Mathematics Subject Classification (2020). 53C42; 35J60; 53C21.
Introduction
Let M M M ( n + 1 ) (n+1) ( n + 1 ) R n + 1 \mathbb{R}^{n+1} R n + 1 κ ( X ) \kappa(X) κ ( X ) H ( X ) H(X) H ( X ) ν ( X ) \nu(X) ν ( X ) M M M X X X ( 0 , 2 ) (0,2) ( 0 , 2 )
η i j = H g i j − h i j , \eta_{ij}=Hg_{ij}-h_{ij}, η ij = H g ij − h ij , where g i j g_{ij} g ij h i j h_{ij} h ij η \eta η h h h g g g λ ( η ) = ( λ 1 , … , λ n ) \lambda(\eta)=(\lambda_1,\ldots,\lambda_n) λ ( η ) = ( λ 1 , … , λ n ) η \eta η
λ i = H − κ i = ∑ j ≠ i κ j , i = 1 , … , n . \lambda_i=H-\kappa_i=\sum_{j\ne i}\kappa_j,\qquad i=1,\ldots,n. λ i = H − κ i = j = i ∑ κ j , i = 1 , … , n . The classical prescribed curvature equation
σ k ( κ ( X ) ) = f ( X , ν ( X ) ) \sigma_k(\kappa(X))=f(X,\nu(X)) σ k ( κ ( X )) = f ( X , ν ( X )) plays a fundamental role in fully nonlinear elliptic equations and geometric analysis. The cases k = 1 k=1 k = 1 k = 2 k=2 k = 2 k = n k=n k = n [1 , 2 ] . In a different but closely related direction, Chu and Jiao [3 ] studied the Hessian-type curvature equation
σ k ( λ ( η ) ) = f ( X , ν ( X ) ) \sigma_k(\lambda(\eta))=f(X,\nu(X)) σ k ( λ ( η )) = f ( X , ν ( X )) for star-shaped hypersurfaces. They established a global curvature estimate under the natural condition λ ( η ) ∈ Γ k \lambda(\eta)\in\Gamma_k λ ( η ) ∈ Γ k [6 , 5 ] . The evolution of star-shaped hypersurfaces by symmetric functions of their principal curvatures was also studied in [15 ] .
Sum Hessian equations constitute another natural class of fully nonlinear equations. In Euclidean domains the model equation is
σ k ( D 2 u ) + α σ k − 1 ( D 2 u ) = f ( x , u , D u ) , α > 0. \sigma_k(D^2u)+\alpha\sigma_{k-1}(D^2u)=f(x,u,Du),\qquad\alpha>0. σ k ( D 2 u ) + α σ k − 1 ( D 2 u ) = f ( x , u , D u ) , α > 0. The operator is a linear combination of two neighboring Hessian operators of different degrees and is therefore nonhomogeneous. Li, Ren, and Wang [9 ] identified the natural admissible region
Γ k − 1 ∩ { λ : σ k ( λ ) + α σ k − 1 ( λ ) > 0 } \Gamma_{k-1}\cap\{\lambda:\sigma_k(\lambda)+\alpha\sigma_{k-1}(\lambda)>0\} Γ k − 1 ∩ { λ : σ k ( λ ) + α σ k − 1 ( λ ) > 0 } and studied curvature estimates for related linear combinations of elementary symmetric functions on convex hypersurfaces. Liu and Ren [12 ] obtained Pogorelov-type C 2 C^2 C 2 [11 ] subsequently developed Pogorelov estimates and rigidity results for ( k − 1 ) (k-1) ( k − 1 ) k k k [10 ] and the interior estimate of Ren and Sun [13 ] for a two-sum Hessian equation with a nonnegative variable coefficient α ( x ) ≥ 0 \alpha(x)\geq0 α ( x ) ≥ 0 D 2 α = β I D^2\alpha=\beta I D 2 α = β I Δ u > 0 \Delta u>0 Δ u > 0
Of particular relevance to the present problem is the work of Ren and Wang [14 ] , who considered
σ k ( μ ) + α σ k − 1 ( μ ) = f ( x , u , D u ) , μ i = ∑ j ≠ i λ j ( D 2 u ) , \sigma_k(\mu)+\alpha\sigma_{k-1}(\mu)=f(x,u,Du),\qquad\mu_i=\sum_{j\ne i}\lambda_j(D^2u), σ k ( μ ) + α σ k − 1 ( μ ) = f ( x , u , D u ) , μ i = j = i ∑ λ j ( D 2 u ) , so that μ \mu μ ( Δ u ) I − D 2 u (\Delta u)I-D^2u ( Δ u ) I − D 2 u C 2 C^2 C 2 0 < k < n 0<k<n 0 < k < n k = n k=n k = n η = H g − h \eta=Hg-h η = H g − h
There are also recent geometric works on sum curvature equations formulated directly in the principal curvatures. For example, Yang and Lu [16 ] study complete hypersurfaces of constant sum Hessian curvature in hyperbolic space, with curvature function σ n − 1 ( κ ) + α σ n ( κ ) \sigma_{n-1}(\kappa)+\alpha\sigma_n(\kappa) σ n − 1 ( κ ) + α σ n ( κ ) [9 ] , are structurally distinct from applying the sum Hessian operator to the eigenvalues of the first Newton transformation H g − h Hg-h H g − h
Motivated by the above developments and by the geometric framework of Chu and Jiao, we study the prescribed sum Hessian curvature equation associated with η \eta η α > 0 \alpha>0 α > 0 1 ≤ j ≤ n 1\le j\le n 1 ≤ j ≤ n
S j , α ( λ ) = σ j ( λ ) + α σ j − 1 ( λ ) , σ 0 ≡ 1. S_{j,\alpha}(\lambda)=\sigma_j(\lambda)+\alpha\sigma_{j-1}(\lambda),\qquad\sigma_0\equiv1. S j , α ( λ ) = σ j ( λ ) + α σ j − 1 ( λ ) , σ 0 ≡ 1. Our equation is
S k , α ( λ ( η ) ) = σ k ( λ ( η ) ) + α σ k − 1 ( λ ( η ) ) = f ( X , ν ( X ) ) , X ∈ M . (1) S_{k,\alpha}(\lambda(\eta))=\sigma_k(\lambda(\eta))+\alpha\sigma_{k-1}(\lambda(\eta))=f(X,\nu(X)),\qquad X\in M. \tag*{(1)} S k , α ( λ ( η )) = σ k ( λ ( η )) + α σ k − 1 ( λ ( η )) = f ( X , ν ( X )) , X ∈ M . (1) The natural admissible region is
Γ ~ k , α : = Γ k − 1 ∩ { λ ∈ R n : S k , α ( λ ) > 0 } , \widetilde{\Gamma}_{k,\alpha}:=\Gamma_{k-1}\cap\{\lambda\in\mathbb{R}^n:S_{k,\alpha}(\lambda)>0\}, Γ k , α := Γ k − 1 ∩ { λ ∈ R n : S k , α ( λ ) > 0 } , where
Γ j = { λ ∈ R n : σ 1 ( λ ) > 0 , … , σ j ( λ ) > 0 } . \Gamma_j=\{\lambda\in\mathbb{R}^n:\sigma_1(\lambda)>0,\ldots,\sigma_j(\lambda)>0\}. Γ j = { λ ∈ R n : σ 1 ( λ ) > 0 , … , σ j ( λ ) > 0 } . A C 2 C^2 C 2 ( η , k , α ) (\eta,k,\alpha) ( η , k , α ) λ ( η ) ∈ Γ ~ k , α \lambda(\eta)\in\widetilde{\Gamma}_{k,\alpha} λ ( η ) ∈ Γ k , α M M M
The key structural observation is obtained by introducing the augmented vector
λ ^ = ( λ 1 , … , λ n , α ) ∈ R n + 1 . \widehat{\lambda}=(\lambda_1,\ldots,\lambda_n,\alpha)\in\mathbb{R}^{n+1}. λ = ( λ 1 , … , λ n , α ) ∈ R n + 1 . Then
S j , α ( λ ) = σ j ( λ ^ ) . S_{j,\alpha}(\lambda)=\sigma_j(\widehat{\lambda}). S j , α ( λ ) = σ j ( λ ) . Thus the nonhomogeneous sum Hessian operator is the restriction of the standard Hessian operator in one higher dimension to the affine hyperplane whose last coordinate equals α \alpha α σ k 1 / k \sigma_k^{1/k} σ k 1/ k
To the best of our knowledge, a global curvature estimate and a corresponding prescribed-curvature existence theorem for (1) on closed star-shaped hypersurfaces under the natural ( η , k , α ) (\eta,k,\alpha) ( η , k , α )
Theorem 1.1. Let 2 ≤ k ≤ n − 1 2\le k\le n-1 2 ≤ k ≤ n − 1 α > 0 \alpha>0 α > 0 M M M ( η , k , α ) (\eta,k,\alpha) ( η , k , α ) R n + 1 \mathbb{R}^{n+1} R n + 1 f > 0 f>0 f > 0 f ∈ C 2 ( U ) f\in C^2(\mathcal{U}) f ∈ C 2 ( U ) U \mathcal{U} U M M M R n + 1 × S n \mathbb{R}^{n+1}\times\mathbb{S}^n R n + 1 × S n
max X ∈ M max 1 ≤ i ≤ n ∣ κ i ( X ) ∣ ≤ C , \max_{X\in M}\max_{1\le i\le n}|\kappa_i(X)|\le C, X ∈ M max 1 ≤ i ≤ n max ∣ κ i ( X ) ∣ ≤ C , where
C = C ( n , k , α , ∥ M ∥ C 1 , inf f , ∥ f ∥ C 2 ) . C=C\left(n,k,\alpha,\|M\|_{C^1},\inf f,\|f\|_{C^2}\right). C = C ( n , k , α , ∥ M ∥ C 1 , inf f , ∥ f ∥ C 2 ) . The restriction k ≤ n − 1 k \le n-1 k ≤ n − 1 § 3 ; the endpoint k = n k=n k = n α σ k − 1 \alpha\sigma_{k-1} α σ k − 1 R α R_\alpha R α
We also prove an existence result. Put
Φ α ( r ) : = ( n k ) ( n − 1 ) k r k + α ( n k − 1 ) ( n − 1 ) k − 1 r k − 1 . (2) \Phi_\alpha(r) := \binom{n}{k}\frac{(n-1)^k}{r^k}+\alpha\binom{n}{k-1}\frac{(n-1)^{k-1}}{r^{k-1}}. \tag*{(2)} Φ α ( r ) := ( k n ) r k ( n − 1 ) k + α ( k − 1 n ) r k − 1 ( n − 1 ) k − 1 . (2) This is the value of the left-hand side of (1) on the sphere of radius r r r k k k (2) .
Theorem 1.2. Let 2 ≤ k ≤ n − 1 2 \le k \le n-1 2 ≤ k ≤ n − 1 α > 0 \alpha> 0 α > 0 0 < r 1 < 1 < r 2 0 < r_1 < 1 < r_2 0 < r 1 < 1 < r 2
f ∈ C 2 ( ( B ‾ r 2 ∖ B r 1 ) × S n ) f \in C^2\left((\overline{B}_{r_2}\setminus B_{r_1})\times\mathbb{S}^n\right) f ∈ C 2 ( ( B r 2 ∖ B r 1 ) × S n ) be positive and assume
f ( X , X ∣ X ∣ ) ≥ Φ α ( r 1 ) , ∣ X ∣ = r 1 , f ( X , X ∣ X ∣ ) ≤ Φ α ( r 2 ) , ∣ X ∣ = r 2 . (3) \begin{aligned}
f\left(X,\frac{X}{|X|}\right)&\ge\Phi_\alpha(r_1), \quad|X|=r_1,\\
f\left(X,\frac{X}{|X|}\right)&\le\Phi_\alpha(r_2), \quad|X|=r_2. \tag*{(3)}
\end{aligned} f ( X , ∣ X ∣ X ) f ( X , ∣ X ∣ X ) ≥ Φ α ( r 1 ) , ∣ X ∣ = r 1 , ≤ Φ α ( r 2 ) , ∣ X ∣ = r 2 . (3) Assume moreover that for every fixed x , ν ∈ S n x,\nu\in\mathbb{S}^n x , ν ∈ S n
∂ ∂ ρ [ ρ k f ( ρ x , ν ) ] ≤ 0 , r 1 ≤ ρ ≤ r 2 . (4) \frac{\partial}{\partial\rho}\left[\rho^k f(\rho x,\nu)\right]\le0,\quad r_1\le\rho\le r_2. \tag*{(4)} ∂ ρ ∂ [ ρ k f ( ρ x , ν ) ] ≤ 0 , r 1 ≤ ρ ≤ r 2 . (4) Then, for every 0 < β < 1 0<\beta<1 0 < β < 1 C 3 , β C^{3,\beta} C 3 , β ( η , k , α ) (\eta,k,\alpha) ( η , k , α ) r 1 ≤ ∣ X ∣ ≤ r 2 r_1\le|X|\le r_2 r 1 ≤ ∣ X ∣ ≤ r 2
When α = 0 \alpha=0 α = 0 α > 0 \alpha>0 α > 0 R α R_\alpha R α
The paper is organized as follows. § 2 records the structural properties of the sum Hessian operator. § 3 proves the global curvature estimate. § 4 establishes the a priori estimates and completes the existence and uniqueness proof by the continuity method.
Preliminaries
We first recall some basic definitions and properties of the elementary symmetric functions. For 1 ≤ k ≤ n 1\le k\le n 1 ≤ k ≤ n k k k
σ k ( λ ) = ∑ 1 ≤ i 1 < ⋯ < i k ≤ n λ i 1 ⋯ λ i k , \sigma_k(\lambda)=\sum_{1\le i_1<\cdots<i_k\le n}\lambda_{i_1}\cdots\lambda_{i_k}, σ k ( λ ) = 1 ≤ i 1 < ⋯ < i k ≤ n ∑ λ i 1 ⋯ λ i k , for λ = ( λ 1 , λ 2 , … , λ n ) ∈ R n \lambda=(\lambda_1,\lambda_2,\ldots,\lambda_n)\in\mathbb{R}^n λ = ( λ 1 , λ 2 , … , λ n ) ∈ R n
σ 0 ( λ ) = 1 , σ k ( λ ) = 0 for k > n or k < 0. \sigma_0(\lambda)=1,\qquad\sigma_k(\lambda)=0\quad\text{for }k>n\text{ or }k<0. σ 0 ( λ ) = 1 , σ k ( λ ) = 0 for k > n or k < 0. We next record the structural properties of the Hessian sum operator.
