On degenerate case of prescribed curvature measure problems

Guohuan Qiu; Jingjing Suo

doi:10.1515/ans-2022-0035

Article Open Access

On degenerate case of prescribed curvature measure problems

Guohuan Qiu and Jingjing Suo

Published/Copyright: January 10, 2023

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From the journal Advanced Nonlinear Studies Volume 23 Issue 1

Abstract

In this article, we prove the C^1,1 estimate for solutions of prescribed curvature measure problems when the prescribed function may touch zero somewhere.

Keywords: prescribed curvature measure problems; degenerate equations

MSC 2010: 35J70

1 Introduction

For a compact C 2 hypersurface M which is the boundary of a bounded domain Ω in R n + 1 , let κ = ( κ 1 , … , κ n ) be the principal curvature of M , and σ k be the k th elementary symmetric function. The n − k th curvature measure of Ω is defined as:

C n − k ( Ω , γ ) ≔ ∫ γ ∩ M σ k ( κ ) d μ g ,

for γ ∈ R n + 1 , where d μ g is the volume element with respect to the induced metric g of M in R n + 1 .

If the surface is star shape at origin, it can be parameterized as a radial graph over S n :

R M : S n → M x → ρ ( x ) x .

For a star-shaped C 2 hypersurface, the n − k curvature measure on each Borel set β on S n can be defined as

C n − k ( Ω , β ) ≔ C n − k ( Ω , R M ( β ) ) = ∫ R M ( β ) σ k ( κ ) d μ g = ∫ β σ k ( κ ) detg d S n .

The following version of prescribed curvature measure problem is formulated as in [3].

Prescribed curvature problem 1

For each 1 ≤ k ≤ n and each given nonnegative function f ∈ C 2 ( S n ) , find a closed hypersurface M as a radial graph over S n , such that C n − k ( Ω , β ) = ∫ β f d μ for every Borel set β in S n , where d μ is the standard volume element on S n .

For k = n , it is the Alexandrov problem which was first obtained by Alexandrov in [1] for existence and uniqueness. He also gave a necessary and sufficient condition for existence of the weak solution. The regularity of solutions of the Alexandrov problem was proved by Pogorelov [9] and Oliker [8]. For k < n , Guan et al. [4] obtained the existence and regularity of convex solutions under certain assumptions on f . And Guan et al. [3] solved the problem for admissible solutions.

All the aforementioned regularity results need to assume f > 0 , even for k = 1 case. It is natural to ask whether the solutions to the prescribed curvature problem are always smooth when the given density function is smooth but only nonnegative. It will cause degeneracy at the point when f = 0 in this fully nonlinear problem. For k = n , Treibergs [11] has proved that a stronger version of the Alexandrov condition is, in fact, a necessary and sufficient condition for uniformly C 0 bound. Then Guan and Li [2] proved C 1 , 1 estimates in this degenerate case. They also constructed counter examples to show that C 1 , 1 estimates are in some sense optimal, see also [7].

In this article, we consider the degenerate case for k < n and prove C 1 , 1 estimates for star-shaped admissible hypersurface under some pointwise conditions.

Theorem 1.1

For 1 ≤ k < n , let f be a nonnegative C 1 , 1 function on S n . We suppose that f satisfies the following conditions:

(1.1) 0 < η = min S n n f 2 k ( n − 1 ) max S n f 1 k + ( n − 1 ) ∫ S n f 1 k n ω n − ∣ ∇ f 1 k ∣

and

(1.2) ∫ S n f 1 k > 0 ,

where ω n = ∣ S n ∣ . Moreover, we assume that

(1.3) Δ f 1 k − 1 ≥ − A

and

(1.4) ∣ ∇ f 1 k − 1 ∣ ≤ A .

Then there exists a C 1 , 1 star-shaped k-admissible hypersurface M as the n–k curvature measure is given by f .

We are going to explain these Conditions (1.1), (1.2), (1.3), and (1.4) as follows. The main difficulty of these estimates are lack of positive lower bound of radial function. By introducing condition (1.1) in the theorem we have an existence result for this family of degenerate equations. This condition which is very similar to the conditions in [4,10] is used to obtain the positive lower bound. The key obeservation is that we can reduce the problem to a homogeneous mean curvature equation (4.20). Because only in this case we can obtain Harnack estimates. For the homogeneous equation (4.27), it can be shown that this condition is sharp for the gradient estimate from the counter-example of Treibergs. The second condition (1.2) is relatively natural. And the C 1 , 1 estimate is due to the last two conditions which are also assumed as the k = n case in [2].

