _l(\eta )- \sigma _k(\eta )\sigma _{l-1}(\eta |p)}{\sigma _l(\eta )^2} \\&= \sum _{p \ne i} \frac{\sigma _{k-1} (\eta |p) \sigma _l(\eta |p)- \sigma _k(\eta |p)\sigma _{l-1}(\eta |p)}{\sigma _l(\eta )^2} \\&\ge \sum _{p \ne i}\frac{\sigma _{k-1} (\eta |p) \sigma _l(\eta |p) - \frac{l(n-k)}{k(n-l)} \sigma _{k-1} (\eta |p) \sigma _l(\eta |p)}{\sigma _l(\eta )^2} \\&=\frac{n(k-l)}{k(n-l)} \sum _{p \ne i}\frac{\sigma _{k-1} (\eta |p) \sigma _l(\eta |p)}{\sigma _l(\eta )^2} >0. \end{aligned}$$

From (2.8) in Proposition 2.5, it follows that

$$\begin{aligned} \sum \limits _{i,j=1}^n\frac{\partial ^2 \Big [\frac{\sigma _k(\eta )}{\sigma _l(\eta )} \Big ]}{\partial \lambda _i \partial \lambda _j} \xi _i \xi _j&= \sum \limits _{i,j,a,b=1}^n \frac{\partial ^2 \Big [\frac{\sigma _k(\eta )}{\sigma _l(\eta )} \Big ]}{\partial \eta _a \partial \eta _b} \frac{\partial \eta _a}{\partial \lambda _i} \frac{\partial \eta _b}{\partial \lambda _j}\xi _i \xi _j \nonumber \\&\le \bigg (1- \frac{1}{k-l}\bigg ) \frac{\left[ \sum \limits _{a=1}^n \frac{\partial \Big [\frac{\sigma _k(\eta )}{\sigma _l(\eta )} \Big ]}{\partial \eta _a}\Big ( \sum \limits _{i=1}^n \frac{\partial \eta _a}{\partial \lambda _i}\xi _i \Big )\right] ^2}{\frac{\sigma _k(\eta )}{\sigma _l(\eta )} } \nonumber \\&=\bigg (1- \frac{1}{k-l}\bigg ) \frac{[\sum \limits _{i=1}^n \frac{\partial \Big [\frac{\sigma _k(\eta )}{\sigma _l(\eta )} \Big ]}{\partial \lambda _i} \xi _i ]^2}{\frac{\sigma _k(\eta )}{\sigma _l(\eta )} }. \end{aligned}$$

(2.16)

$\square$

Proposition 2.7

If $\lambda =(\lambda _1,\dots ,\lambda _n)$ are eigenvalues of matrix $\{ b_{ij} \}$ and $\lambda \in \widetilde{\Gamma _k}$, then

• (1) $\frac{\sigma _k(\eta )}{\sigma _l(\eta )}$ is elliptic with respect to $\{ b_{ij} \}$, that is

  $$\begin{aligned} \left\{ \frac{\partial \big [\frac{\sigma _k(\eta )}{\sigma _l(\eta )} \big ]}{\partial b_{ij}} \right\} >0, \end{aligned}$$

  (2.17)

• (2) $\Big [\frac{\sigma _k(\eta )}{\sigma _l(\eta )}\Big ]^{\frac{1}{k-l}}$ is concave w.r.t. $\{ b_{ij} \}$, i.e., for any $n \times n$ symmetric matrix $\{ \xi _{ij} \}$,

  $$\begin{aligned} \sum _{i,j,r,s=1}^n\frac{\partial ^2 \Big [\frac{\sigma _k(\eta )}{\sigma _l(\eta )} \Big ]}{\partial b_{ij}\partial b_{rs}}\xi _{ij}\xi _{rs} \le (1- \frac{1}{k-l}) \frac{\Big [\sum \limits _{i,j=1}^n \frac{\partial \Big [\frac{\sigma _k(\eta )}{\sigma _l(\eta )} \Big ]}{\partial b_{ij}} \xi _{ij} \Big ]^2}{\frac{\sigma _k(\eta )}{\sigma _l(\eta )} }. \end{aligned}$$