Lemma 2.1. Let α > 0 \alpha> 0 α > 0 λ ^ = ( λ , α ) ∈ R n + 1 \widehat{\lambda} = (\lambda,\alpha) \in\mathbb{R}^{n+1} λ = ( λ , α ) ∈ R n + 1
S j , α ( λ ) = σ j ( λ ^ ) , 1 ≤ j ≤ n . S_{j,\alpha}(\lambda) = \sigma_j(\widehat{\lambda}), \qquad1 \le j \le n. S j , α ( λ ) = σ j ( λ ) , 1 ≤ j ≤ n . Moreover,
λ ∈ Γ ~ k , α ⟺ λ ^ ∈ Γ k n + 1 . \lambda\in\widetilde{\Gamma}_{k,\alpha} \Longleftrightarrow\widehat{\lambda} \in\Gamma_k^{n+1}. λ ∈ Γ k , α ⟺ λ ∈ Γ k n + 1 . Consequently, Γ ~ k , α \widetilde{\Gamma}_{k,\alpha} Γ k , α
G ( λ ) : = S k , α ( λ ) 1 / k G(\lambda) := S_{k,\alpha}(\lambda)^{1/k} G ( λ ) := S k , α ( λ ) 1/ k is elliptic and concave on Γ ~ k , α \widetilde{\Gamma}_{k,\alpha} Γ k , α
Proof. The identity
σ j ( λ 1 , … , λ n , α ) = σ j ( λ ) + α σ j − 1 ( λ ) \sigma_j(\lambda_1,\ldots,\lambda_n,\alpha) = \sigma_j(\lambda) + \alpha\sigma_{j-1}(\lambda) σ j ( λ 1 , … , λ n , α ) = σ j ( λ ) + α σ j − 1 ( λ ) is immediate from the definition of the elementary symmetric functions. If λ ∈ Γ ~ k , α \lambda\in\widetilde{\Gamma}_{k,\alpha} λ ∈ Γ k , α 1 ≤ j ≤ k − 1 1 \le j \le k-1 1 ≤ j ≤ k − 1
σ j ( λ ^ ) = σ j ( λ ) + α σ j − 1 ( λ ) > 0 , \sigma_j(\widehat{\lambda}) = \sigma_j(\lambda) + \alpha\sigma_{j-1}(\lambda) > 0, σ j ( λ ) = σ j ( λ ) + α σ j − 1 ( λ ) > 0 , and σ k ( λ ^ ) = S k , α ( λ ) > 0 \sigma_k(\widehat{\lambda}) = S_{k,\alpha}(\lambda) > 0 σ k ( λ ) = S k , α ( λ ) > 0 λ ^ ∈ Γ k n + 1 \widehat{\lambda} \in\Gamma_k^{n+1} λ ∈ Γ k n + 1 λ ^ ∈ Γ k n + 1 \widehat{\lambda} \in\Gamma_k^{n+1} λ ∈ Γ k n + 1 1 ≤ j ≤ k − 1 1 \le j \le k-1 1 ≤ j ≤ k − 1 σ j + 1 \sigma_{j+1} σ j + 1 Γ j + 1 \Gamma_{j+1} Γ j + 1
σ j ( λ ) = ∂ σ j + 1 ( λ ^ ) ∂ λ ^ n + 1 > 0. \sigma_j(\lambda) = \frac{\partial\sigma_{j+1}(\widehat{\lambda})}{\partial\widehat{\lambda}_{n+1}} > 0. σ j ( λ ) = ∂ λ n + 1 ∂ σ j + 1 ( λ ) > 0. Thus λ ∈ Γ k − 1 \lambda\in\Gamma_{k-1} λ ∈ Γ k − 1 S k , α ( λ ) = σ k ( λ ^ ) > 0 S_{k,\alpha}(\lambda) = \sigma_k(\widehat{\lambda}) > 0 S k , α ( λ ) = σ k ( λ ) > 0 Γ k n + 1 \Gamma_k^{n+1} Γ k n + 1 λ ^ n + 1 = α \widehat{\lambda}_{n+1} = \alpha λ n + 1 = α G ( λ ) = σ k ( λ ^ ) 1 / k G(\lambda) = \sigma_k(\widehat{\lambda})^{1/k} G ( λ ) = σ k ( λ ) 1/ k σ k 1 / k \sigma_k^{1/k} σ k 1/ k Γ k \Gamma_k Γ k [1 ] .
For later use, if λ \lambda λ
S k i i = ∂ S k , α ∂ λ i , G i i = ∂ G ∂ λ i . S_k^{ii} = \frac{\partial S_{k,\alpha}}{\partial\lambda_i}, \qquad G^{ii} = \frac{\partial G}{\partial\lambda_i}. S k ii = ∂ λ i ∂ S k , α , G ii = ∂ λ i ∂ G . Then
S k i i = σ k − 1 ( λ ∣ i ) + α σ k − 2 ( λ ∣ i ) , S_k^{ii} = \sigma_{k-1}(\lambda|i) + \alpha\sigma_{k-2}(\lambda|i), S k ii = σ k − 1 ( λ ∣ i ) + α σ k − 2 ( λ ∣ i ) , ∑ i S k i i = ( n − k + 1 ) σ k − 1 ( λ ) + α ( n − k + 2 ) σ k − 2 ( λ ) (5) \sum_i S_k^{ii} = (n-k+1)\sigma_{k-1}(\lambda) + \alpha(n-k+2)\sigma_{k-2}(\lambda) \tag*{(5)} i ∑ S k ii = ( n − k + 1 ) σ k − 1 ( λ ) + α ( n − k + 2 ) σ k − 2 ( λ ) (5) and
∑ i λ i S k i i = k S k , α ( λ ) − α σ k − 1 ( λ ) . \sum_i \lambda_i S_k^{ii} = kS_{k,\alpha}(\lambda) - \alpha\sigma_{k-1}(\lambda). i ∑ λ i S k ii = k S k , α ( λ ) − α σ k − 1 ( λ ) . Thus, on a solution of (1) , if
f ~ = f 1 / k , R α : = α k f 1 / k − 1 σ k − 1 ( λ ) (6) \widetilde{f} = f^{1/k}, \qquad R_\alpha:= \frac{\alpha}{k}f^{1/k-1}\sigma_{k-1}(\lambda) \tag*{(6)} f = f 1/ k , R α := k α f 1/ k − 1 σ k − 1 ( λ ) (6) then
∑ i G i i λ i = f ~ − R α . \sum_i G^{ii}\lambda_i = \widetilde{f} - R_\alpha. i ∑ G ii λ i = f − R α . For F ( κ ) = G ( η ( κ ) ) F(\kappa)=G(\eta(\kappa)) F ( κ ) = G ( η ( κ )) g i j = δ i j g_{ij}=\delta_{ij} g ij = δ ij h i j = κ i δ i j h_{ij}=\kappa_i\delta_{ij} h ij = κ i δ ij
F i i = ∂ F ∂ κ i = ∑ p ≠ i G p p . (7) F^{ii}=\frac{\partial F}{\partial\kappa_i}=\sum_{p\ne i}G^{pp}. \tag*{(7)} F ii = ∂ κ i ∂ F = p = i ∑ G pp . (7) Consequently,
∑ i F i i h i i = ∑ i G i i η i i = f ~ − R α . (8) \sum_i F^{ii}h_{ii}=\sum_i G^{ii}\eta_{ii}=\widetilde f-R_\alpha. \tag*{(8)} i ∑ F ii h ii = i ∑ G ii η ii = f − R α . (8) Lemma 2.2. For λ ∈ Γ ~ k , α \lambda\in\widetilde{\Gamma}_{k,\alpha} λ ∈ Γ k , α
∑ i F i i ≥ c n , k > 0 , (9) \sum_i F^{ii}\ge c_{n,k}>0, \tag*{(9)} i ∑ F ii ≥ c n , k > 0 , (9) and
0 ≤ R α ≤ α ( n − 1 ) ( n − k + 1 ) ∑ i F i i . (10) 0\le R_\alpha\le\frac{\alpha}{(n-1)(n-k+1)}\sum_i F^{ii}. \tag*{(10)} 0 ≤ R α ≤ ( n − 1 ) ( n − k + 1 ) α i ∑ F ii . (10) Proof. By (7) and (5) ,
∑ i F i i = ( n − 1 ) ∑ i G i i = n − 1 k f 1 / k − 1 [ ( n − k + 1 ) σ k − 1 ( λ ) + α ( n − k + 2 ) σ k − 2 ( λ ) ] ≥ ( n − 1 ) ( n − k + 1 ) k f 1 / k − 1 σ k − 1 ( λ ^ ) . \begin{aligned}
\sum_i F^{ii}&=(n-1)\sum_i G^{ii}\\
&=\frac{n-1}{k}f^{1/k-1}\left[(n-k+1)\sigma_{k-1}(\lambda)+\alpha(n-k+2)\sigma_{k-2}(\lambda)\right]\\
&\ge\frac{(n-1)(n-k+1)}{k}f^{1/k-1}\sigma_{k-1}(\widehat{\lambda}).
\end{aligned} i ∑ F ii = ( n − 1 ) i ∑ G ii = k n − 1 f 1/ k − 1 [ ( n − k + 1 ) σ k − 1 ( λ ) + α ( n − k + 2 ) σ k − 2 ( λ ) ] ≥ k ( n − 1 ) ( n − k + 1 ) f 1/ k − 1 σ k − 1 ( λ ) . Since λ ^ ∈ Γ k n + 1 \widehat{\lambda}\in\Gamma_k^{n+1} λ ∈ Γ k n + 1
σ k − 1 ( λ ^ ) ≥ c n , k f ( k − 1 ) / k , \sigma_{k-1}(\widehat{\lambda})\ge c_{n,k}f^{(k-1)/k}, σ k − 1 ( λ ) ≥ c n , k f ( k − 1 ) / k , which proves (9). Formula (10) follows immediately by comparing (6) with the displayed formula for ∑ i F i i \sum_i F^{ii} ∑ i F ii
We now recall the standard geometric formulas for star-shaped radial graphs that will be used in the sequel.
Let M M M ρ \rho ρ S n \mathbb{S}^n S n
X ( x ) = ρ ( x ) x , x ∈ S n . X(x)=\rho(x)x,\qquad x\in\mathbb{S}^n. X ( x ) = ρ ( x ) x , x ∈ S n . If g ^ \widehat g g ∇ \nabla ∇ S n \mathbb{S}^n S n
g i j = ρ 2 g ^ i j + ρ i ρ j , h i j = ρ 2 g ^ i j + 2 ρ i ρ j − ρ ρ i j ρ 2 + ∣ ∇ ρ ∣ 2 . (11) g_{ij}=\rho^2\widehat g_{ij}+\rho_i\rho_j,\qquad
h_{ij}=\frac{\rho^2\widehat g_{ij}+2\rho_i\rho_j-\rho\rho_{ij}}{\sqrt{\rho^2+|\nabla\rho|^2}}. \tag*{(11)} g ij = ρ 2 g ij + ρ i ρ j , h ij = ρ 2 + ∣∇ ρ ∣ 2 ρ 2 g ij + 2 ρ i ρ j − ρ ρ ij . (11) and
ν = ρ x − ∇ ρ ρ 2 + ∣ ∇ ρ ∣ 2 . \nu=\frac{\rho x-\nabla\rho}{\sqrt{\rho^2+|\nabla\rho|^2}}. ν = ρ 2 + ∣∇ ρ ∣ 2 ρ x − ∇ ρ . In particular, with
u = ⟨ X , ν ⟩ , u=\langle X,\nu\rangle, u = ⟨ X , ν ⟩ , we have
u = ρ 2 ρ 2 + ∣ ∇ ρ ∣ 2 . (12) u=\frac{\rho^2}{\sqrt{\rho^2+|\nabla\rho|^2}}. \tag*{(12)} u = ρ 2 + ∣∇ ρ ∣ 2 ρ 2 . (12) Fix X 0 ∈ M X_0 \in M X 0 ∈ M { e 1 , … , e n } \{e_1,\ldots,e_n\} { e 1 , … , e n } X 0 X_0 X 0
X i j = − h i j ν , ν i = h i j e j (13) \begin{aligned}
X_{ij} &= -h_{ij}\nu,\\
\nu_i &= h_{ij}e_j \tag*{(13)}
\end{aligned} X ij ν i = − h ij ν , = h ij e j (13) h i j k = h i k j , R i j k l = h i k h j l − h i l h j k . (14) \begin{aligned}
h_{ijk} &= h_{ikj},\\
R_{ijkl} &= h_{ik}h_{jl}-h_{il}h_{jk}. \tag*{(14)}
\end{aligned} h ij k R ij k l = h ik j , = h ik h j l − h i l h j k . (14) We shall also use
h i j k l = h k l i j + ( h m j h i l − h m l h i j ) h m k + ( h m j h k l − h m l h k j ) h m i . (15) h_{ijkl}=h_{klij}+(h_{mj}h_{il}-h_{ml}h_{ij})h_{mk}+(h_{mj}h_{kl}-h_{ml}h_{kj})h_{mi}. \tag*{(15)} h ij k l = h k l ij + ( h mj h i l − h m l h ij ) h mk + ( h mj h k l − h m l h k j ) h mi . (15) Curvature estimate
We prove Theorem 1.1. Since λ ( η ) ∈ Γ k − 1 ⊂ Γ 1 \lambda(\eta)\in\Gamma_{k-1}\subset\Gamma_1 λ ( η ) ∈ Γ k − 1 ⊂ Γ 1
0 < σ 1 ( λ ( η ) ) = ( n − 1 ) H , 0<\sigma_1(\lambda(\eta))=(n-1)H, 0 < σ 1 ( λ ( η )) = ( n − 1 ) H , so H > 0 H>0 H > 0 κ max \kappa_{\max} κ m a x
By (12) there is a constant C C C C 1 C^1 C 1
C − 1 ≤ u ≤ C . (16) C^{-1}\leq u\leq C. \tag*{(16)} C − 1 ≤ u ≤ C . (16) Set
a = 1 2 inf M u > 0. a=\frac{1}{2}\inf_M u>0. a = 2 1 M inf u > 0. On the set D = { κ max > 0 } D=\{\kappa_{\max}>0\} D = { κ m a x > 0 }
Q = log κ max − log ( u − a ) + A 2 ∣ X ∣ 2 , Q=\log\kappa_{\max}-\log(u-a)+\frac{A}{2}|X|^2, Q = log κ m a x − log ( u − a ) + 2 A ∣ X ∣ 2 , where A > 1 A>1 A > 1 Q Q Q − ∞ -\infty − ∞ ∂ D \partial D ∂ D X 0 ∈ D X_0\in D X 0 ∈ D X 0 X_0 X 0
h i j = δ i j h i i , h 11 ≥ h 22 ≥ ⋯ ≥ h n n h_{ij}=\delta_{ij}h_{ii},\qquad h_{11}\geq h_{22}\geq\cdots\geq h_{nn} h ij = δ ij h ii , h 11 ≥ h 22 ≥ ⋯ ≥ h nn at X 0 X_0 X 0
η 11 ≤ η 22 ≤ ⋯ ≤ η n n . \eta_{11}\leq\eta_{22}\leq\cdots\leq\eta_{nn}. η 11 ≤ η 22 ≤ ⋯ ≤ η nn . If κ max ( X 0 ) \kappa_{\max}(X_0) κ m a x ( X 0 ) X 0 X_0 X 0 h 11 h_{11} h 11 X 0 X_0 X 0
Q ^ = log h 11 − log ( u − a ) + A 2 ∣ X ∣ 2 . \widehat{Q}=\log h_{11}-\log(u-a)+\frac{A}{2}|X|^2. Q = log h 11 − log ( u − a ) + 2 A ∣ X ∣ 2 . It has a local maximum at X 0 X_0 X 0 X 0 X_0 X 0
Since G G G
G 11 ≥ G 22 ≥ ⋯ ≥ G n n , F 11 ≤ F 22 ≤ ⋯ ≤ F n n . G^{11} \ge G^{22} \ge\cdots\ge G^{nn}, \qquad F^{11} \le F^{22} \le\cdots\le F^{nn}. G 11 ≥ G 22 ≥ ⋯ ≥ G nn , F 11 ≤ F 22 ≤ ⋯ ≤ F nn . The first- and second-derivative conditions are
0 = Q ^ i = h 11 i h 11 − u i u − a + A ⟨ X , e i ⟩ , (17) 0 = \widehat{Q}_i = \frac{h_{11i}}{h_{11}} - \frac{u_i}{u-a} + A\langle X,e_i\rangle, \tag*{(17)} 0 = Q i = h 11 h 11 i − u − a u i + A ⟨ X , e i ⟩ , (17) and
0 ≥ F i i Q ^ i i = F i i ( log h 11 ) i i − F i i ( log ( u − a ) ) i i + A 2 F i i ( ∣ X ∣ 2 ) i i . (18) 0 \ge F^{ii}\widehat{Q}_{ii} = F^{ii}(\log h_{11})_{ii} - F^{ii}(\log(u-a))_{ii} + \frac{A}{2}F^{ii}(|X|^2)_{ii}. \tag*{(18)} 0 ≥ F ii Q ii = F ii ( log h 11 ) ii − F ii ( log ( u − a ) ) ii + 2 A F ii ( ∣ X ∣ 2 ) ii . (18) Lemma 3.1. At X 0 X_0 X 0 h 11 ≥ 1 h_{11} \ge1 h 11 ≥ 1