2 Preliminary

We give some notations and introduce our theorem more specifically. For λ = ( λ 1 , … , λ n ) ∈ R n , σ k ( λ ) is defined as

σ k ( λ ) = ∑ λ i 1 ⋯ λ i k ,

where the sum is taken over for all increasing sequences i 1 , … , i k of the indices chosen from the set { 1 , … , n } . The definition can be extended to symmetric matrices.

Definition 2.1

For 1 ≤ k ≤ n , let Γ k be a cone in R n determined by

Γ k = { λ ∈ R n : σ 1 ( λ ) > 0 , … , σ k ( λ ) > 0 } .

A C 2 surface M is called k -admissible if at every point X ∈ M , ( κ 1 , κ 2 , … , κ n ) ∈ Γ ¯ k . Moreover, for any symmetric matrix h i j , F i j ≔ ∂ σ k ( λ ( h i j ) ) ∂ h i j is semi-positive if λ ( h i j ) ∈ Γ ¯ k . For C 1 , 1 surface the notion of curvature and k -admissible are understood almost everywhere on M .

Then we write our theorem into a more precise from. In terms of ρ , we have

detg = ρ n − 1 ρ 2 + ∣ ∇ ρ ∣ 2 .

Therefore, prescribed curvature measure problem can be reduced to solving a fully nonlinear partial differential equation on S n for a given function f :

(2.1) σ k ( κ 1 , … , κ n ) = f ( x ) ρ 1 − n ( ρ 2 + ∣ ∇ ρ ∣ 2 ) − 1 2 .

The metric g i j and its inverse g i j on M are given by:

g i j = ρ 2 δ i j + ρ i ρ j

and

g i j = ρ − 2 ( δ i j − ρ i ρ j ρ 2 + ∣ ∇ ρ ∣ 2 ) .

The unit outer normal and the second fundamental form of M are

ν = − ∇ ρ + ρ x ρ 2 + ∣ ∇ ρ ∣ 2 , h i j = ρ 2 δ i j + 2 ρ i ρ j − ρ ρ i j ρ 2 + ∣ ∇ ρ ∣ 2 .

So the principal curvature κ of M are the solutions of

det ( A i j − κ δ i j ) = 0 ,

where { A i j } is a symmetric matrix

{ A i j } ≔ { g i p } 1 2 { h p q } { g q j } 1 2 ,

and { g i j } 1 2 the positive square root of { g i j } given by

{ g i j } 1 2 ≔ ρ − 1 δ i j − ρ i ρ j ρ 2 + ∣ ∇ ρ ∣ 2 ( ρ + ρ 2 + ∣ ∇ ρ ∣ 2 ) .

Theorem 2.1

For 1 ≤ k < n , let f be a nonnegative C 1 , 1 function on S n . Suppose that f satisfies conditions (1.1), (1.2), (1.3), and (1.4), then there exists a unique C 1 , 1 star-shaped and k-admissible hypersurface M, such that it satisfies (2.1). Moreover, there is a constant C depending only on k , n , η , A , f 1 k C 1 ( S n ) , and inf ∫ S n f 1 k , but independent of positive lower bound of f such that

∣ ∣ ρ ∣ ∣ C 1 , 1 ( S n ) ≤ C .

3 Upper bound of ρ

At the maximum point x 0 ∈ S n of function ρ , we have

∇ ρ = 0

and

∇ 2 ρ ≤ 0 .

So using equation (2.1), we obtain

max ρ = ρ ( x 0 ) ≤ f ( x 0 ) C n k 1 n − k ≤ max f C n k 1 n − k .

Because we deal with the case when f can attain zero somewhere, this argument fails about the positive lower bound of ρ .

4 Harnack estimate

It is well known that Harnack estimate follows from gradient estimate for logarithm of positive function. So without loss of generality assuming ρ > 0 , let us consider v = − log ρ . We can write the first and second fundamental form into

g i j = e − 2 v ( δ i j + v i v j )

and

h i j = e − v ( 1 + ∣ ∇ v ∣ 2 ) − 1 2 [ δ i j + v i v j + v i j ] .