  (2.18)

Proof

Without loss of generality, we assume $\{ b_{ij} \}$ is diagonal, then by Proposition 2.3 we have

$$\begin{aligned} \frac{\partial \Big [\frac{\sigma _k(\eta )}{\sigma _l(\eta )} \Big ]}{\partial b_{ij}}= \sum _{p=1}^n\frac{\partial \Big [\frac{\sigma _k(\eta )}{\sigma _l(\eta )} \Big ]}{\partial \lambda _{p}}\frac{\partial \lambda _{p}}{\partial b_{ij}}= \left\{ \begin{array}{l} \begin{array}{*{20}c} {\frac{\partial \Big [\frac{\sigma _k(\eta )}{\sigma _l(\eta )} \Big ]}{\partial \lambda _{i}},} &{} {i = j} \\ \end{array} \\ \begin{array}{*{20}c} {0,} &{} {i \ne j.} \\ \end{array} \\ \end{array} \right. \end{aligned}$$

(2.19)

So (2.17) holds thanks to (2.14).

Using Proposition 2.3 again, we have

$$\begin{aligned} \frac{\partial ^2 \Big [\frac{\sigma _k(\eta )}{\sigma _l(\eta )} \Big ]}{\partial b_{ij}\partial b_{rs}}=&\sum _{p,q=1}^n\frac{\partial ^2 \Big [\frac{\sigma _k(\eta )}{\sigma _l(\eta )} \Big ]}{\partial \lambda _{p}\partial \lambda _{q}} \frac{\partial \lambda _{p}}{\partial b_{ij}}\frac{\partial \lambda _{q}}{\partial b_{rs}} + \sum _{p=1}^n\frac{\partial \Big [\frac{\sigma _k(\eta )}{\sigma _l(\eta )} \Big ]}{\partial \lambda _{p}}\frac{\partial ^2 \lambda _{p}}{\partial b_{ij} \partial b_{rs}} \nonumber \\ =&\left\{ \begin{array}{l} \begin{array}{*{20}c} {\frac{\partial ^2 \Big [\frac{\sigma _k(\eta )}{\sigma _l(\eta )} \Big ]}{\partial \lambda _{i}\partial \lambda _{r}} ,} &{} {i = j, r = s} \\ \end{array} \\ \begin{array}{*{20}c} { \frac{\frac{\partial \Big [\frac{\sigma _k(\eta )}{\sigma _l(\eta )} \Big ]}{\partial \lambda _{i}}-\frac{\partial \Big [\frac{\sigma _k(\eta )}{\sigma _l(\eta )} \Big ]}{\partial \lambda _{i}}}{\lambda _{i} -\lambda _{j}},} &{} {s= i \ne j = r} \\ \end{array} \\ \begin{array}{*{20}c} {0,} &{} {otherwise} \\ \end{array} \\ \end{array}. \right. \end{aligned}$$

(2.20)

The Lemma 2.7 in [9] reads: If $\lambda _1 \ge \lambda _2 \ge \cdots \ge \lambda _n$, then

$$\begin{aligned} \frac{\partial \Big [\frac{\sigma _k(\eta )}{\sigma _l(\eta )} \Big ]}{\partial \lambda _{1}} \le \frac{\partial \Big [\frac{\sigma _k(\eta )}{\sigma _l(\eta )} \Big ]}{\partial \lambda _{2}} \le \cdots \le \frac{\partial \Big [\frac{\sigma _k(\eta )}{\sigma _l(\eta )} \Big ]}{\partial \lambda _{n}}. \end{aligned}$$

(2.21)