0 ≥ − 2 h 11 ∑ i ≥ 2 G 1 i , 1 i h 11 i 2 − F i i h 11 i 2 h 11 2 + a u − a F i i h i i 2 + F i i u i 2 ( u − a ) 2 − ( C + R α ) h 11 + A ∑ i F i i − C A . (19) \begin{aligned}
0 \ge{}&-\frac{2}{h_{11}}\sum_{i\ge2}G^{1i,1i}h_{11i}^{2}
-\frac{F^{ii}h_{11i}^{2}}{h_{11}^{2}}
+\frac{a}{u-a}F^{ii}h_{ii}^{2}
+\frac{F^{ii}u_i^{2}}{(u-a)^{2}}\\
&-(C+R_\alpha)h_{11}+A\sum_iF^{ii}-CA. \tag*{(19)}
\end{aligned} 0 ≥ − h 11 2 i ≥ 2 ∑ G 1 i , 1 i h 11 i 2 − h 11 2 F ii h 11 i 2 + u − a a F ii h ii 2 + ( u − a ) 2 F ii u i 2 − ( C + R α ) h 11 + A i ∑ F ii − C A . (19) Proof. From (13), (8),
A 2 F i i ( ∣ X ∣ 2 ) i i = A ∑ i F i i ( 1 − h i i u ) = A ∑ i F i i − A u f ~ + A u R α . (20) \begin{aligned}
\frac{A}{2}F^{ii}(|X|^2)_{ii}
=A\sum_iF^{ii}(1-h_{ii}u) \\
&=A\sum_iF^{ii}-Au\widetilde f+AuR_\alpha. \tag*{(20)}
\end{aligned} 2 A F ii ( ∣ X ∣ 2 ) ii = A i ∑ F ii ( 1 − h ii u ) = A i ∑ F ii − A u f + A u R α . (20) Moreover, by (13)–(14),
u i = h i i ⟨ X , e i ⟩ , u i i = ∑ k h i i k ⟨ X , e k ⟩ − u h i i 2 + h i i . (21) u_i=h_{ii}\langle X,e_i\rangle,\qquad
u_{ii}=\sum_k h_{iik}\langle X,e_k\rangle-uh_{ii}^{2}+h_{ii}. \tag*{(21)} u i = h ii ⟨ X , e i ⟩ , u ii = k ∑ h iik ⟨ X , e k ⟩ − u h ii 2 + h ii . (21) Thus
− F i i ( log ( u − a ) ) i i = − 1 u − a ∑ k F i i h i i k ⟨ X , e k ⟩ + u u − a F i i h i i 2 -F^{ii}(\log(u-a))_{ii}
=-\frac{1}{u-a}\sum_kF^{ii}h_{iik}\langle X,e_k\rangle
+\frac{u}{u-a}F^{ii}h_{ii}^{2} − F ii ( log ( u − a ) ) ii = − u − a 1 k ∑ F ii h iik ⟨ X , e k ⟩ + u − a u F ii h ii 2 − f ~ − R α u − a + F i i u i 2 ( u − a ) 2 . (22) -\frac{\widetilde f-R_\alpha}{u-a}
+\frac{F^{ii}u_i^{2}}{(u-a)^{2}}. \tag*{(22)} − u − a f − R α + ( u − a ) 2 F ii u i 2 . (22) Since η i i , k = H k − h i i , k \eta_{ii,k}=H_k-h_{ii,k} η ii , k = H k − h ii , k
F i i h i i , k = G i i η i i , k . F^{ii}h_{ii,k}=G^{ii}\eta_{ii,k}. F ii h ii , k = G ii η ii , k . Differentiating G ( η ) = f ~ G(\eta)=\widetilde f G ( η ) = f e k e_k e k
G i i η i i , k = f ~ k . G^{ii}\eta_{ii,k}=\widetilde f_k. G ii η ii , k = f k . At the diagonal point X 0 X_0 X 0 ν k = h k k e k \nu_k=h_{kk}e_k ν k = h k k e k f ~ = f ~ ( X , ν ) \widetilde f=\widetilde f(X,\nu) f = f ( X , ν )
f ~ k = ( d X f ~ ) ( e k ) + h k k ( d ν f ~ ) ( e k ) . \widetilde f_k=(d_X\widetilde f)(e_k)+h_{kk}(d_\nu\widetilde f)(e_k). f k = ( d X f ) ( e k ) + h k k ( d ν f ) ( e k ) . On the other hand,
∑ i F i i h i i , k = ∑ i ∑ p ≠ i G p p h i i , k = ∑ p G p p ( H k − h p p , k ) = ∑ p G p p η p p , k . \sum_iF^{ii}h_{ii,k}
=\sum_i\sum_{p\ne i}G^{pp}h_{ii,k}
=\sum_pG^{pp}(H_k-h_{pp,k})
=\sum_pG^{pp}\eta_{pp,k}. i ∑ F ii h ii , k = i ∑ p = i ∑ G pp h ii , k = p ∑ G pp ( H k − h pp , k ) = p ∑ G pp η pp , k . Hence
F i i h i i , k = ( d X f ~ ) ( e k ) + h k k ( d ν f ~ ) ( e k ) . (23) F^{ii}h_{ii,k}=(d_X\widetilde{f})(e_k)+h_{kk}(d_\nu\widetilde{f})(e_k). \tag*{(23)} F ii h ii , k = ( d X f ) ( e k ) + h k k ( d ν f ) ( e k ) . (23) Indeed, by (23),
− 1 u − a ∑ k F i i h i i , k ⟨ X , e k ⟩ = − 1 u − a ∑ k h k k ( d ν f ~ ) ( e k ) ⟨ X , e k ⟩ − 1 u − a ( d X f ~ ) ( X T ) , -\frac{1}{u-a}\sum_k F^{ii}h_{ii,k}\langle X,e_k\rangle
=-\frac{1}{u-a}\sum_k h_{kk}(d_\nu\widetilde{f})(e_k)\langle X,e_k\rangle
-\frac{1}{u-a}(d_X\widetilde{f})(X^T), − u − a 1 k ∑ F ii h ii , k ⟨ X , e k ⟩ = − u − a 1 k ∑ h k k ( d ν f ) ( e k ) ⟨ X , e k ⟩ − u − a 1 ( d X f ) ( X T ) , where X T = ∑ k ⟨ X , e k ⟩ e k X^T=\sum_k\langle X,e_k\rangle e_k X T = ∑ k ⟨ X , e k ⟩ e k ∣ X T ∣ ≤ ∣ X ∣ ≤ C |X^T|\leq|X|\leq C ∣ X T ∣ ≤ ∣ X ∣ ≤ C u − a ≥ a > 0 u-a\geq a>0 u − a ≥ a > 0 ∣ d X f ~ ∣ ≤ C |d_X\widetilde{f}|\leq C ∣ d X f ∣ ≤ C d X f ~ = 1 k f 1 / k − 1 d X f d_X\widetilde{f}=\frac{1}{k}f^{1/k-1}d_Xf d X f = k 1 f 1/ k − 1 d X f inf f > 0 \inf f>0 inf f > 0 ∥ f ∥ C 1 ≤ C \|f\|_{C^1}\leq C ∥ f ∥ C 1 ≤ C
− 1 u − a ∑ k F i i h i i , k ⟨ X , e k ⟩ ≥ − 1 u − a ∑ k h k k ( d ν f ~ ) ( e k ) ⟨ X , e k ⟩ − C . -\frac{1}{u-a}\sum_k F^{ii}h_{ii,k}\langle X,e_k\rangle
\geq-\frac{1}{u-a}\sum_k h_{kk}(d_\nu\widetilde{f})(e_k)\langle X,e_k\rangle-C. − u − a 1 k ∑ F ii h ii , k ⟨ X , e k ⟩ ≥ − u − a 1 k ∑ h k k ( d ν f ) ( e k ) ⟨ X , e k ⟩ − C . Using (8), we therefore obtain
− F i i ( log ( u − a ) ) i i ≥ − 1 u − a ∑ k h k k ( d ν f ~ ) ( e k ) ⟨ X , e k ⟩ + u u − a F i i h i i 2 + R α u − a + F i i u i 2 ( u − a ) 2 − C . (24) -F^{ii}(\log(u-a))_{ii}
\geq-\frac{1}{u-a}\sum_k h_{kk}(d_\nu\widetilde{f})(e_k)\langle X,e_k\rangle
+\frac{u}{u-a}F^{ii}h_{ii}^{2}
+\frac{R_\alpha}{u-a}
+\frac{F^{ii}u_i^2}{(u-a)^2}-C. \tag*{(24)} − F ii ( log ( u − a ) ) ii ≥ − u − a 1 k ∑ h k k ( d ν f ) ( e k ) ⟨ X , e k ⟩ + u − a u F ii h ii 2 + u − a R α + ( u − a ) 2 F ii u i 2 − C . (24) For the first term in (18),
F i i ( log h 11 ) i i = F i i h 11 i i h 11 − F i i h 11 i 2 h 11 2 . (25) F^{ii}(\log h_{11})_{ii}
=\frac{F^{ii}h_{11ii}}{h_{11}}-\frac{F^{ii}h_{11i}^{2}}{h_{11}^{2}}. \tag*{(25)} F ii ( log h 11 ) ii = h 11 F ii h 11 ii − h 11 2 F ii h 11 i 2 . (25) The commutation identity (15) and (8) yield
F i i h 11 i i = F i i h i i 11 − F i i h i i 2 h 11 + ( f ~ − R α ) h 11 2 . (26) F^{ii}h_{11ii}=F^{ii}h_{ii11}-F^{ii}h_{ii}^{2}h_{11}+(\widetilde{f}-R_\alpha)h_{11}^{2}. \tag*{(26)} F ii h 11 ii = F ii h ii 11 − F ii h ii 2 h 11 + ( f − R α ) h 11 2 . (26) Differentiating G ( η ) = f ~ G(\eta)=\widetilde{f} G ( η ) = f e 1 e_1 e 1
F i i h i i 11 = − G i j , k l η i j , 1 η k l , 1 + f ~ 11 . F^{ii}h_{ii11}=-G^{ij,kl}\eta_{ij,1}\eta_{kl,1}+\widetilde{f}_{11}. F ii h ii 11 = − G ij , k l η ij , 1 η k l , 1 + f 11 . We next make the estimate for f ~ 11 \widetilde{f}_{11} f 11 f f f S n \mathbb{S}^{n} S n C 2 C^2 C 2 X 0 X_0 X 0 h h h f ~ ( X , ν ) \widetilde{f}(X,\nu) f ( X , ν )
∑ k h k 11 ( d ν f ~ ) ( e k ) \sum_k h_{k11}(d_\nu\widetilde{f})(e_k) k ∑ h k 11 ( d ν f ) ( e k ) from the tangential part of ν 11 \nu_{11} ν 11 f f f h 1 j h_{1j} h 1 j h 1 j = 0 h_{1j}=0 h 1 j = 0 j ≠ 1 j\neq1 j = 1 X 0 X_0 X 0 − C ( 1 + h 11 2 ) -C(1+h_{11}^{2}) − C ( 1 + h 11 2 )
F i i h i i 11 ≥ − G i j , k l η i j , 1 η k l , 1 + ∑ k h k 11 ( d ν f ~ ) ( e k ) − C h 11 2 − C . (27) F^{ii}h_{ii11}\geq-G^{ij,kl}\eta_{ij,1}\eta_{kl,1}+\sum_k h_{k11}(d_\nu\widetilde{f})(e_k)-Ch_{11}^{2}-C. \tag*{(27)} F ii h ii 11 ≥ − G ij , k l η ij , 1 η k l , 1 + k ∑ h k 11 ( d ν f ) ( e k ) − C h 11 2 − C . (27) For completeness, at a diagonal matrix the standard second-derivative formula for a symmetric spectral function gives
G i j , k l ξ i j ξ k l = ∑ p , q G p p , q q ξ p p ξ q q + 2 ∑ p < q G p p − G q q η p p − η q q ξ p q 2 . G^{ij,kl}\xi_{ij}\xi_{kl}=\sum_{p,q}G^{pp,qq}\xi_{pp}\xi_{qq}+2\sum_{p<q}\frac{G^{pp}-G^{qq}}{\eta_{pp}-\eta_{qq}}\xi_{pq}^{2}. G ij , k l ξ ij ξ k l = p , q ∑ G pp , q q ξ pp ξ q q + 2 p < q ∑ η pp − η q q G pp − G q q ξ pq 2 . with the quotient understood by continuity when two eigenvalues coincide. Concavity makes the whole quadratic form nonpositive and, since η 11 ≤ η i i \eta_{11}\leq\eta_{ii} η 11 ≤ η ii G 11 ≥ G i i G^{11}\geq G^{ii} G 11 ≥ G ii G 1 i , i 1 ≤ 0 G^{1i,i1}\leq0 G 1 i , i 1 ≤ 0
η 1 i , 1 = − h 1 i , 1 = − h 11 i \eta_{1i,1}=-h_{1i,1}=-h_{11i} η 1 i , 1 = − h 1 i , 1 = − h 11 i by Codazzi, we obtain
− G i j , k l η i j , 1 η k l , 1 ≥ − 2 ∑ i ≥ 2 G 1 i , i 1 h 11 i 2 . (28) -G^{ij,kl}\eta_{ij,1}\eta_{kl,1}\geq-2\sum_{i\geq2}G^{1i,i1}h_{11i}^{2}. \tag*{(28)} − G ij , k l η ij , 1 η k l , 1 ≥ − 2 i ≥ 2 ∑ G 1 i , i 1 h 11 i 2 . (28) Substituting (27) –(28) into (26) and then (25) gives
F i i ( log h 11 ) i i ≥ − 2 h 11 ∑ i ≥ 2 G 1 i , i 1 h 11 i 2 − F i i h 11 i 2 h 11 2 − F i i h i i 2 + 1 h 11 ∑ k h k 11 ( d ν f ~ ) ( e k ) − ( C + R α ) h 11 . (29) F^{ii}(\log h_{11})_{ii}\geq-\frac{2}{h_{11}}\sum_{i\geq2}G^{1i,i1}h_{11i}^{2}-\frac{F^{ii}h_{11i}^{2}}{h_{11}^{2}}-F^{ii}h_{ii}^{2}
+\frac{1}{h_{11}}\sum_k h_{k11}(d_\nu\widetilde f)(e_k)-(C+R_\alpha)h_{11}. \tag*{(29)} F ii ( log h 11 ) ii ≥ − h 11 2 i ≥ 2 ∑ G 1 i , i 1 h 11 i 2 − h 11 2 F ii h 11 i 2 − F ii h ii 2 + h 11 1 k ∑ h k 11 ( d ν f ) ( e k ) − ( C + R α ) h 11 . (29) The two d ν f ~ d_\nu\widetilde f d ν f (24) and (29) actually cancel up to the harmless A A A h k 11 = h 11 k h_{k11}=h_{11k} h k 11 = h 11 k (17) and u k = h k k ⟨ X , e k ⟩ u_k=h_{kk}\langle X,e_k\rangle u k = h k k ⟨ X , e k ⟩
h k 11 h 11 = h k k ⟨ X , e k ⟩ u − a − A ⟨ X , e k ⟩ . \frac{h_{k11}}{h_{11}}=\frac{h_{kk}\langle X,e_k\rangle}{u-a}-A\langle X,e_k\rangle. h 11 h k 11 = u − a h k k ⟨ X , e k ⟩ − A ⟨ X , e k ⟩ . Hence
1 h 11 ∑ k h k 11 ( d ν f ~ ) ( e k ) − 1 u − a ∑ k h k k ( d ν f ~ ) ( e k ) ⟨ X , e k ⟩ = − A ∑ k ⟨ X , e k ⟩ ( d ν f ~ ) ( e k ) ≥ − C A , \frac{1}{h_{11}}\sum_k h_{k11}(d_\nu\widetilde f)(e_k)-\frac{1}{u-a}\sum_k h_{kk}(d_\nu\widetilde f)(e_k)\langle X,e_k\rangle
=-A\sum_k\langle X,e_k\rangle(d_\nu\widetilde f)(e_k)\geq-CA, h 11 1 k ∑ h k 11 ( d ν f ) ( e k ) − u − a 1 k ∑ h k k ( d ν f ) ( e k ) ⟨ X , e k ⟩ = − A k ∑ ⟨ X , e k ⟩ ( d ν f ) ( e k ) ≥ − C A , where the last inequality uses the C 1 C^{1} C 1 f f f C 0 C^{0} C 0 ∣ X ∣ |X| ∣ X ∣ (20) , (24) , and (29) in (18) , and discarding the nonnegative terms R α / ( u − a ) R_\alpha/(u-a) R α / ( u − a ) A u R α AuR_\alpha A u R α (19) . □
We next absorb the unfavorable term in (19) .