Then we introduce some new notations for convenience

h ¯ i j ≔ δ i j + v i v j + v i j , g ¯ i j ≔ δ i j − v i v j 1 + ∣ ∇ v ∣ 2 ( 1 + 1 + ∣ ∇ v ∣ 2 ) ,

and

a i j ≔ g ¯ i l h ¯ l m g ¯ m j .

So equation (2.1) becomes

(4.1) σ k ( λ ( a i j ) ) = f ( x ) e ( n − k ) v ( 1 + ∣ ∇ v ∣ 2 ) k − 1 2 .

Theorem 4.1

For 1 ≤ k < n , suppose f is a nonnegative function on S n satisfying

0 < η = inf S n n f 2 k ( n − 1 ) max S n f 1 k + ( n − 1 ) ∫ S n f 1 k n ω n − ∣ ∇ f 1 k ∣ ,

and

∫ S n f 1 k > 0 ,

where ω n = ∣ S n ∣ . There exists a constant C which depends on k , n , η , f 1 k C 1 ( S n ) , and inf ∫ S n f 1 k . Such that the solution of the equation (2.1) has gradient estimate

sup S n ∣ ∇ ρ ∣ ≤ C .

And there also exists a positive constant c depending on k , n , inf ∫ S n f 1 k , ∣ ∇ f 1 k ∣ , and η such that

ρ ≥ c .

For later usage, we consider the equation

(4.2) σ k ( λ ( a i j ) ) = f ( x ) e θ v h ( ∣ ∇ v ∣ 2 ) ,

where h ( ∣ ∇ v ∣ 2 ) = ( 1 + ∣ ∇ v ∣ 2 ) ϑ 2 , for constants θ ≥ 0 and ϑ < k .

The following lemma is obtained in [10] and [5]. For completeness, we give its proof here.

Lemma 4.1

For 1 ≤ k < n , suppose f is a nonnegative function on S n . In the case that θ > 0 , there exists a constant C which depends on k , n , ∣ ∣ f 1 k ∣ ∣ C 1 ( S n ) , and upper bound of v. The solution of equation (4.2) has gradient estimate

max S n ∣ ∇ v ∣ ≤ C .

In particular, when θ = 0 , k = 1 , and ϑ = 1 we need to further assume f satisfies

(4.3) η ≔ min x ∈ S n f 2 n − 1 + ( n − 1 ) − ∣ ∇ f ∣ > 0 .

And in this case gradient estimate of v is independent on upper bound of v.

Remark 4.1

The counter-example of Treibergs in [10] tells us that this condition (4.3) is sharp for the gradient estimate of equation (4.2) when k = 1 , θ = 0 , and ϑ = 1 .

Proof

We consider a test function

P ≔ ∣ ∇ v ∣ 2 2 .

At the maximum point of P

(4.4) 0 = P i = ∑ l v l v l i ,

(4.5) 0 ≥ P i j = ∑ l v l i v l j + ∑ l v l v l i j .

We denote F l m ≔ ∂ σ k ∂ a l m .

(4.6) 0 ≥ F l m g ¯ l i g ¯ m j P i j = F l m g ¯ l i g ¯ m j v q j v q i + F l m g ¯ l i g ¯ m j v q v q i j .

For convenience, at the maximum point we assume ∣ ∇ v ∣ = v 1 , and v i = 0 for i ≥ 2 . So we have from equality (4.4)

v 11 = 0 , v 1 i = 0 , i ≥ 2 .

We can also assume { v i j } i , j ≥ 2 is diagonal at this point. So the matrixes { g ¯ i j } , { h ¯ i j } , and { a i j } are all diagonal at this point, and g ¯ 11 = 1 1 + v 1 2 , h ¯ 11 = 1 + v 1 2 , a 11 = 1 ; and for all i > 1 , g ¯ i i = 1 , h ¯ i i = 1 + v i i , a i i = 1 + v i i .

So inequality (4.6) becomes

(4.7) ∑ i ≥ 1 F i i ( g ¯ i i ) 2 v 1 v 1 i i ≤ − ∑ i ≥ 2 F i i v i i 2 .