Then combining (2.20), (2.21) and Proposition 2.6 gives that

$$\begin{aligned} \sum _{i,j,r,s=1}^n\frac{\partial ^2 \Big [\frac{\sigma _k(\eta )}{\sigma _l(\eta )} \Big ]}{\partial b_{ij}\partial b_{rs}}\xi _{ij}\xi _{rs}&= \sum _{i, r=1}^n \frac{\partial ^2 \Big [\frac{\sigma _k(\eta )}{\sigma _l(\eta )} \Big ]}{\partial b_{ii}\partial b_{rr}}\xi _{ii}\xi _{rr} + \sum _{i, j=1}^n \frac{\partial ^2 \Big [\frac{\sigma _k(\eta )}{\sigma _l(\eta )} \Big ]}{\partial b_{ij}\partial b_{ji}}\xi _{ij}^2 \nonumber \\&\le \sum _{i, r=1}^n \frac{\partial ^2 \Big [\frac{\sigma _k(\eta )}{\sigma _l(\eta )} \Big ]}{\partial \lambda _{i}\partial \lambda _{r}}\xi _{ii}\xi _{rr} \nonumber \\&\le (1- \frac{1}{k-l}) \frac{\Big [\sum \limits _{i,j=1}^n \frac{\partial \Big [\frac{\sigma _k(\eta )}{\sigma _l(\eta )} \Big ]}{\partial b_{ij}} \xi _{ij} \Big ]^2}{\frac{\sigma _k(\eta )}{\sigma _l(\eta )} }. \end{aligned}$$

(2.22)

$\square$

3 Proof of Theorem 1.1

In this section, we prove Theorem 1.1 following the idea of Brendle–Choi–Daskalopoulos [3] and Gao–Li–Wang [14].

Denote $b_{ij} = u_{ij} + u\delta _{ij}$ for $1 \le i, j \le n$, and let $\lambda = (\lambda _1,\lambda _2, \ldots ,\lambda _n )$ be the eigenvalues of $\{ b_{ij} \}$. To prove Theorem 1.1, we just need to prove $\lambda _1 = \lambda _2 = \cdots = \lambda _n$ for any $x \in {\mathbb {S}}^n$.

As in Brendle–Choi–Daskapoulous [3], we consider the auxiliary function

$$\begin{aligned} W(x) = u\lambda _{\max } (b_{ij} ) - \beta \big (u^2 + |\nabla u|^2\big ), \end{aligned}$$

(3.1)

where $\lambda _{\max } (b_{ij} )$ is the maximum eigenvalue of $\{ b_{ij} \}$, and $\beta = \frac{k-l+p-1}{2(k-l)} \in (0, \frac{1}{2})$. We denote

$$\begin{aligned} \Psi =\{ x \in {\mathbb {S}}^n:W(x) = \mathop {\max }\limits _{{\mathbb {S}}^n } W\}. \end{aligned}$$

Next, We divide the proof into two steps.

Step 1 We will prove $\lambda _1 = \lambda _2 = \cdots = \lambda _n$ for any $x_0 \in \Psi$.

For any $x_0 \in \Psi$, we choose local coordinates at $x_0$ such that

$$\begin{aligned} \{ b_{ij} (x_0 )\} \text { is diagonal, and } \lambda _i = b_{ii} (x_0 ),~~ \lambda _1 \ge \lambda _2 \ge \cdots \ge \lambda _n. \end{aligned}$$

(3.2)

Moreover, if the eigenvalues are not the same, we assume $\lambda _1 = \cdots = \lambda _\mu > \lambda _{\mu + 1} \ge \cdots \ge \lambda _n$ for $1 \le \mu < n$. In addition, we can rotate $\{e_1,e_2, \ldots ,e_\mu \}$, such that

$$\begin{aligned} u_1 (x_0 ) = |(u_1 (x_0 ), \ldots ,u_\mu (x_0 ))|, \end{aligned}$$

(3.3)

that is $u_2 (x_0 ) = \cdots = u_\mu (x_0 ) = 0$. As in Brendle–Choi–Daskapoulous [3], we can define a smooth function $\varphi$ on ${\mathbb {S}}^n$ such that

$$\begin{aligned} {\widetilde{W}} := u\varphi - \beta (u^2 + |\nabla u|^2 )\equiv W(x_0). \end{aligned}$$