Lemma 3.2. If h 11 h_{11} h 11 A A A Theorem 1.1 , then
( C + R α ) h 11 ≤ a 2 ( u − a ) ∑ i F i i h i i 2 + A 2 ∑ i F i i . (30) (C+R_\alpha)h_{11}\leq\frac{a}{2(u-a)}\sum_i F^{ii}h_{ii}^{2}+\frac{A}{2}\sum_i F^{ii}. \tag*{(30)} ( C + R α ) h 11 ≤ 2 ( u − a ) a i ∑ F ii h ii 2 + 2 A i ∑ F ii . (30) Proof. Fix a small number δ > 0 \delta>0 δ > 0
Case 1. Suppose ∣ h i i ∣ ≤ δ h 11 |h_{ii}|\leq\delta h_{11} ∣ h ii ∣ ≤ δ h 11 i ≥ 2 i\geq2 i ≥ 2
∣ η 11 ∣ ≤ ( n − 1 ) δ h 11 |\eta_{11}|\leq(n-1)\delta h_{11} ∣ η 11 ∣ ≤ ( n − 1 ) δ h 11 and
[ 1 − ( n − 2 ) δ ] h 11 ≤ η 22 ≤ ⋯ ≤ η n n ≤ [ 1 + ( n − 2 ) δ ] h 11 . (31) [1-(n-2)\delta]h_{11}\leq\eta_{22}\leq\cdots\leq\eta_{nn}\leq[1+(n-2)\delta]h_{11}. \tag*{(31)} [ 1 − ( n − 2 ) δ ] h 11 ≤ η 22 ≤ ⋯ ≤ η nn ≤ [ 1 + ( n − 2 ) δ ] h 11 . (31) Introduce the normalized eigenvalues μ i = η i i / h 11 \mu_i=\eta_{ii}/h_{11} μ i = η ii / h 11 (31) , under the present assumption,
∣ μ 1 ∣ ≤ ( n − 1 ) δ , 1 − ( n − 2 ) δ ≤ μ i ≤ 1 + ( n − 2 ) δ ( i ≥ 2 ) . |\mu_1|\leq(n-1)\delta,\qquad1-(n-2)\delta\leq\mu_i\leq1+(n-2)\delta\quad(i\geq2). ∣ μ 1 ∣ ≤ ( n − 1 ) δ , 1 − ( n − 2 ) δ ≤ μ i ≤ 1 + ( n − 2 ) δ ( i ≥ 2 ) . Thus μ \mu μ ( 0 , 1 , … , 1 ) (0,1,\ldots,1) ( 0 , 1 , … , 1 )
σ k − 1 ( 0 , 1 , … , 1 ) = ( n − 1 k − 1 ) > 0 , σ k − 2 ( 0 , 1 , … , 1 ) = ( n − 1 k − 2 ) > 0 , \sigma_{k-1}(0,1,\ldots,1)=\binom{n-1}{k-1}>0,\qquad\sigma_{k-2}(0,1,\ldots,1)=\binom{n-1}{k-2}>0, σ k − 1 ( 0 , 1 , … , 1 ) = ( k − 1 n − 1 ) > 0 , σ k − 2 ( 0 , 1 , … , 1 ) = ( k − 2 n − 1 ) > 0 , continuity gives, after fixing δ > 0 \delta>0 δ > 0
σ k − 1 ( η ) = h 11 k − 1 σ k − 1 ( μ ) ≥ c h 11 k − 1 , σ k − 2 ( η ) = h 11 k − 2 σ k − 2 ( μ ) ≥ c h 11 k − 2 . \sigma_{k-1}(\eta)=h_{11}^{k-1}\sigma_{k-1}(\mu)\geq ch_{11}^{k-1},\qquad\sigma_{k-2}(\eta)=h_{11}^{k-2}\sigma_{k-2}(\mu)\geq ch_{11}^{k-2}. σ k − 1 ( η ) = h 11 k − 1 σ k − 1 ( μ ) ≥ c h 11 k − 1 , σ k − 2 ( η ) = h 11 k − 2 σ k − 2 ( μ ) ≥ c h 11 k − 2 . Using the exact trace formula from Lemma 2.2,
∑ i F i i = n − 1 k f 1 / k − 1 [ ( n − k + 1 ) σ k − 1 ( η ) + α ( n − k + 2 ) σ k − 2 ( η ) ] . \sum_i F^{ii}=\frac{n-1}{k}f^{1/k-1}\left[(n-k+1)\sigma_{k-1}(\eta)+\alpha(n-k+2)\sigma_{k-2}(\eta)\right]. i ∑ F ii = k n − 1 f 1/ k − 1 [ ( n − k + 1 ) σ k − 1 ( η ) + α ( n − k + 2 ) σ k − 2 ( η ) ] . and the positive upper and lower bounds for f f f
∑ i F i i ≥ c h 11 k − 1 . \sum_i F^{ii}\geq ch_{11}^{k-1}. i ∑ F ii ≥ c h 11 k − 1 . Since k ≥ 2 k\geq2 k ≥ 2 h 11 ≥ 1 h_{11}\geq1 h 11 ≥ 1 h 11 k − 1 ≥ h 11 h_{11}^{k-1}\geq h_{11} h 11 k − 1 ≥ h 11 A A A
C h 11 ≤ A 2 ∑ i F i i . Ch_{11}\leq\frac{A}{2}\sum_i F^{ii}. C h 11 ≤ 2 A i ∑ F ii . It remains to estimate R α h 11 R_\alpha h_{11} R α h 11 μ \mu μ δ → 0 \delta\to0 δ → 0
μ → ( 0 , 1 , … , 1 ) . \mu\to(0,1,\ldots,1). μ → ( 0 , 1 , … , 1 ) . Because k ≤ n − 1 k\leq n-1 k ≤ n − 1 j ≥ 2 j\geq2 j ≥ 2
σ k − 1 ( ( 0 , 1 , … , 1 ) ∣ j ) = ( n − 2 k − 1 ) > 0 , \sigma_{k-1}((0,1,\ldots,1)|j)=\binom{n-2}{k-1}>0, σ k − 1 (( 0 , 1 , … , 1 ) ∣ j ) = ( k − 1 n − 2 ) > 0 , and
σ k − 2 ( ( 0 , 1 , … , 1 ) ∣ j ) = ( n − 2 k − 2 ) > 0. \sigma_{k-2}((0,1,\ldots,1)|j)=\binom{n-2}{k-2}>0. σ k − 2 (( 0 , 1 , … , 1 ) ∣ j ) = ( k − 2 n − 2 ) > 0. For j ≥ 2 j\geq2 j ≥ 2
G j j = 1 k f 1 / k − 1 [ σ k − 1 ( η ∣ j ) + α σ k − 2 ( η ∣ j ) ] . G^{jj}=\frac{1}{k}f^{1/k-1}\left[\sigma_{k-1}(\eta|j)+\alpha\sigma_{k-2}(\eta|j)\right]. G j j = k 1 f 1/ k − 1 [ σ k − 1 ( η ∣ j ) + α σ k − 2 ( η ∣ j ) ] . The two displayed limiting values above are strictly positive, so, after decreasing the already fixed δ \delta δ
σ k − 1 ( η ∣ j ) ≥ c h 11 k − 1 , σ k − 2 ( η ∣ j ) ≥ c h 11 k − 2 , j ≥ 2. \sigma_{k-1}(\eta|j)\geq ch_{11}^{k-1},\qquad\sigma_{k-2}(\eta|j)\geq ch_{11}^{k-2},\qquad j\geq2. σ k − 1 ( η ∣ j ) ≥ c h 11 k − 1 , σ k − 2 ( η ∣ j ) ≥ c h 11 k − 2 , j ≥ 2. Hence
G j j ≥ c f 1 / k − 1 ( h 11 k − 1 + α h 11 k − 2 ) , j ≥ 2. G^{jj}\geq cf^{1/k-1}\left(h_{11}^{k-1}+\alpha h_{11}^{k-2}\right),\qquad j\geq2. G j j ≥ c f 1/ k − 1 ( h 11 k − 1 + α h 11 k − 2 ) , j ≥ 2. On the other hand, all components of μ \mu μ G i i G^{ii} G ii
∑ i G i i ≤ C f 1 / k − 1 ( h 11 k − 1 + α h 11 k − 2 ) . \sum_i G^{ii} \leq C f^{1/k-1}\left(h_{11}^{k-1}+\alpha h_{11}^{k-2}\right). i ∑ G ii ≤ C f 1/ k − 1 ( h 11 k − 1 + α h 11 k − 2 ) . Since ∑ i F i i = ( n − 1 ) ∑ i G i i \sum_i F^{ii}=(n-1)\sum_i G^{ii} ∑ i F ii = ( n − 1 ) ∑ i G ii
F 11 = ∑ j ≠ 1 G j j ≥ c ∑ i F i i . F^{11}=\sum_{j\ne1}G^{jj}\geq c\sum_i F^{ii}. F 11 = j = 1 ∑ G j j ≥ c i ∑ F ii . It follows that
∑ i F i i h i i 2 ≥ F 11 h 11 2 ≥ c h 11 2 ∑ i F i i . (32) \sum_i F^{ii}h_{ii}^2\geq F^{11}h_{11}^2\geq ch_{11}^2\sum_i F^{ii}. \tag*{(32)} i ∑ F ii h ii 2 ≥ F 11 h 11 2 ≥ c h 11 2 i ∑ F ii . (32) By (10), R α ≤ C ∑ i F i i R_\alpha\leq C\sum_i F^{ii} R α ≤ C ∑ i F ii
∑ i F i i ≤ C h 11 2 ∑ i F i i h i i 2 . \sum_i F^{ii}\leq\frac{C}{h_{11}^2}\sum_i F^{ii}h_{ii}^2. i ∑ F ii ≤ h 11 2 C i ∑ F ii h ii 2 . Consequently
R α h 11 ≤ C h 11 ∑ i F i i h i i 2 . R_\alpha h_{11}\leq\frac{C}{h_{11}}\sum_i F^{ii}h_{ii}^2. R α h 11 ≤ h 11 C i ∑ F ii h ii 2 . Because a > 0 a>0 a > 0 u u u a / [ 4 ( u − a ) ] a/[4(u-a)] a / [ 4 ( u − a )] h 11 h_{11} h 11
R α h 11 ≤ a 4 ( u − a ) ∑ i F i i h i i 2 . R_\alpha h_{11}\leq\frac{a}{4(u-a)}\sum_i F^{ii}h_{ii}^2. R α h 11 ≤ 4 ( u − a ) a i ∑ F ii h ii 2 . This proves (30) in Case 1.