It follows from the expression of g ¯ i j and (4.4)

(4.8) v l g ¯ l i p = 0 .

So we have (4.8) and the expression of h ¯ i j

F i j v l a i j l = 2 F i j v l g ¯ l i p h ¯ p m g ¯ m j + F i j v l g ¯ i p h ¯ p m l g ¯ m j = F i j v l g ¯ i p g ¯ m j ( 2 v p l v m + v p m l ) = F i i ( g ¯ i i ) 2 v 1 v i i 1 .

Using the Ricci identity on sphere v i i 1 = v 1 i i + v i δ 1 i − v 1 δ i i , we have

(4.9) F i j v l a i j l = F i i ( g ¯ i i ) 2 v 1 v 1 i i + F 11 v 1 2 1 + v 1 2 − F i i ( g ¯ i i ) 2 v 1 2 .

On the other hand, we take the first derivative of equation (4.2),

(4.10) F i j v l a i j l = v 1 f 1 e θ v h + θ v 1 2 h f e θ v .

From (4.9) and (4.10), we have

(4.11) F i i ( g ¯ i i ) 2 v 1 v 1 i i = v 1 f 1 e θ v h + θ v 1 2 h f e θ v − F 11 v 1 2 1 + v 1 2 + F i i ( g ¯ i i ) 2 v 1 2 .

We combine (4.7) and (4.11) to be

(4.12) 0 ≥ v 1 f 1 e θ v h + θ v 1 2 h f e θ v − F 11 v 1 2 1 + v 1 2 + F i i ( g i i ) 2 v 1 2 + ∑ i = 2 n F i i v i i 2 ≥ v 1 f 1 e θ v h + θ v 1 2 h f e θ v + ∑ i = 2 n F i i v 1 2 + ∑ i = 2 n F i i v i i 2 .

In order to handle the last two terms of (4.12), we denote λ = ( λ 1 , … , λ n ) to be the eigenvalue of the symmetric matrix of { a i j } . So we have λ 1 = 1 and λ i = 1 + v i i for i > 1 .

Due to the following elementary equalities:

∑ i = 1 n F i i = ( n − k + 1 ) σ k − 1 , ∑ i = 1 n F i i λ i = k σ k , ∑ i = 1 n F i i λ i 2 = σ 1 σ k − ( k + 1 ) σ k + 1 ,

we have

v 1 2 ∑ i = 2 n F i i = v 1 2 [ ( n − k + 1 ) σ k − 1 − σ k − 1 ( λ ∣ 1 ) ]

and

∑ i = 2 n F i i v i i 2 = ∑ i = 2 n F i i ( λ i − 1 ) 2 = ∑ i = 2 n F i i λ i 2 − 2 ∑ i = 2 n F i i λ i + ∑ i = 2 n F i i = σ 1 σ k − ( k + 1 ) σ k + 1 − 2 k σ k + ( n − k + 1 ) σ k − 1 .

Now inequality (4.12) becomes

0 ≥ v 1 f 1 e θ v h + θ v 1 2 f e θ v h − 2 k σ k + σ 1 σ k − ( k + 1 ) σ k + 1 + ( v 1 2 + 1 ) ( n − k + 1 ) σ k − 1 − σ k − 1 ( λ ∣ 1 ) v 1 2 .

From λ ( a i j ) ∈ Γ ¯ k and Newton-MacLaurin inequality, we have

σ k − 1 − σ k − 1 ( λ ∣ 1 ) = λ 1 σ k − 2 ( λ ∣ 1 ) ≥ 0

and

σ 1 σ k − ( k + 1 ) σ k + 1 ≥ 0 .

We obtain for k < n

(4.13) v 1 2 σ k − 1 ≤ C ( n , k ) ∣ ∇ v ∣ ∣ ∇ f ∣ f σ k + σ k .

Here we assume ∣ ∇ v ∣ sufficiently large, because we are doing the estimate of ∣ ∇ v ∣ . Due to the Newton-MacLaurin inequality, we have

(4.14) σ k − 1 ≥ c ( n , k ) σ k k − 1 k .