(3.4)

Since W attains its maximum at $x_0$, we obtain $\varphi (x_0 ) = b_{11} (x_0 )$, $\varphi (x) \ge \lambda _{\max } (b_{ij}(x) )$ everywhere. Moreover, by a similar lemma to Lemma 5 in [3] (or see Lemma 2.2 in [5]), we can prove

$$\begin{aligned} \varphi _i (x_0 ) =&b_{11i} (x_0 ), \end{aligned}$$

(3.5)

$$\begin{aligned} \varphi _{ii} (x_0 ) \ge&b_{11ii} (x_0 ) + 2\sum \limits _{j > \mu } {\frac{{b_{1ij} (x_0 )^2 }}{{\lambda _1 - \lambda _j }}}. \end{aligned}$$

(3.6)

By assumption, ${\widetilde{W}}$ is constant. Consequently,

$$\begin{aligned} 0 = \widetilde{W_i }&= u_i \varphi + u\varphi _i - \beta \big (2uu_i + 2\sum _ju_j u_{ji}\big ) \nonumber \\&= u_i b_{11} + ub_{11i} - 2\beta u_i b_{ii}, \end{aligned}$$

(3.7)

which yields

$$\begin{aligned} b_{11i} = \frac{{u_i }}{u}(2\beta b_{ii} - b_{11} ), \quad i = 1,2, \ldots , n. \end{aligned}$$

(3.8)

Denote

$$\begin{aligned} F = \left[ {\frac{{\sigma _k (\eta )}}{{\sigma _l (\eta )}}} \right] ^{\frac{1}{{k - l}}}, \qquad F^{ij} = \frac{\partial F}{\partial b_{ij} }, \qquad F^{ij,rs} = \frac{\partial ^2 F}{\partial b_{ij} \partial b_{rs} }. \end{aligned}$$

(3.9)

$\{F^{ij} \} >0$ because of (2.17). Moreover, $\{F^{ij} \}$ is diagonal at $x_0$, and

$$\begin{aligned} \sum _i F^{ii} b_{ii} = F. \end{aligned}$$

(3.10)

If we rewrite the equation (1.5) as

$$\begin{aligned} F = u^\alpha , \qquad \alpha = \frac{{p - 1}}{{k - l}} = 2 \beta -1 \le 0. \end{aligned}$$

(3.11)

Then at $x_0$, we have for any $m=1,...,n$

$$\begin{aligned} \sum _i F^{ii} b_{iim} = \alpha u^{\alpha - 1} u_m, \end{aligned}$$

(3.12)

and

$$\begin{aligned} \sum _i F^{ii} b_{iimm} = \alpha u^{\alpha - 2} [uu_{mm} + (\alpha - 1)u_m ^2 ] - \sum _{i,j,r,s} F^{ij,rs} b_{ijm} b_{rsm}. \end{aligned}$$

(3.13)

In fact, by Proposition 2.3 and (2.15)

$$\begin{aligned} \sum _{i,j,r,s}F^{ij,rs} b_{ij1} b_{rs1}&=\sum _{i,j} F^{ii,jj} b_{ii1} b_{jj1} + \sum \limits _{i \ne j} {F^{ij,ji} b_{ij1} ^2 } \nonumber \\&\le \sum \limits _{i \ne j} {F^{ij,ji} b_{ij1} ^2 } \nonumber \\&= 2\sum \limits _{\lambda _i> \lambda _j} \frac{F^{ii}-F^{jj}}{\lambda _i -\lambda _j} b_{ij1} ^2 \nonumber \\ \le&2\sum \limits _{j > \mu } {\frac{{F^{11} - F^{jj} }}{{\lambda _1 - \lambda _j }}b_{11j} ^2 }, \end{aligned}$$