Case 2. Suppose h 22 > δ h 11 h_{22}>\delta h_{11} h 22 > δ h 11 h n n < − δ h 11 h_{nn}<-\delta h_{11} h nn < − δ h 11 G 11 ≥ G i i G^{11}\geq G^{ii} G 11 ≥ G ii i ≥ 2 i\geq2 i ≥ 2
F i i ≥ G 11 ≥ 1 n ∑ j G j j = 1 n ( n − 1 ) ∑ j F j j . (33) F^{ii}\geq G^{11}\geq\frac{1}{n}\sum_jG^{jj}=\frac{1}{n(n-1)}\sum_jF^{jj}. \tag*{(33)} F ii ≥ G 11 ≥ n 1 j ∑ G j j = n ( n − 1 ) 1 j ∑ F j j . (33) Therefore
∑ i F i i h i i 2 ≥ c δ 2 h 11 2 ∑ i F i i . (34) \sum_i F^{ii}h_{ii}^2\geq c\delta^2h_{11}^2\sum_iF^{ii}. \tag*{(34)} i ∑ F ii h ii 2 ≥ c δ 2 h 11 2 i ∑ F ii . (34) Indeed, Lemma 2.2 gives ∑ i F i i ≥ c > 0 \sum_iF^{ii}\geq c>0 ∑ i F ii ≥ c > 0
C h 11 ≤ C h 11 ∑ i F i i , R α h 11 ≤ C h 11 ∑ i F i i Ch_{11}\leq Ch_{11}\sum_iF^{ii},\qquad R_\alpha h_{11}\leq Ch_{11}\sum_iF^{ii} C h 11 ≤ C h 11 i ∑ F ii , R α h 11 ≤ C h 11 i ∑ F ii by (10). In view of (34), the right-hand sides are bounded by
C δ 2 h 11 ∑ i F i i h i i 2 . \frac{C}{\delta^2h_{11}}\sum_iF^{ii}h_{ii}^2. δ 2 h 11 C i ∑ F ii h ii 2 . For sufficiently large h 11 h_{11} h 11 a [ 4 ( u − a ) ] − 1 ∑ i F i i h i i 2 a[4(u-a)]^{-1}\sum_iF^{ii}h_{ii}^2 a [ 4 ( u − a ) ] − 1 ∑ i F ii h ii 2 □ \square □
Combining Lemmas 3.1 and 3.2,
0 ≥ − 2 h 11 ∑ i ≥ 2 G 1 i , i 1 h 11 i 2 − F i i h 11 i 2 h 11 2 + a 2 ( u − a ) F i i h i i 2 + F i i u i 2 ( u − a ) 2 + A 2 ∑ i F i i − C A . (35) \begin{aligned}
0 \ge{}& -\frac{2}{h_{11}}\sum_{i\ge2}G^{1i,i1}h_{11i}^{2}-\frac{F^{ii}h_{11i}^{2}}{h_{11}^{2}}+\frac{a}{2(u-a)}F^{ii}h_{ii}^{2}\\
&+\frac{F^{ii}u_i^{2}}{(u-a)^{2}}+\frac{A}{2}\sum_i F^{ii}-CA. \tag*{(35)}
\end{aligned} 0 ≥ − h 11 2 i ≥ 2 ∑ G 1 i , i 1 h 11 i 2 − h 11 2 F ii h 11 i 2 + 2 ( u − a ) a F ii h ii 2 + ( u − a ) 2 F ii u i 2 + 2 A i ∑ F ii − C A . (35) Lemma 3.3. If h 11 h_{11} h 11 A A A X 0 X_0 X 0
∣ h i i ∣ ≤ C A , i ≥ 2. |h_{ii}| \le CA,\qquad i\ge2. ∣ h ii ∣ ≤ C A , i ≥ 2. Proof. From (17) , for any 0 < ε < 1 0<\varepsilon<1 0 < ε < 1
F i i h 11 i 2 h 11 2 ≤ 1 + ε ( u − a ) 2 F i i u i 2 + C ( 1 + 1 ε ) A 2 F i i ⟨ X , e i ⟩ 2 . \frac{F^{ii}h_{11i}^{2}}{h_{11}^{2}}\le\frac{1+\varepsilon}{(u-a)^{2}}F^{ii}u_i^{2}+C\left(1+\frac{1}{\varepsilon}\right)A^{2}F^{ii}\langle X,e_i\rangle^{2}. h 11 2 F ii h 11 i 2 ≤ ( u − a ) 2 1 + ε F ii u i 2 + C ( 1 + ε 1 ) A 2 F ii ⟨ X , e i ⟩ 2 . Substitute this into (35) and discard the nonnegative concavity term − 2 h 11 − 1 ∑ i ≥ 2 G 1 i , i 1 h 11 i 2 -2h_{11}^{-1}\sum_{i\ge2}G^{1i,i1}h_{11i}^{2} − 2 h 11 − 1 ∑ i ≥ 2 G 1 i , i 1 h 11 i 2 u i = h i i ⟨ X , e i ⟩ u_i=h_{ii}\langle X,e_i\rangle u i = h ii ⟨ X , e i ⟩ u i 2 u_i^2 u i 2
ε ( u − a ) 2 F i i u i 2 ≤ C ε F i i h i i 2 , \frac{\varepsilon}{(u-a)^{2}}F^{ii}u_i^{2}\le C\varepsilon F^{ii}h_{ii}^{2}, ( u − a ) 2 ε F ii u i 2 ≤ C ε F ii h ii 2 , where we used ∣ X ∣ ≤ C |X|\le C ∣ X ∣ ≤ C u − a ≥ a > 0 u-a\ge a>0 u − a ≥ a > 0 ε > 0 \varepsilon>0 ε > 0 a [ 2 ( u − a ) ] − 1 F i i h i i 2 a[2(u-a)]^{-1}F^{ii}h_{ii}^{2} a [ 2 ( u − a ) ] − 1 F ii h ii 2 A A A
C ( 1 + 1 ε ) A 2 F i i ⟨ X , e i ⟩ 2 ≤ C A 2 ∑ i F i i . C\left(1+\frac{1}{\varepsilon}\right)A^{2}F^{ii}\langle X,e_i\rangle^{2}\le CA^{2}\sum_i F^{ii}. C ( 1 + ε 1 ) A 2 F ii ⟨ X , e i ⟩ 2 ≤ C A 2 i ∑ F ii . We therefore obtain
0 ≥ c ∑ i F i i h i i 2 − C A 2 ∑ i F i i − C A . (36) 0\ge c\sum_i F^{ii}h_{ii}^{2}-CA^{2}\sum_i F^{ii}-CA. \tag*{(36)} 0 ≥ c i ∑ F ii h ii 2 − C A 2 i ∑ F ii − C A . (36) Starting from (36) , and using (9) , ∑ i F i i ≥ c 0 > 0 \sum_i F^{ii}\ge c_0>0 ∑ i F ii ≥ c 0 > 0 A ≥ 1 A\ge1 A ≥ 1 C A ≤ C A 2 ∑ i F i i CA\le CA^{2}\sum_i F^{ii} C A ≤ C A 2 ∑ i F ii
c ∑ i F i i h i i 2 ≤ C A 2 ∑ i F i i . c\sum_i F^{ii}h_{ii}^{2}\le CA^{2}\sum_i F^{ii}. c i ∑ F ii h ii 2 ≤ C A 2 i ∑ F ii . For each i ≥ 2 i\ge2 i ≥ 2 (33) gives F i i ≥ c ∑ j F j j F^{ii}\ge c\sum_j F^{jj} F ii ≥ c ∑ j F j j
c ( ∑ j F j j ) ∑ i ≥ 2 h i i 2 ≤ C A 2 ∑ j F j j . c\left(\sum_j F^{jj}\right)\sum_{i\ge2}h_{ii}^{2}\le CA^{2}\sum_j F^{jj}. c ( j ∑ F j j ) i ≥ 2 ∑ h ii 2 ≤ C A 2 j ∑ F j j . Cancelling the positive trace gives
∑ i ≥ 2 h i i 2 ≤ C A 2 , \sum_{i\ge2}h_{ii}^{2}\le CA^{2}, i ≥ 2 ∑ h ii 2 ≤ C A 2 , and therefore ∣ h i i ∣ ≤ C A |h_{ii}|\le CA ∣ h ii ∣ ≤ C A i ≥ 2 i\ge2 i ≥ 2
Proof of Theorem 1.1. If h 11 h_{11} h 11 A A A
∣ h i i ∣ ≤ δ h 11 , i ≥ 2 , (37) |h_{ii}| \le\delta h_{11}, \qquad i \ge2, \tag*{(37)} ∣ h ii ∣ ≤ δ h 11 , i ≥ 2 , (37) where δ > 0 \delta> 0 δ > 0
For the i = 1 i=1 i = 1 (17) gives
h 111 h 11 = u 1 u − a − A ⟨ X , e 1 ⟩ . \frac{h_{111}}{h_{11}}=\frac{u_1}{u-a}-A\langle X,e_1\rangle. h 11 h 111 = u − a u 1 − A ⟨ X , e 1 ⟩ . For any ε > 0 \varepsilon> 0 ε > 0
F 11 h 111 2 h 11 2 ≤ ( 1 + ε ) F 11 u 1 2 ( u − a ) 2 + C ε A 2 F 11 ⟨ X , e 1 ⟩ 2 . \frac{F^{11}h_{111}^2}{h_{11}^2}\le(1+\varepsilon)\frac{F^{11}u_1^2}{(u-a)^2}+C_\varepsilon A^2F^{11}\langle X,e_1\rangle^2. h 11 2 F 11 h 111 2 ≤ ( 1 + ε ) ( u − a ) 2 F 11 u 1 2 + C ε A 2 F 11 ⟨ X , e 1 ⟩ 2 . Since u 1 = h 11 ⟨ X , e 1 ⟩ u_1=h_{11}\langle X,e_1\rangle u 1 = h 11 ⟨ X , e 1 ⟩ ε F 11 u 1 2 / ( u − a ) 2 \varepsilon F^{11}u_1^2/(u-a)^2 ε F 11 u 1 2 / ( u − a ) 2 C ε F 11 h 11 2 C\varepsilon F^{11}h_{11}^2 C ε F 11 h 11 2 ε \varepsilon ε h 11 ≫ A h_{11}\gg A h 11 ≫ A A 2 F 11 A^2F^{11} A 2 F 11 a ( u − a ) − 1 ∑ i F i i h i i 2 a(u-a)^{-1}\sum_i F^{ii}h_{ii}^2 a ( u − a ) − 1 ∑ i F ii h ii 2
F 11 h 111 2 h 11 2 ≤ F 11 u 1 2 ( u − a ) 2 + a 8 ( u − a ) F i i h i i 2 . (38) \frac{F^{11}h_{111}^2}{h_{11}^2}\le\frac{F^{11}u_1^2}{(u-a)^2}+\frac{a}{8(u-a)}F^{ii}h_{ii}^2. \tag*{(38)} h 11 2 F 11 h 111 2 ≤ ( u − a ) 2 F 11 u 1 2 + 8 ( u − a ) a F ii h ii 2 . (38) For i ≥ 2 i\ge2 i ≥ 2
− G 1 i , i 1 = G 11 − G i i η i i − η 11 = F i i − F 11 h 11 − h i i . -G^{1i,i1}=\frac{G^{11}-G^{ii}}{\eta_{ii}-\eta_{11}}=\frac{F^{ii}-F^{11}}{h_{11}-h_{ii}}. − G 1 i , i 1 = η ii − η 11 G 11 − G ii = h 11 − h ii F ii − F 11 . Set d i = h 11 − h i i d_i=h_{11}-h_{ii} d i = h 11 − h ii (37) ,
( 1 − δ ) h 11 ≤ d i ≤ ( 1 + δ ) h 11 , i ≥ 2. (1-\delta)h_{11}\le d_i\le(1+\delta)h_{11},\qquad i\ge2. ( 1 − δ ) h 11 ≤ d i ≤ ( 1 + δ ) h 11 , i ≥ 2. The spectral second-derivative formula therefore gives
− 2 h 11 G 1 i , i 1 h 11 i 2 = 2 ( F i i − F 11 ) h 11 d i h 11 i 2 ≥ 2 1 + δ ( F i i − F 11 ) h 11 i 2 h 11 2 . -\frac{2}{h_{11}}G^{1i,i1}h_{11i}^2=\frac{2(F^{ii}-F^{11})}{h_{11}d_i}h_{11i}^2\ge\frac{2}{1+\delta}\frac{(F^{ii}-F^{11})h_{11i}^2}{h_{11}^2}. − h 11 2 G 1 i , i 1 h 11 i 2 = h 11 d i 2 ( F ii − F 11 ) h 11 i 2 ≥ 1 + δ 2 h 11 2 ( F ii − F 11 ) h 11 i 2 . Since
F i i ≤ 2 1 + δ ( F i i − F 11 ) + 2 1 + δ F 11 , F^{ii}\le\frac{2}{1+\delta}(F^{ii}-F^{11})+\frac{2}{1+\delta}F^{11}, F ii ≤ 1 + δ 2 ( F ii − F 11 ) + 1 + δ 2 F 11 , we obtain, after summing over i ≥ 2 i\ge2 i ≥ 2
∑ i ≥ 2 F i i h 11 i 2 h 11 2 ≤ − 2 h 11 ∑ i ≥ 2 G 1 i , i 1 h 11 i 2 + C F 11 ∑ i ≥ 2 h 11 i 2 h 11 2 . \sum_{i\ge2}\frac{F^{ii}h_{11i}^2}{h_{11}^2}\le-\frac{2}{h_{11}}\sum_{i\ge2}G^{1i,i1}h_{11i}^2+CF^{11}\sum_{i\ge2}\frac{h_{11i}^2}{h_{11}^2}. i ≥ 2 ∑ h 11 2 F ii h 11 i 2 ≤ − h 11 2 i ≥ 2 ∑ G 1 i , i 1 h 11 i 2 + C F 11 i ≥ 2 ∑ h 11 2 h 11 i 2 . For i ≥ 2 i\ge2 i ≥ 2 (17) , u i = h i i ⟨ X , e i ⟩ u_i=h_{ii}\langle X,e_i\rangle u i = h ii ⟨ X , e i ⟩
∣ h 11 i h 11 ∣ ≤ ∣ h i i ∣ u − a ∣ ⟨ X , e i ⟩ ∣ + A ∣ ⟨ X , e i ⟩ ∣ ≤ C A ∣ ⟨ X , e i ⟩ ∣ . \left|\frac{h_{11i}}{h_{11}}\right|\le\frac{|h_{ii}|}{u-a}|\langle X,e_i\rangle|+A|\langle X,e_i\rangle|\le CA|\langle X,e_i\rangle|. h 11 h 11 i ≤ u − a ∣ h ii ∣ ∣ ⟨ X , e i ⟩ ∣ + A ∣ ⟨ X , e i ⟩ ∣ ≤ C A ∣ ⟨ X , e i ⟩ ∣. Hence
F 11 ∑ i ≥ 2 h 11 i 2 h 11 2 ≤ C A 2 F 11 . F^{11}\sum_{i\ge2}\frac{h_{11i}^2}{h_{11}^2}\le CA^2F^{11}. F 11 i ≥ 2 ∑ h 11 2 h 11 i 2 ≤ C A 2 F 11 . If h 11 h_{11} h 11 A A A a [ 8 ( u − a ) ] − 1 F 11 h 11 2 a[8(u-a)]^{-1}F^{11}h_{11}^{2} a [ 8 ( u − a ) ] − 1 F 11 h 11 2
∑ i ≥ 2 F i i h i i 2 h 11 2 ≤ 2 h 11 ∑ i ≥ 2 G i i h 11 i 2 + a 8 ( u − a ) F 11 h 11 2 . (39) \sum_{i\ge2}\frac{F^{ii}h_{ii}^{2}}{h_{11}^{2}}\le\frac{2}{h_{11}}\sum_{i\ge2}G^{ii}h_{11i}^{2}+\frac{a}{8(u-a)}F^{11}h_{11}^{2}. \tag*{(39)} i ≥ 2 ∑ h 11 2 F ii h ii 2 ≤ h 11 2 i ≥ 2 ∑ G ii h 11 i 2 + 8 ( u − a ) a F 11 h 11 2 . (39) Substituting (38) and (39) into (35) yields