From (4.14) and (4.13), we obtain

v 1 2 σ k k − 1 k ≤ C ( n , k ) σ k ∣ ∇ v ∣ ∣ ∇ f ∣ f + 1 v 1 2 f − 1 k e − θ v k ( 1 + v 1 2 ) − ϑ 2 k ≤ C ( n , k ) ∣ ∇ v ∣ ∣ ∇ f ∣ f + 1 ∣ ∇ v ∣ ≤ C k , n , ∣ ∇ f 1 k ∣ e θ k − ϑ v .

When k = 1 , θ = 0 , and ϑ = 1 , inequality (4.12) becomes

(4.15) 0 ≥ v 1 f 1 1 + v 1 2 + ( n − 1 ) v 1 2 + ∑ i = 2 n v i i 2 .

In order to obtain the optimal gradient estimate, we use geometric-arithmetic mean inequality

(4.16) ∑ i = 2 n v i i 2 ≥ ∑ i = 2 n v i i 2 n − 1 .

Then we insert (4.16) into (4.15)

(4.17) f 2 ( 1 + v 1 2 ) n − 1 + ( n − 1 ) v 1 2 ≤ − v 1 f 1 1 + v 1 2 + 2 n f 1 + v 1 2 n − 1 − n 2 n − 1 .

So if we assume

η = min x ∈ S n f 2 n − 1 + ( n − 1 ) − ∣ ∇ f ∣ > 0 ,

we obtain the gradient estimate

□ max S n ∣ ∇ v ∣ ≤ 1 η C ( n , ∣ ∣ f ∣ ∣ C 1 ( S n ) ) .

We first obtain a existence result about homogeneous mean curvature equation of the following form

Theorem 4.2

Let g be a nonnegative, nontrivial smooth function on S n , and further assume that g satisfies

(4.18) 0 < η = min S n n g 2 ( n − 1 ) max S n g + ( n − 1 ) ∫ S n g n ω n − ∣ ∇ g ∣ , ∫ S n g > 0 ,

where ω n = ∣ S n ∣ .

There exists a unique constant λ with

(4.19) n max S n g ≤ λ ≤ n ω n ∫ S n g .

And there is a smooth solution v unique up to an additive constant satisfying the following mean curvature equation on sphere

(4.20) 1 + ∣ ∇ v ∣ 2 div ∇ v 1 + ∣ ∇ v ∣ 2 + n = λ g ( x ) 1 + ∣ ∇ v ∣ 2 .

Moreover, there is constant C depending only on n, ∣ ∣ g ∣ ∣ C 1 , η , and inf ∫ S n g such that

(4.21) ∣ ∣ v − ⨍ S n v ∣ ∣ C 1 ( S n ) ≤ C .

Remark 4.2

We emphasize that g can touch zero somewhere on sphere. This theorem is a modified version of Treibergs [10].

In order to obtain the existence of this equation we need several lemmas.

Lemma 4.2

Given a function w ∈ C 1 , γ ( S n ) and any positive constant α , we define λ ( w ) satisfying

(4.22) λ ( w ) ∫ S n g = ∫ S n n 1 + ∣ ∇ w ∣ 2 .

Then for any 0 ≤ t ≤ 1 and any C γ function g , there exists a unique C 2 , γ solution v satisfying equation on S n

Δ v 1 + ∣ ∇ w ∣ 2 − w i w j v i j ( 1 + ∣ ∇ w ∣ 2 ) 3 2 = α v + t [ − n 1 + ∣ ∇ w ∣ 2 + λ ( w ) g ] .

In particular, there is only a trivial solution when t = 0 .

Proof

By the maximum principle and the Schauder estimate, we have

∣ ∣ v ∣ ∣ C 2 , γ ( S n ) ≤ C ( ∣ ∣ w ∣ ∣ C 1 , γ ( S n ) , 1 α , n , ∣ ∣ g ∣ ∣ C 1 ( S n ) ) .

Then the method of continuity (see [Theorem 5.2, in [6]]), gives the existence of this equation.□

Lemma 4.3

For any fixed α > 0 , we suppose g ∈ C 1 ( S n ) satisfies condition (4.18). There is a unique solution v ∈ C 2 , γ ( S n ) satisfying the following equation on S n

(4.23) Δ v 1 + ∣ ∇ v ∣ 2 − v i v j v i j ( 1 + ∣ ∇ v ∣ 2 ) 3 2 = α v + − n 1 + ∣ ∇ v ∣ 2 + λ ( v ) g ,

where λ ( v ) satisfies (4.22). Moreover, we have λ ( v ) satisfies inequality (4.19), and a Schauder estimate of v which is independent of 1 α

(4.24) ∣ ∣ v ∣ ∣ C 2 , γ ( S n ) ≤ C n , ∣ ∣ g ∣ ∣ C 1 ( S n ) , η , inf ∫ S n g .