(3.14)

where the last inequality holds by the fact that $F^{ii} \le F^{jj}$ if $\lambda _i \ge \lambda _j$ (see Lemma 2.7 in [9]). So combining (3.13) with (3.14), we know that

$$\begin{aligned} \sum _i F^{ii} b_{ii11} \ge \alpha u^{\alpha - 2} [ub_{11} - u^2 + (\alpha - 1)u_1 ^2 ] - 2\sum \limits _{j > \mu } {\frac{{F^{11} - F^{jj} }}{{\lambda _1 - \lambda _j }}b_{11j} ^2 }. \end{aligned}$$

(3.15)

At $x_0$, we also have $\{\widetilde{W_{ii} }(x_0 )\} \le 0$ and thus

$$\begin{aligned} \sum _i F^{ii} \widetilde{W_{ii} }(x_0 ) \le 0. \end{aligned}$$

(3.16)

However, in view of (3.1)–(3.15) and the Ricci identity (see [21]), we compute that

$$\begin{aligned} \sum _i F^{ii} \widetilde{W_{ii} }(x_0 ) =&\sum _i F^{ii} [u_{ii} \varphi + 2u_i \varphi _i + u\varphi _{ii} - \beta (2u_i ^2 \nonumber \\&+2uu_{ii} + 2\sum _j u_{ji} ^2 + 2\sum _j u_j u_{jii} )] \nonumber \\ \ge&\sum _i F^{ii} \left[ {u_{ii} b_{11} + 2u_i b_{11i} + u\left( {b_{11ii} + 2\sum \limits _{j> \mu } {\frac{{b_{1ij} ^2 }}{{\lambda _1 - \lambda _j }}} } \right) } \right. \nonumber \\&\left. { - 2\beta (u_{ii} b_{ii} + \sum _j u_j b_{jii} )} \right] \nonumber \\ =&\,\sum _i F^{ii} \left[ {(b_{ii} - u)b_{11} + 2\frac{{u_i ^2 }}{u}(2\beta b_{ii} - b_{11} )} \right. \nonumber \\&+\, u\left( {b_{ii11} - b_{ii} + b_{11} + 2\sum \limits _{j> \mu } {\frac{{b_{1ij} ^2 }}{{\lambda _1 - \lambda _j }}} } { - 2\beta (b_{ii} ^2 - ub_{ii} +\sum _j u_j b_{jii} )} \right] \nonumber \\ \ge&Fb_{11} + 2\sum _i F^{ii} \frac{{u_i ^2 }}{u}(2\beta b_{ii} - b_{11} ) - uF \nonumber \\&+ \,u\left[ {\alpha u^{\alpha - 2} [ub_{11} - u^2 + (\alpha - 1)u_1 ^2 ] - 2\sum \limits _{j> \mu } {\frac{{F^{11} - F^{jj} }}{{\lambda _1 - \lambda _j }}b_{11j} ^2 } } \right] \nonumber \\&+ 2uF^{11} \sum \limits _{j> \mu } {\frac{{b_{11j} ^2 }}{{\lambda _1 - \lambda _j }}} - 2\beta (\sum _i F^{ii} b_{ii} ^2 - uF + \alpha u^{\alpha - 1} \sum _j u_j ^2 ) \nonumber \\ =&\, b_{11} [F + \alpha u^\alpha ] - 2\beta \sum _i F^{ii} b_{ii} ^2 + uF[ - 1 - \alpha + 2\beta ] \nonumber \\&+ \,\frac{{u_1 ^2 }}{u}\left[ {2F^{11} (2\beta b_{11} - b_{11} ) + \alpha (\alpha - 1)u^\alpha - 2\beta \alpha u^\alpha } \right] \nonumber \\&+ \,\sum \limits _{i> \mu } {\frac{{u_i ^2 }}{u}\left[ {2F^{ii} (2\beta b_{ii} - b_{11} ) + 2F^{ii} \frac{{(2\beta b_{ii} - b_{11} )^2 }}{{b_{11} - b_{ii} }} - 2\beta \alpha u^\alpha } \right] } \nonumber \\ =&\, 2\beta \sum _i F^{ii} b_{ii} (b_{11} - b_{ii} ) + \frac{{u_1 ^2 }}{u}2\alpha \left[ {F^{11} b_{11} - u^\alpha } \right] \nonumber \\&+\, \sum \limits _{i> \mu } {\frac{{u_i ^2 }}{u}\left[ {2F^{ii} \frac{{(2\beta - 1)^2 b_{ii} ^2 }}{{b_{11} - b_{ii} }} - 2\beta \alpha u^\alpha } \right] } \nonumber \\ >&0. \end{aligned}$$