0 ≥ a 4 ( u − a ) ∑ i F i i h i i 2 + A 2 ∑ i F i i − C A . (40) 0\ge\frac{a}{4(u-a)}\sum_i F^{ii}h_{ii}^{2}+\frac{A}{2}\sum_i F^{ii}-CA. \tag*{(40)} 0 ≥ 4 ( u − a ) a i ∑ F ii h ii 2 + 2 A i ∑ F ii − C A . (40) Dropping the first nonnegative term in (40) gives
A 2 ∑ i F i i ≤ C A . \frac{A}{2}\sum_i F^{ii}\le CA. 2 A i ∑ F ii ≤ C A . Since A > 0 A>0 A > 0 ∑ i F i i ≤ C \sum_i F^{ii}\le C ∑ i F ii ≤ C ∑ i F i i = ( n − 1 ) ∑ i G i i \sum_i F^{ii}=(n-1)\sum_i G^{ii} ∑ i F ii = ( n − 1 ) ∑ i G ii
∑ i G i i ≤ C . (41) \sum_i G^{ii}\le C. \tag*{(41)} i ∑ G ii ≤ C . (41) It remains to obtain a positive lower bound for F 11 F^{11} F 11
η ^ = ( η 11 , … , η n n , α ) ∈ Γ k n + 1 . \hat{\eta}=(\eta_{11},\ldots,\eta_{nn},\alpha)\in\Gamma_k^{n+1}. η ^ = ( η 11 , … , η nn , α ) ∈ Γ k n + 1 . Let G ^ = σ k ( η ^ ) 1 / k \hat{G}=\sigma_k(\hat{\eta})^{1/k} G ^ = σ k ( η ^ ) 1/ k n n n G ^ i i = G i i \hat{G}^{ii}=G^{ii} G ^ ii = G ii
G ^ n + 1 , n + 1 = 1 k F 1 / k − 1 σ k − 1 ( η ) ≤ 1 n − k + 1 ∑ i = 1 n G i i . \hat{G}^{n+1,n+1}=\frac{1}{k}F^{1/k-1}\sigma_{k-1}(\eta)\le\frac{1}{n-k+1}\sum_{i=1}^{n}G^{ii}. G ^ n + 1 , n + 1 = k 1 F 1/ k − 1 σ k − 1 ( η ) ≤ n − k + 1 1 i = 1 ∑ n G ii . Thus (41) implies
∑ a = 1 n + 1 G ^ a a ≤ C . (42) \sum_{a=1}^{n+1}\hat{G}^{aa}\le C. \tag*{(42)} a = 1 ∑ n + 1 G ^ aa ≤ C . (42) Indeed, writing N = n + 1 N=n+1 N = n + 1
σ k ( η ^ ) 1 / k − 1 σ k − 1 ( η ^ ) = k N − k + 1 ∑ a = 1 N G ^ a a ≤ C \sigma_k(\hat{\eta})^{1/k-1}\sigma_{k-1}(\hat{\eta})=\frac{k}{N-k+1}\sum_{a=1}^{N}\hat{G}^{aa}\le C σ k ( η ^ ) 1/ k − 1 σ k − 1 ( η ^ ) = N − k + 1 k a = 1 ∑ N G ^ aa ≤ C verifies the hypothesis of part (2) of Lemma 2.2 in Hou–Ma–Wu [7 ] . Applied to η ^ ∈ Γ k N \hat{\eta}\in\Gamma_k^N η ^ ∈ Γ k N G ^ \hat{G} G ^ η ^ \hat{\eta} η ^
η 11 = ∑ j ≥ 2 h j j = O ( A ) , \eta_{11}=\sum_{j\ge2}h_{jj}=O(A), η 11 = j ≥ 2 ∑ h j j = O ( A ) , whereas, for i ≥ 2 i\ge2 i ≥ 2
η i i = h 11 + O ( A ) . \eta_{ii}=h_{11}+O(A). η ii = h 11 + O ( A ) . Hence, if h 11 h_{11} h 11 η ^ \hat{\eta} η ^ η p p \eta_{pp} η pp p ≥ 2 p\ge2 p ≥ 2
G p p ≥ c > 0 , G^{pp}\ge c>0, G pp ≥ c > 0 , and therefore
F 11 = ∑ p ≠ 1 G p p ≥ c > 0. (43) F^{11}=\sum_{p\ne1}G^{pp}\ge c>0. \tag*{(43)} F 11 = p = 1 ∑ G pp ≥ c > 0. (43) Using (43) in the first term of (40), and using the uniform bounds for u − a u-a u − a
0 ≥ c F 11 h 11 2 − C A ≥ c h 11 2 − C A . 0\ge cF^{11}h_{11}^{2}-CA\ge ch_{11}^{2}-CA. 0 ≥ c F 11 h 11 2 − C A ≥ c h 11 2 − C A . Hence h 11 ≤ C h_{11}\le C h 11 ≤ C H > 0 H>0 H > 0 h 11 ≥ h 22 ≥ ⋯ ≥ h n n h_{11}\ge h_{22}\ge\cdots\ge h_{nn} h 11 ≥ h 22 ≥ ⋯ ≥ h nn
0 < H = ∑ i = 1 n h i i ≤ ( n − 1 ) h 11 + h n n , 0<H=\sum_{i=1}^{n}h_{ii}\le(n-1)h_{11}+h_{nn}, 0 < H = i = 1 ∑ n h ii ≤ ( n − 1 ) h 11 + h nn , so h n n ≥ − ( n − 1 ) h 11 h_{nn}\ge-(n-1)h_{11} h nn ≥ − ( n − 1 ) h 11
A priori estimates and proof of Theorem 1.2
We prove Theorem 1.2 by the continuity method. In order to make the argument transparent, we first establish the C 0 C^{0} C 0 C 1 C^{1} C 1 C 2 C^{2} C 2
Set
A n , k = ( n k ) ( n − 1 ) k , B n , k , α = α ( n k − 1 ) ( n − 1 ) k − 1 . A_{n,k}=\binom{n}{k}(n-1)^{k},\qquad B_{n,k,\alpha}=\alpha\binom{n}{k-1}(n-1)^{k-1}. A n , k = ( k n ) ( n − 1 ) k , B n , k , α = α ( k − 1 n ) ( n − 1 ) k − 1 . so that
Φ α ( r ) = A n , k r k + B n , k , α r k − 1 . \Phi_{\alpha}(r)=\frac{A_{n,k}}{r^{k}}+\frac{B_{n,k,\alpha}}{r^{k-1}}. Φ α ( r ) = r k A n , k + r k − 1 B n , k , α . Let
C n , k , α = A n , k + B n , k , α . C_{n,k,\alpha}=A_{n,k}+B_{n,k,\alpha}. C n , k , α = A n , k + B n , k , α . Choose
0 < ε < 1 r 2 k − 1 0<\varepsilon<\frac{1}{r_{2}^{k}-1} 0 < ε < r 2 k − 1 1 and define
f 0 ( X , ν ) = C n , k , α ( 1 + ε ∣ X ∣ k − ε ) . f_{0}(X,\nu)=C_{n,k,\alpha}\left(\frac{1+\varepsilon}{\lvert X\rvert^{k}}-\varepsilon\right). f 0 ( X , ν ) = C n , k , α ( ∣ X ∣ k 1 + ε − ε ) . The choice of ε \varepsilon ε
1 + ε r 2 k − ε = 1 − ε ( r 2 k − 1 ) r 2 k > 0. \frac{1+\varepsilon}{r_{2}^{k}}-\varepsilon
=\frac{1-\varepsilon(r_{2}^{k}-1)}{r_{2}^{k}}>0. r 2 k 1 + ε − ε = r 2 k 1 − ε ( r 2 k − 1 ) > 0. Since ρ ↦ ( 1 + ε ) ρ − k − ε \rho\mapsto(1+\varepsilon)\rho^{-k}-\varepsilon ρ ↦ ( 1 + ε ) ρ − k − ε f 0 > 0 f_{0}>0 f 0 > 0 r 1 ≤ ∣ X ∣ ≤ r 2 r_{1}\le\lvert X\rvert\le r_{2} r 1 ≤ ∣ X ∣ ≤ r 2 λ i ( η ) = n − 1 \lambda_{i}(\eta)=n-1 λ i ( η ) = n − 1 i i i
S k , α ( λ ( η ) ) = A n , k + B n , k , α = C n , k , α = f 0 . S_{k,\alpha}(\lambda(\eta))=A_{n,k}+B_{n,k,\alpha}=C_{n,k,\alpha}=f_{0}. S k , α ( λ ( η )) = A n , k + B n , k , α = C n , k , α = f 0 . Thus the unit sphere is a solution of the t = 0 t=0 t = 0
∂ ∂ ρ ( ρ k f 0 ) = − k ε C n , k , α ρ k − 1 < 0. \frac{\partial}{\partial\rho}\left(\rho^{k}f_{0}\right)=-k\varepsilon C_{n,k,\alpha}\rho^{k-1}<0. ∂ ρ ∂ ( ρ k f 0 ) = − k ε C n , k , α ρ k − 1 < 0. A direct calculation gives
ρ k ( f 0 ( ρ ) − Φ α ( ρ ) ) = B n , k , α ( 1 − ρ ) + ε C n , k , α ( 1 − ρ k ) . \rho^k\left(f_0(\rho)-\Phi_\alpha(\rho)\right)=B_{n,k,\alpha}(1-\rho)+\varepsilon C_{n,k,\alpha}(1-\rho^k). ρ k ( f 0 ( ρ ) − Φ α ( ρ ) ) = B n , k , α ( 1 − ρ ) + ε C n , k , α ( 1 − ρ k ) . Thus
f 0 ( ρ ) > Φ α ( ρ ) ( ρ < 1 ) , f 0 ( ρ ) < Φ α ( ρ ) ( ρ > 1 ) . f_0(\rho)>\Phi_\alpha(\rho)\quad(\rho<1),\qquad f_0(\rho)<\Phi_\alpha(\rho)\quad(\rho>1). f 0 ( ρ ) > Φ α ( ρ ) ( ρ < 1 ) , f 0 ( ρ ) < Φ α ( ρ ) ( ρ > 1 ) . For t ∈ [ 0 , 1 ] t\in[0,1] t ∈ [ 0 , 1 ]
f t = t f + ( 1 − t ) f 0 . f_t=tf+(1-t)f_0. f t = t f + ( 1 − t ) f 0 . Because the path is a convex combination, f t f_t f t ∂ ρ ( ρ k f t ) \partial_\rho(\rho^k f_t) ∂ ρ ( ρ k f t ) ∂ ρ ( ρ k f ) \partial_\rho(\rho^k f) ∂ ρ ( ρ k f ) ∂ ρ ( ρ k f 0 ) \partial_\rho(\rho^k f_0) ∂ ρ ( ρ k f 0 ) (4) holds for every t t t f f f (3) non-strictly while f 0 f_0 f 0 r 1 r_1 r 1 r 2 r_2 r 2 t < 1 t<1 t < 1 f t f_t f t t t t
C 0 C^0 C 0 Let M t M_t M t ρ \rho ρ ρ \rho ρ
∇ ρ = 0 , ∇ 2 ρ ≤ 0. \nabla\rho=0,\qquad\nabla^2\rho\leq0. ∇ ρ = 0 , ∇ 2 ρ ≤ 0. By (11) ,
g i j = ρ 2 g ^ i j , h i j = ρ g ^ i j − ρ i j ≥ ρ g ^ i j , g_{ij}=\rho^2\widehat{g}_{ij},\qquad h_{ij}=\rho\widehat{g}_{ij}-\rho_{ij}\geq\rho\widehat{g}_{ij}, g ij = ρ 2 g ij , h ij = ρ g ij − ρ ij ≥ ρ g ij , and hence
κ i ≥ 1 ρ , λ i ( η ) ≥ n − 1 ρ . \kappa_i\geq\frac{1}{\rho},\qquad\lambda_i(\eta)\geq\frac{n-1}{\rho}. κ i ≥ ρ 1 , λ i ( η ) ≥ ρ n − 1 . Since S k , α S_{k,\alpha} S k , α λ ( η ) \lambda(\eta) λ ( η )
f t ( X , ν ) = S k , α ( λ ( η ) ) ≥ Φ α ( ρ ) . (44) f_t(X,\nu)=S_{k,\alpha}(\lambda(\eta))\geq\Phi_\alpha(\rho). \tag*{(44)} f t ( X , ν ) = S k , α ( λ ( η )) ≥ Φ α ( ρ ) . (44) At such an extremum ν = x = X / ∣ X ∣ \nu=x=X/|X| ν = x = X /∣ X ∣ ρ = r 2 \rho=r_2 ρ = r 2 (44) would give f t ( r 2 x , x ) ≥ Φ α ( r 2 ) f_t(r_2x,x)\geq\Phi_\alpha(r_2) f t ( r 2 x , x ) ≥ Φ α ( r 2 ) t < 1 t<1 t < 1
At a minimum point of ρ \rho ρ
h i j ≤ ρ g ^ i j , λ i ( η ) ≤ n − 1 ρ , h_{ij}\leq\rho\widehat{g}_{ij},\qquad\lambda_i(\eta)\leq\frac{n-1}{\rho}, h ij ≤ ρ g ij , λ i ( η ) ≤ ρ n − 1 , and therefore
f t ( X , ν ) ≤ Φ α ( ρ ) . f_t(X,\nu)\leq\Phi_\alpha(\rho). f t ( X , ν ) ≤ Φ α ( ρ ) . Again ν = x \nu=x ν = x ρ = r 1 \rho=r_1 ρ = r 1 f t ( r 1 x , x ) > Φ α ( r 1 ) f_t(r_1x,x)>\Phi_\alpha(r_1) f t ( r 1 x , x ) > Φ α ( r 1 ) t < 1 t<1 t < 1
r 1 < ρ < r 2 ( 0 ≤ t < 1 ) , (45) r_1<\rho<r_2\qquad(0\leq t<1), \tag*{(45)} r 1 < ρ < r 2 ( 0 ≤ t < 1 ) , (45) with the corresponding non-strict bounds at t = 1 t=1 t = 1
C1 ^{1} 1
By (12) , it suffices to prove a positive lower bound for the support function u = ⟨ X , ν ⟩ u=\langle X,\nu\rangle u = ⟨ X , ν ⟩
w = − log u + γ ( ∣ X ∣ 2 ) , γ ( s ) = L s , w=-\log u+\gamma(|X|^{2}),\qquad\gamma(s)=\frac{L}{s}, w = − log u + γ ( ∣ X ∣ 2 ) , γ ( s ) = s L , where L > 1 L>1 L > 1 w w w X 0 X_{0} X 0 X 0 X_{0} X 0 ν \nu ν u = ∣ X ∣ ≥ r 1 u=|X|\ge r_{1} u = ∣ X ∣ ≥ r 1 X T ≠ 0 X^{T}\ne0 X T = 0 e 1 = X T / ∣ X T ∣ e_{1}=X^{T}/|X^{T}| e 1 = X T /∣ X T ∣
⟨ X , e 1 ⟩ ≠ 0 , ⟨ X , e i ⟩ = 0 ( i ≥ 2 ) . \langle X,e_{1}\rangle\ne0,\qquad\langle X,e_{i}\rangle=0\quad(i\ge2). ⟨ X , e 1 ⟩ = 0 , ⟨ X , e i ⟩ = 0 ( i ≥ 2 ) . At the maximum point,