(3.17)

This contradicts to (3.16), which implies $b_{11} = b_{22} = \cdots = b_{nn}$ at $x_0$.

Step 2 We will prove that $\Psi$ is an open set.

As in Gao–Li–Wang [14], we take the auxiliary function

$$\begin{aligned} Z(x) = u g(b_{ij} ) - \beta (u^2 + |\nabla u|^2 ), \end{aligned}$$

(3.18)

where

$$\begin{aligned} g(b_{ij} ) =: \frac{{\sum \limits _{i = 1}^n {\lambda _i ^2 } }}{{\sigma _1 (\lambda )}} = \sigma _1 (\lambda ) - 2\frac{{\sigma _2 (\lambda )}}{{\sigma _1 (\lambda )}}. \end{aligned}$$

(3.19)

We can easily see that g is 1-homogenous and convex in $\lambda$.

For any fixed $x_0 \in \Psi$, we consider a small neighborhood $N_{x_0}$ of $x_0$. $\forall x \in N_{x_0 }$, we choose local coordinates such that

$$\begin{aligned} \{ b_{ij} (x)\} \text { is diagonal, and } \lambda _i = b_{ii} (x). \end{aligned}$$

(3.20)

Then at x, we have for any $i=1,...,n$

$$\begin{aligned} Z_i&= u_i g + ug_i - \beta \bigg (2uu_i + 2\sum _j u_j u_{ji} \bigg ) \nonumber \\&= u_i g + ug_i - 2\beta u_i b_{ii}, \end{aligned}$$

(3.21)

which yields

$$\begin{aligned} g_i = \frac{{Z_i }}{u} + \frac{{u_i }}{u}(2\beta b_{ii} - g). \end{aligned}$$

(3.22)

Moreover, $\{F^{ij}\}$ is diagonal at x, and we have

$$\begin{aligned} \sum _i F^{ii} Z_{ii}&= \sum _i F^{ii} [u_{ii} g + 2u_i g_i + ug_{ii} - \beta (2u_i ^2 + 2uu_{ii} + 2\sum _j u_{ji} ^2 + 2\sum _ju_j u_{jii} )] \nonumber \\&\ge \sum _i F^{ii} \left[ {u_{ii} g + 2u_i g_i + u\left( \sum _j {\frac{{\partial g}}{{\partial b_{jj} }}b_{jjii} + \sum _{j,m,r,s}\frac{{\partial ^2 g}}{{\partial b_{jm} \partial b_{rs} }}b_{jmi} b_{rsi} } \right) }\right. \nonumber \\&\quad \left. { - \,2\beta (u_{ii} b_{ii} +\sum _j u_j b_{jii} )} \right] \ge \sum _iF^{ii} \left[ {(b_{ii} - u)g + 2u_i \frac{{Z_i }}{u} + 2\frac{{u_i ^2 }}{u}(2\beta b_{ii} - g)} \right. \nonumber \\&\quad + u\sum _j\frac{{\partial g}}{{\partial b_{jj} }}\left( {b_{iijj} - b_{ii} + b_{jj} } \right) \left. { - 2\beta (b_{ii} ^2 - ub_{ii} + \sum _ju_j b_{jii} )} \right] \nonumber \\&\ge Fg + 2\sum _iF^{ii} u_i \frac{{Z_i }}{u} + 2\sum _iF^{ii} \frac{{u_i ^2 }}{u}(2\beta b_{ii} - g) - u\sum _j\frac{{\partial g}}{{\partial b_{jj} }}F \nonumber \\&\quad + u\sum _j\frac{{\partial g}}{{\partial b_{jj} }}\alpha u^{\alpha - 2} [ub_{jj} - u^2 + (\alpha - 1)u_j ^2 ] - 2\beta (\sum _iF^{ii} b_{ii} ^2 \nonumber \\&\quad - uF + \alpha u^{\alpha - 1} \sum _ju_j ^2 ) = 2\sum _iF^{ii} u_i \frac{{Z_i }}{u} + g[F + \alpha u^\alpha ] \nonumber \\&\quad -\,2\beta \sum _iF^{ii} b_{ii} ^2 + uF[ -\sum _j \frac{{\partial g}}{{\partial b_{jj} }} -\sum _j \alpha \frac{{\partial g}}{{\partial b_{jj} }} + 2\beta ] \nonumber \\&\quad + \sum \limits _{i} {\frac{{u_i ^2 }}{u}\left[ {2F^{ii} (2\beta b_{ii} - g) + \alpha (\alpha - 1)u^\alpha \frac{{\partial g}}{{\partial b_{ii} }} - 2\beta \alpha u^\alpha } \right] }, \end{aligned}$$

(3.23)

where we use (3.13) and (2.18) to deal with $\sum _iF^{ii}b_{iijj}$.

In the following, we give three Claims.

Claim 1 $g[F + \alpha u^\alpha ] - 2\beta \sum _i F^{ii} b_{ii} ^2 \ge 0$.

Obviously,

$$\begin{aligned} g[F + \alpha u^\alpha ] - 2\beta \sum _iF^{ii} b_{ii} ^2&= 2\beta [gF - \sum _iF^{ii} b_{ii} ^2 ] \nonumber \\&= \frac{{2\beta }}{{\sigma _1 (\lambda )}}[\sum \limits _j {b_{jj} ^2 } \sum \limits _i {F^{ii} b_{ii} } - \sum \limits _i {F^{ii} b_{ii} ^2 } \sum \limits _j {b_{jj} } ] \nonumber \\&= \frac{{2\beta }}{{\sigma _1 (\lambda )}}\sum \limits _{i,j} {b_{ii} b_{jj} F^{ii} (b_{jj} - b_{ii} )}\nonumber \\&= \frac{\beta }{{\sigma _1 (\lambda )}}\sum \limits _{i,j} {b_{ii} b_{jj} [F^{ii} (b_{jj} - b_{ii} ) + F^{jj} (b_{ii} - b_{jj} )]} \nonumber \\&= \frac{\beta }{{\sigma _1 (\lambda )}}\sum \limits _{i,j} {b_{ii} b_{jj} (F^{ii} - F^{jj} )(b_{jj} - b_{ii} )} \nonumber \\&\ge 0, \end{aligned}$$

(3.24)

which is desired.

Claim 2 $- \sum _j\frac{{\partial g}}{{\partial b_{jj} }} - \alpha \sum _j\frac{{\partial g}}{{\partial b_{jj} }} + 2\beta \ge 0$.

In fact, by the Newton–MacLaurin inequality (2.6)

$$\begin{aligned} \sum \limits _j {\frac{{\partial g}}{{\partial b_{jj} }}}&= \sum \limits _j {\frac{{2\lambda _j \sigma _1 (\lambda ) - \sum \limits _i {\lambda _i ^2 } }}{{\sigma _1 (\lambda )^2 }}} = \frac{{2\sigma _1 (\lambda )^2 - n[\sigma _1 (\lambda )^2 - 2\sigma _2 (\lambda )]}}{{\sigma _1 (\lambda )^2 }} \nonumber \\&= \frac{{2n\sigma _2 (\lambda ) - (n - 2)\sigma _1 (\lambda )^2 }}{{\sigma _1 (\lambda )^2 }} \nonumber \\&\le \frac{{(n - 1)\sigma _1 (\lambda )^2 - (n - 2)\sigma _1 (\lambda )^2 }}{{\sigma _1 (\lambda )^2 }} = 1, \end{aligned}$$

(3.25)

and then

$$\begin{aligned} -\sum _j \frac{{\partial g}}{{\partial b_{jj} }} - \alpha \sum _j\frac{{\partial g}}{{\partial b_{jj} }} + 2\beta = 2\beta [ - \sum \limits _j {\frac{{\partial g}}{{\partial b_{jj} }}} + 1] \ge 0. \end{aligned}$$

(3.26)

This proves the assertion.