0 = w i = − u i u + 2 γ ′ ⟨ X , e i ⟩ . 0=w_{i}=-\frac{u_{i}}{u}+2\gamma'\langle X,e_{i}\rangle. 0 = w i = − u u i + 2 γ ′ ⟨ X , e i ⟩ . Since u i = h i j ⟨ X , e j ⟩ u_{i}=h_{ij}\langle X,e_{j}\rangle u i = h ij ⟨ X , e j ⟩ e 1 e_{1} e 1 X T X^{T} X T u i = h i 1 ⟨ X , e 1 ⟩ u_{i}=h_{i1}\langle X,e_{1}\rangle u i = h i 1 ⟨ X , e 1 ⟩ i = 1 i=1 i = 1 h 11 = 2 γ ′ u h_{11}=2\gamma'u h 11 = 2 γ ′ u i ≥ 2 i\ge2 i ≥ 2 h 1 i = 0 h_{1i}=0 h 1 i = 0 e 2 , … , e n e_{2},\ldots,e_{n} e 2 , … , e n ( n − 1 ) × ( n − 1 ) (n-1)\times(n-1) ( n − 1 ) × ( n − 1 ) e 1 e_{1} e 1 γ ′ < 0 \gamma'<0 γ ′ < 0 h 11 < 0 h_{11}<0 h 11 < 0 H > 0 H>0 H > 0 p ≥ 2 p\ge2 p ≥ 2 h p p > 0 h_{pp}>0 h pp > 0 η p p = H − h p p < H − h 11 = η 11 \eta_{pp}=H-h_{pp}<H-h_{11}=\eta_{11} η pp = H − h pp < H − h 11 = η 11 G p p ≥ G 11 G^{pp}\ge G^{11} G pp ≥ G 11 G p p G^{pp} G pp F 11 = ∑ j ≠ 1 G j j F^{11}=\sum_{j\ne1}G^{jj} F 11 = ∑ j = 1 G j j F 11 ≥ G p p ≥ G 11 F^{11}\ge G^{pp}\ge G^{11} F 11 ≥ G pp ≥ G 11 ∑ j G j j = G 11 + F 11 ≤ 2 F 11 \sum_{j}G^{jj}=G^{11}+F^{11}\le2F^{11} ∑ j G j j = G 11 + F 11 ≤ 2 F 11
F 11 = ∑ j ≠ 1 G j j ≥ 1 2 ∑ j G j j = 1 2 ( n − 1 ) ∑ i F i i . (46) F^{11}=\sum_{j\ne1}G^{jj}\ge\frac{1}{2}\sum_{j}G^{jj}=\frac{1}{2(n-1)}\sum_{i}F^{ii}. \tag*{(46)} F 11 = j = 1 ∑ G j j ≥ 2 1 j ∑ G j j = 2 ( n − 1 ) 1 i ∑ F ii . (46) At X 0 X_{0} X 0
w i i = − u i i u + u i 2 u 2 + γ ′ ′ ( ( ∣ X ∣ 2 ) i ) 2 + γ ′ ( ∣ X ∣ 2 ) i i . w_{ii}=-\frac{u_{ii}}{u}+\frac{u_{i}^{2}}{u^{2}}+\gamma''\bigl((|X|^{2})_{i}\bigr)^{2}+\gamma'(|X|^{2})_{ii}. w ii = − u u ii + u 2 u i 2 + γ ′′ ( ( ∣ X ∣ 2 ) i ) 2 + γ ′ ( ∣ X ∣ 2 ) ii . Because ( ∣ X ∣ 2 ) i = 2 ⟨ X , e i ⟩ (|X|^{2})_{i}=2\langle X,e_{i}\rangle ( ∣ X ∣ 2 ) i = 2 ⟨ X , e i ⟩ i = 1 i=1 i = 1
u 1 2 u 2 = 4 γ ′ 2 ⟨ X , e 1 ⟩ 2 , u i = 0 ( i ≥ 2 ) . \frac{u_{1}^{2}}{u^{2}}=4\gamma'^{2}\langle X,e_{1}\rangle^{2},\qquad u_{i}=0\quad(i\ge2). u 2 u 1 2 = 4 γ ′ 2 ⟨ X , e 1 ⟩ 2 , u i = 0 ( i ≥ 2 ) . Multiplying by F i i F^{ii} F ii F i i w i i ≤ 0 F^{ii}w_{ii}\le0 F ii w ii ≤ 0
0 ≥ − F i i u i i u + 4 ( γ ′ 2 + γ ′ ′ ) F 11 ⟨ X , e 1 ⟩ 2 + γ ′ F i i ( ∣ X ∣ 2 ) i i . (47) 0\ge-\frac{F^{ii}u_{ii}}{u}+4(\gamma'^{2}+\gamma'')F^{11}\langle X,e_{1}\rangle^{2}+\gamma'F^{ii}(|X|^{2})_{ii}. \tag*{(47)} 0 ≥ − u F ii u ii + 4 ( γ ′ 2 + γ ′′ ) F 11 ⟨ X , e 1 ⟩ 2 + γ ′ F ii ( ∣ X ∣ 2 ) ii . (47) By (21) ,
F i i u i i = ∑ k ⟨ X , e k ⟩ F i i h i i , k − u F i i h i i 2 + ∑ i F i i h i i . F^{ii}u_{ii}=\sum_{k}\langle X,e_{k}\rangle F^{ii}h_{ii,k}-uF^{ii}h_{ii}^{2}+\sum_{i}F^{ii}h_{ii}. F ii u ii = k ∑ ⟨ X , e k ⟩ F ii h ii , k − u F ii h ii 2 + i ∑ F ii h ii . Only k = 1 k=1 k = 1 (23) and (8) therefore gives
F i i u i i = ⟨ X , e 1 ⟩ ( d X f ~ ) ( e 1 ) + h 11 ⟨ X , e 1 ⟩ ( d ν f ~ ) ( e 1 ) − u F i i h i i 2 + f ~ − R α . F^{ii}u_{ii}=\langle X,e_{1}\rangle(d_{X}\tilde f)(e_{1})+h_{11}\langle X,e_{1}\rangle(d_{\nu}\tilde f)(e_{1})-uF^{ii}h_{ii}^{2}+\tilde f-R_{\alpha}. F ii u ii = ⟨ X , e 1 ⟩ ( d X f ~ ) ( e 1 ) + h 11 ⟨ X , e 1 ⟩ ( d ν f ~ ) ( e 1 ) − u F ii h ii 2 + f ~ − R α . Similarly, from ( ∣ X ∣ 2 ) i i = 2 ( 1 − u h i i ) (|X|^{2})_{ii}=2(1-uh_{ii}) ( ∣ X ∣ 2 ) ii = 2 ( 1 − u h ii )
F i i ( ∣ X ∣ 2 ) i i = 2 ∑ i F i i − 2 u ( f ~ − R α ) . F^{ii}(|X|^{2})_{ii}=2\sum_{i}F^{ii}-2u(\tilde f-R_{\alpha}). F ii ( ∣ X ∣ 2 ) ii = 2 i ∑ F ii − 2 u ( f ~ − R α ) . The radial condition for f t f_t f t k f t + ( d X f t ) ( X ) ≤ 0 k f_t+(d_Xf_t)(X)\le0 k f t + ( d X f t ) ( X ) ≤ 0 f ~ = f t 1 / k \tilde f=f_t^{1/k} f ~ = f t 1/ k
f ~ + ( d X f ~ ) ( X ) = 1 k f t 1 / k − 1 ( k f t + ( d X f t ) ( X ) ) ≤ 0. \tilde f+(d_X\tilde f)(X)=\frac{1}{k}f_t^{1/k-1}\bigl(kf_t+(d_Xf_t)(X)\bigr)\le0. f ~ + ( d X f ~ ) ( X ) = k 1 f t 1/ k − 1 ( k f t + ( d X f t ) ( X ) ) ≤ 0. Since at X 0 X_0 X 0
X = ⟨ X , e 1 ⟩ e 1 + u ν , X=\langle X,e_1\rangle e_1+u\nu, X = ⟨ X , e 1 ⟩ e 1 + uν , the preceding radial inequality gives
− ⟨ X , e 1 ⟩ ( d X f ~ ) ( e 1 ) + f ~ u ≥ ( d X f ~ ) ( ν ) . -\frac{\langle X,e_1\rangle(d_X\tilde f)(e_1)+\tilde f}{u}\ge(d_X\tilde f)(\nu). − u ⟨ X , e 1 ⟩ ( d X f ~ ) ( e 1 ) + f ~ ≥ ( d X f ~ ) ( ν ) . Also h 11 = 2 γ ′ u h_{11}=2\gamma'u h 11 = 2 γ ′ u
− h 11 ⟨ X , e 1 ⟩ u ( d ν f ~ ) ( e 1 ) = − 2 γ ′ ⟨ X , e 1 ⟩ ( d ν f ~ ) ( e 1 ) , -\frac{h_{11}\langle X,e_1\rangle}{u}(d_\nu\tilde f)(e_1)=-2\gamma'\langle X,e_1\rangle(d_\nu\tilde f)(e_1), − u h 11 ⟨ X , e 1 ⟩ ( d ν f ~ ) ( e 1 ) = − 2 γ ′ ⟨ X , e 1 ⟩ ( d ν f ~ ) ( e 1 ) , which is bounded below by − C ∣ γ ′ ∣ -C|\gamma'| − C ∣ γ ′ ∣ F i i h i i 2 F^{ii}h_{ii}^2 F ii h ii 2 R α / u R_\alpha/u R α / u − F i i u i i / u -F^{ii}u_{ii}/u − F ii u ii / u
γ ′ F i i ( ∣ X ∣ 2 ) i i = 2 γ ′ ∑ i F i i − 2 u γ ′ f ~ + 2 u γ ′ R α . \gamma'F^{ii}(|X|^2)_{ii}=2\gamma'\sum_iF^{ii}-2u\gamma'\tilde f+2u\gamma'R_\alpha. γ ′ F ii ( ∣ X ∣ 2 ) ii = 2 γ ′ i ∑ F ii − 2 u γ ′ f ~ + 2 u γ ′ R α . Since γ ′ < 0 \gamma'<0 γ ′ < 0 (10) bounds the last term below by − C α ∣ γ ′ ∣ ∑ i F i i -C\alpha|\gamma'|\sum_iF^{ii} − C α ∣ γ ′ ∣ ∑ i F ii f ~ \tilde f f ~ f t f_t f t C 1 C^1 C 1 (47) yields
0 ≥ 4 ( γ ′ 2 + γ ′ ′ ) F 11 ⟨ X , e 1 ⟩ 2 − C ∣ γ ′ ∣ ( 1 + α ) ∑ i F i i − C ( 1 + ∣ γ ′ ∣ ) . (48) 0\ge4(\gamma'^2+\gamma'')F^{11}\langle X,e_1\rangle^2-C|\gamma'|(1+\alpha)\sum_iF^{ii}-C(1+|\gamma'|). \tag*{(48)} 0 ≥ 4 ( γ ′2 + γ ′′ ) F 11 ⟨ X , e 1 ⟩ 2 − C ∣ γ ′ ∣ ( 1 + α ) i ∑ F ii − C ( 1 + ∣ γ ′ ∣ ) . (48) If
⟨ X , e 1 ⟩ 2 < 1 2 r 1 2 , \langle X,e_1\rangle^2<\frac{1}{2}r_1^2, ⟨ X , e 1 ⟩ 2 < 2 1 r 1 2 , then u 2 = ∣ X ∣ 2 − ⟨ X , e 1 ⟩ 2 ≥ r 2 2 / 2 u^2=|X|^2-\langle X,e_1\rangle^2\ge r_2^2/2 u 2 = ∣ X ∣ 2 − ⟨ X , e 1 ⟩ 2 ≥ r 2 2 /2 s = ∣ X ∣ s=|X| s = ∣ X ∣ [ r 1 , r 2 ] [r_1,r_2] [ r 1 , r 2 ] γ ( s ) = L / s \gamma(s)=L/s γ ( s ) = L / s
γ ′ = − L s 2 , γ ′ ′ = 2 L s 3 , \gamma'=-\frac{L}{s^2},\qquad\gamma''=\frac{2L}{s^3}, γ ′ = − s 2 L , γ ′′ = s 3 2 L , so
γ ′ 2 + γ ′ ′ ≥ c L 2 , ∣ γ ′ ∣ ≤ C L . \gamma'^2+\gamma''\ge cL^2,\qquad|\gamma'|\le CL. γ ′2 + γ ′′ ≥ c L 2 , ∣ γ ′ ∣ ≤ C L . Using (46) and Lemma 2.2, (48) becomes
0 ≥ [ c L 2 − C L ( 1 + α ) ] ∑ i F i i − C ( 1 + L ) . 0\ge[cL^2-CL(1+\alpha)]\sum_iF^{ii}-C(1+L). 0 ≥ [ c L 2 − C L ( 1 + α )] i ∑ F ii − C ( 1 + L ) . Because ∑ i F i i ≥ c n , k > 0 \sum_iF^{ii}\ge c_{n,k}>0 ∑ i F ii ≥ c n , k > 0 L L L u ≥ c > 0 u\ge c>0 u ≥ c > 0
u = ρ 2 ρ 2 + ∣ ∇ ρ ∣ 2 u=\frac{\rho^2}{\sqrt{\rho^2+|\nabla\rho|^2}} u = ρ 2 + ∣∇ ρ ∣ 2 ρ 2 and the C 0 C^0 C 0 ρ \rho ρ
∣ ∇ ρ ∣ 2 = ρ 4 u 2 − ρ 2 ≤ C . |\nabla\rho|^2=\frac{\rho^4}{u^2}-\rho^2\le C. ∣∇ ρ ∣ 2 = u 2 ρ 4 − ρ 2 ≤ C . Hence
∣ ∇ ρ ∣ ≤ C . (49) |\nabla\rho|\le C. \tag*{(49)} ∣∇ ρ ∣ ≤ C . (49) C2 estimate
The estimates (45) and (49) give uniform upper and lower bounds for ρ \rho ρ ∣ ∇ ρ ∣ |\nabla\rho| ∣∇ ρ ∣ (11) this controls the first fundamental form, its inverse, the support function, and the C 1 C^1 C 1 Theorem 1.1 . Moreover, since f t = t f + ( 1 − t ) f 0 f_t=tf+(1-t)f_0 f t = t f + ( 1 − t ) f 0 r 1 ≤ ∣ X ∣ ≤ r 2 r_1\leq|X|\leq r_2 r 1 ≤ ∣ X ∣ ≤ r 2
inf f t ≥ c > 0 , ∥ f t ∥ C 2 ≤ C \inf f_t\geq c>0,\qquad\|f_t\|_{C^2}\leq C inf f t ≥ c > 0 , ∥ f t ∥ C 2 ≤ C uniformly for t ∈ [ 0 , 1 ] t\in[0,1] t ∈ [ 0 , 1 ] Theorem 1.1 therefore yields
max M t max 1 ≤ i ≤ n ∣ κ i ∣ ≤ C . \max_{M_t}\max_{1\leq i\leq n}|\kappa_i|\leq C. M t max 1 ≤ i ≤ n max ∣ κ i ∣ ≤ C . Thus the second fundamental form of M t M_t M t
It remains to translate this geometric curvature bound into a C 2 C^2 C 2
W = ρ 2 + ∣ ∇ ρ ∣ 2 , W=\sqrt{\rho^2+|\nabla\rho|^2}, W = ρ 2 + ∣∇ ρ ∣ 2 , the radial graph formula (11) gives
ρ i j = ρ g ^ i j + 2 ρ ρ i ρ j − W ρ h i j . \rho_{ij}=\rho\widehat{g}_{ij}+\frac{2}{\rho}\rho_i\rho_j-\frac{W}{\rho}h_{ij}. ρ ij = ρ g ij + ρ 2 ρ i ρ j − ρ W h ij . The C 0 C^0 C 0 C 1 C^1 C 1 g g g g ^ \widehat{g} g g g g g ^ \widehat{g} g W / ρ W/\rho W / ρ 2 / ρ 2/\rho 2/ ρ
∣ ∇ 2 ρ ∣ ≤ C , |\nabla^2\rho|\leq C, ∣ ∇ 2 ρ ∣ ≤ C , and therefore
∥ ρ ∥ C 2 ( S n ) ≤ C \|\rho\|_{C^2(S^n)}\leq C ∥ ρ ∥ C 2 ( S n ) ≤ C uniformly for all admissible solutions along the path.
We also record why this gives uniform ellipticity. The curvature bound makes λ ^ = ( λ ( η ) , α ) \widehat{\lambda}=(\lambda(\eta),\alpha) λ = ( λ ( η ) , α ) σ k ( λ ^ ) = f t ≥ c > 0 \sigma_k(\widehat{\lambda})=f_t\geq c>0 σ k ( λ ) = f t ≥ c > 0 λ ^ ∈ Γ k n + 1 \widehat{\lambda}\in\Gamma_k^{n+1} λ ∈ Γ k n + 1 σ j ( λ ^ ) \sigma_j(\widehat{\lambda}) σ j ( λ ) 1 ≤ j < k 1\leq j<k 1 ≤ j < k σ k \sigma_k σ k λ ^ \widehat{\lambda} λ ∂ Γ k n + 1 \partial\Gamma_k^{n+1} ∂ Γ k n + 1 Γ k n + 1 \Gamma_k^{n+1} Γ k n + 1
Existence
Let
I = { t ∈ [ 0 , 1 ] : S k , α ( λ ( η ) ) = f t ( X , ν ) has an admissible star-shaped solution } . \mathcal{I}=\{t\in[0,1]:S_{k,\alpha}(\lambda(\eta))=f_t(X,\nu)\text{ has an admissible star-shaped solution}\}. I = { t ∈ [ 0 , 1 ] : S k , α ( λ ( η )) = f t ( X , ν ) has an admissible star-shaped solution } . By construction, 0 ∈ I 0\in\mathcal{I} 0 ∈ I
We first prove openness. Write ρ = e v \rho=e^v ρ = e v
κ i = e − v κ ‾ i ( D v , D 2 v ) , \kappa_i=e^{-v}\overline{\kappa}_i(Dv,D^2v), κ i = e − v κ i ( D v , D 2 v ) , and therefore
λ i ( η ) = e − v λ ‾ i ( D v , D 2 v ) . \lambda_i(\eta)=e^{-v}\overline{\lambda}_i(Dv,D^2v). λ i ( η ) = e − v λ i ( D v , D 2 v ) . After multiplying the equation by e k v e^{kv} e k v
σ k ( λ ‾ ) + α e v σ k − 1 ( λ ‾ ) = e k v f t ( e v x , ν ( D v ) ) . \sigma_k(\overline{\lambda})+\alpha e^v\sigma_{k-1}(\overline{\lambda})=e^{kv}f_t(e^v x,\nu(Dv)). σ k ( λ ) + α e v σ k − 1 ( λ ) = e k v f t ( e v x , ν ( D v )) . Define
P [ v ] : = e k v f t ( e v x , ν ( D v ) ) − σ k ( λ ‾ ) − α e v σ k − 1 ( λ ‾ ) . \mathcal{P}[v]:=e^{kv}f_t(e^v x,\nu(Dv))-\sigma_k(\overline{\lambda})-\alpha e^v\sigma_{k-1}(\overline{\lambda}). P [ v ] := e k v f t ( e v x , ν ( D v )) − σ k ( λ ) − α e v σ k − 1 ( λ ) . For fixed ( v , D v ) (v,Dv) ( v , D v ) − D 2 v -D^2v − D 2 v S k , α 1 / k S_{k,\alpha}^{1/k} S k , α 1/ k P \mathcal{P} P D 2 v D^2v D 2 v D v Dv D v D 2 v D^2v D 2 v
P v = e k v ( k f t + ( d X f t ) ( X ) ) − α e v σ k − 1 ( λ ‾ ) = ρ ∂ ∂ ρ ( ρ k f t ) − α e v σ k − 1 ( λ ‾ ) < 0. (50) \begin{aligned}
\mathcal{P}_v&=e^{kv}\bigl(kf_t+(d_Xf_t)(X)\bigr)-\alpha e^v\sigma_{k-1}(\overline{\lambda})\\
&=\rho\frac{\partial}{\partial\rho}(\rho^k f_t)-\alpha e^v\sigma_{k-1}(\overline{\lambda})<0. \tag*{(50)}
\end{aligned} P v = e k v ( k f t + ( d X f t ) ( X ) ) − α e v σ k − 1 ( λ ) = ρ ∂ ρ ∂ ( ρ k f t ) − α e v σ k − 1 ( λ ) < 0. (50) The strict inequality follows from (4) , α > 0 \alpha>0 α > 0 Γ k − 1 \Gamma_{k-1} Γ k − 1 σ k − 1 ( λ ‾ ) > 0 \sigma_{k-1}(\overline{\lambda})>0 σ k − 1 ( λ ) > 0
L φ = a i j ∇ i j φ + b i ∇ i φ + c φ , ( a i j ) > 0 , c < 0. L\varphi=a^{ij}\nabla_{ij}\varphi+b^i\nabla_i\varphi+c\varphi,\qquad(a^{ij})>0,\qquad c<0. L φ = a ij ∇ ij φ + b i ∇ i φ + c φ , ( a ij ) > 0 , c < 0. The coefficients are uniformly elliptic at that solution because admissibility is an open condition. If L φ = 0 L\varphi=0 L φ = 0 φ \varphi φ φ ≡ 0 \varphi\equiv0 φ ≡ 0 S n S^n S n L : C 2 , β → C 0 , β L:C^{2,\beta}\to C^{0,\beta} L : C 2 , β → C 0 , β I \mathcal{I} I
To prove closedness, let t j ∈ I t_j\in\mathcal{I} t j ∈ I t j → t ∞ t_j\to t_\infty t j → t ∞ ρ j = e v j \rho_j=e^{v_j} ρ j = e v j C 2 C^2 C 2 v j v_j v j Γ k n + 1 \Gamma_k^{n+1} Γ k n + 1
[ σ k ( λ ‾ ) + α e v σ k − 1 ( λ ‾ ) ] 1 / k = e v f t ( e v x , ν ( D v ) ) 1 / k . \left[\sigma_k(\overline{\lambda})+\alpha e^v\sigma_{k-1}(\overline{\lambda})\right]^{1/k}=e^v f_t(e^v x,\nu(Dv))^{1/k}. [ σ k ( λ ) + α e v σ k − 1 ( λ ) ] 1/ k = e v f t ( e v x , ν ( D v ) ) 1/ k . For fixed ( x , v , D v ) (x,v,Dv) ( x , v , D v ) D 2 v D^2v D 2 v σ k 1 / k \sigma_k^{1/k} σ k 1/ k ( λ ‾ , α e v ) (\overline{\lambda},\alpha e^v) ( λ , α e v ) α e v \alpha e^v α e v − D 2 v -D^2v − D 2 v D 2 v D^2v D 2 v D 2 v D^2v D 2 v Q [ v ] = 0 \mathcal{Q}[v]=0 Q [ v ] = 0
Q [ v ] : = e v f t ( e v x , ν ( D v ) ) 1 / k − [ σ k ( λ ‾ ) + α e v σ k − 1 ( λ ‾ ) ] 1 / k . \mathcal{Q}[v]:=e^v f_t(e^v x,\nu(Dv))^{1/k}-\left[\sigma_k(\overline{\lambda})+\alpha e^v\sigma_{k-1}(\overline{\lambda})\right]^{1/k}. Q [ v ] := e v f t ( e v x , ν ( D v ) ) 1/ k − [ σ k ( λ ) + α e v σ k − 1 ( λ ) ] 1/ k . The operator Q \mathcal{Q} Q D 2 v D^2v D 2 v [4 , 8 ] therefore gives
∥ v j ∥ C 2 , θ ( S n ) ≤ C \lVert v_j\rVert_{C^{2,\theta}(S^n)}\le C ∥ v j ∥ C 2 , θ ( S n ) ≤ C for some $0<\theta<1$. Differentiating the equation gives a uniformly elliptic linear equation for the first derivatives of v j v_j v j C θ C^\theta C θ C 3 , θ C^{3,\theta} C 3 , θ C 2 C^2 C 2 f t f_t f t C 3 , β C^{3,\beta} C 3 , β C 3 , β ′ C^{3,\beta'} C 3 , β ′ β ′ < β \beta' < \beta β ′ < β t = t ∞ t = t_{\infty} t = t ∞ I \mathcal{I} I
I = [ 0 , 1 ] , \mathcal{I} = [0,1], I = [ 0 , 1 ] , and in particular a solution exists for t = 1 t = 1 t = 1
Uniqueness
Let v 1 v_1 v 1 v 2 v_2 v 2 t = 1 t = 1 t = 1 w = v 1 − v 2 w = v_1 - v_2 w = v 1 − v 2 x 0 x_0 x 0
D v 1 ( x 0 ) = D v 2 ( x 0 ) , D 2 v 1 ( x 0 ) ≤ D 2 v 2 ( x 0 ) Dv_1(x_0) = Dv_2(x_0), \qquad D^2v_1(x_0) \leq D^2v_2(x_0) D v 1 ( x 0 ) = D v 2 ( x 0 ) , D 2 v 1 ( x 0 ) ≤ D 2 v 2 ( x 0 ) as quadratic forms. Keeping the common gradient and the value v 1 ( x 0 ) v_1(x_0) v 1 ( x 0 ) P \mathcal{P} P
P ( v 1 , D v 1 , D 2 v 1 ) ≤ P ( v 1 , D v 2 , D 2 v 2 ) . \mathcal{P}(v_1,Dv_1,D^2v_1) \leq\mathcal{P}(v_1,Dv_2,D^2v_2). P ( v 1 , D v 1 , D 2 v 1 ) ≤ P ( v 1 , D v 2 , D 2 v 2 ) . Now keep ( D v 2 , D 2 v 2 ) (Dv_2,D^2v_2) ( D v 2 , D 2 v 2 ) λ ‾ ( D v 2 , D 2 v 2 ) \overline{\lambda}(Dv_2,D^2v_2) λ ( D v 2 , D 2 v 2 ) v v v Γ k − 1 \Gamma_{k-1} Γ k − 1 v 2 v_2 v 2 v v v v 2 ( x 0 ) v_2(x_0) v 2 ( x 0 ) v 1 ( x 0 ) v_1(x_0) v 1 ( x 0 ) v 1 ( x 0 ) > v 2 ( x 0 ) v_1(x_0) > v_2(x_0) v 1 ( x 0 ) > v 2 ( x 0 )
P ( v 1 , D v 2 , D 2 v 2 ) < P ( v 2 , D v 2 , D 2 v 2 ) . \mathcal{P}(v_1,Dv_2,D^2v_2) < \mathcal{P}(v_2,Dv_2,D^2v_2). P ( v 1 , D v 2 , D 2 v 2 ) < P ( v 2 , D v 2 , D 2 v 2 ) . Both endpoints are solutions, so altogether
0 = P [ v 1 ] ( x 0 ) < P [ v 2 ] ( x 0 ) = 0 , 0 = \mathcal{P}[v_1](x_0) < \mathcal{P}[v_2](x_0) = 0, 0 = P [ v 1 ] ( x 0 ) < P [ v 2 ] ( x 0 ) = 0 , a contradiction. Therefore v 1 ≤ v 2 v_1 \leq v_2 v 1 ≤ v 2 v 2 ≤ v 1 v_2 \leq v_1 v 2 ≤ v 1 v 1 = v 2 v_1 = v_2 v 1 = v 2
Acknowledgments. Peihe Wang is supported by National Natural Science Foundation of Shandong Province(ZR2026MS0039). Hua Zhu is supported by the National Natural Science Foundation of China (Grant Nos. 12501273 and 12661041).