Claim 3 Choosing $N_{x_0 }$ sufficiently small such that x close to $x_0$, consequently we have

$$\begin{aligned} 2F^{ii} (2\beta b_{ii} - g) + \alpha (\alpha - 1)u^\alpha \frac{{\partial g}}{{\partial b_{ii} }} - 2\beta \alpha u^\alpha \ge 0, \quad \forall i =1, 2, \ldots , n. \end{aligned}$$

(3.27)

We know $\lambda _1 = \lambda _2 = \cdots = \lambda _n$ for any $x_0 \in \Psi$. Hence for any x close to $x_0$, we have $b_{ii} = b_{11} [1 + o(1)]$, and then

$$\begin{aligned} g&= b_{ii} [1 + o(1)], \end{aligned}$$

(3.28)

$$\begin{aligned} F^{ii} b_{ii}&= \frac{1}{n}\sum \limits _j {F^{jj} b_{jj} } [1 + o(1)] = u^\alpha [\frac{1}{n} + o(1)],\quad \forall i =1, 2, \ldots , n, \end{aligned}$$

(3.29)

$$\begin{aligned} \frac{{\partial g}}{{\partial b_{ii} }}&= \frac{{2\lambda _i \sigma _1 (\lambda ) - \sum \limits _j {\lambda _j ^2 } }}{{\sigma _1 (\lambda )^2 }} = \frac{1}{n} + o(1), \quad \forall i =1, 2, \ldots , n. \end{aligned}$$

(3.30)

Thus

$$\begin{aligned}&2F^{ii} (2\beta b_{ii} - g) + \alpha (\alpha - 1)u^\alpha \frac{{\partial g}}{{\partial b_{ii} }} - 2\beta \alpha u^\alpha \nonumber \\&\quad = 2(2\beta - 1)\frac{1}{n}u^\alpha + \alpha (\alpha - 1)u^\alpha \frac{1}{n} - 2\beta \alpha u^\alpha + o(1) \nonumber \\&\quad = - 2\beta \alpha \frac{{n - 1}}{n}u^\alpha + o(1) \nonumber \\&\quad \ge 0. \end{aligned}$$

(3.31)

Claim 3 is finished.

From (3.23) and the Claims, we conclude that

$$\begin{aligned} F^{ij} Z_{ij} - 2F^{ii} u_i \frac{{Z_j }}{u} \ge 0, \quad \forall x \in N_{x_0}. \end{aligned}$$

(3.32)

Together with $Z(x_0) = \mathop {\max }\limits _{{\mathbb {S}}^n } Z$, we know $Z(x) = Z(x_0)$ for any $x \in N_{x_0}$. We also note that $Z(x_0) = W(x_0) \ge W(x) \ge Z(x)$, thus $W(x) = W(x_0)$ for any $x \in N_{x_0}$. Hence $\Psi$ is an open set.

Combining Step 1 with Step 2, since $\Psi$ is a closed set, we know $\Psi =\{ x \in {\mathbb {S}}^n: W(x) = \mathop {\max }\limits _{{\mathbb {S}}^n } W\} = {\mathbb {S}}^n$ and $\lambda _1 = \lambda _2 = \cdots = \lambda _n$ for any $x\in {\mathbb {S}}^n$, which implies $u \equiv \big [ \frac{C_n^l (n-1)^l}{C_n^k (n-1)^k} \big ]^{\frac{1}{k-l-(p-1)}}$.