Proof

By Lemma 4.2, there exists a compact mapping T t : C 1 , γ ( S n ) → C 1 , γ ( S n ) for each 0 ≤ t ≤ 1 , taking w ∈ C 1 , γ ( S n ) to the solution v ∈ C 2 , γ ( S n ) of

Δ v 1 + ∣ ∇ w ∣ 2 − w i w j v i j ( 1 + ∣ ∇ w ∣ 2 ) 3 2 = α v + t [ − n 1 + ∣ ∇ w ∣ 2 + λ ( w ) g ] .

And T 0 w = 0 for all w ∈ C 1 , γ ( S n ) .

We shall apply the Leray-Schauder fixed point theorem ([Theorem 11.6 in [6]]) to solve this equation (4.23). Then we need only to prove a C 1 , γ estimate of the equation

(4.25) Δ v 1 + ∣ ∇ v ∣ 2 − v i v j v i j ( 1 + ∣ ∇ v ∣ 2 ) 3 2 = α v + − n 1 + ∣ ∇ v ∣ 2 + λ ( v ) g .

We note here the adding term α v in equation (4.23) will not cause trouble in our gradient estimate. In fact, inequality (4.17) becomes

λ 2 g 2 ( 1 + v 1 2 ) n − 1 ≤ − α v 1 2 1 + ∣ ∇ v ∣ 2 − λ t v 1 g 1 1 + v 1 2 − ( n − 1 ) v 1 2 − ( 1 + v 1 2 ) α 2 v 2 n − 1 − 2 n 1 + v 1 2 α v n − 1 + 2 α v λ g 1 + v 1 2 n − 1 − n 2 n − 1 + 2 n λ g 1 + v 1 2 n − 1 ≤ − λ t v 1 g 1 1 + v 1 2 + 2 n λ g 1 + v 1 2 n − 1 − n 2 n − 1 .

And we can easily estimate λ ( v ) by (4.22) to obtain

λ ≤ n ω n ∫ S n g .

On the other hand, v attain its minimum at x 0 . Moreover, if we integrate equation (4.25) over S n , we have from (4.22)

(4.26) ∫ S n v = 0 .

This infers

v ( x 0 ) ≤ 0 .

At this point, we have

0 ≤ α v ( x 0 ) + t ( λ g ( x 0 ) − n ) ≤ t ( λ g ( x 0 ) − n ) ,

and we can assume t ≠ 0 here without loss of generality. We the obtain estimate of λ ( v ) .

n max S n g ≤ λ ≤ n ω n ∫ S n g .

So let inf S n n g 2 ( n − 1 ) max S n g + ( n − 1 ) ∫ S n g n ω n − ∣ ∇ g ∣ = η > 0 and throw away the term contained α , we obtain a gradient estimate independent of α and t

sup S n ∣ ∇ v ∣ ≤ C n , ∣ ∣ g ∣ ∣ C 1 ( S n ) , η , inf ∫ S n g .

Finally, the C 1 , γ estimate follows from [Theorem 13.1, [6]] and (4.26), and C 2 , γ estimate (4.24) follows from the standard Schauder estimate.□

Proof of the Theorem 4.2

From Lemma 4.3, we have a solution ( v , λ ) of (4.27) by letting α → 0 from (4.23). And estimate (4.21) follows from (4.24). The uniqueness of λ is easy as in the following comparison argument.□

The main Theorem 4.1 of this section follows from the following lemma.

Lemma 4.4

Suppose f satisfies the same condition as in Theorem 4.1, then the radial function ρ has a positive lower bound.

Proof

Suppose v ˜ satisfies the following mean curvature equation:

(4.27) σ 1 ( λ ( a i j ) ) = 1 + ∣ ∇ v ˜ ∣ 2 div ∇ v ˜ 1 + ∣ ∇ v ˜ ∣ 2 + n = λ f ( x ) 1 k 1 + ∣ ∇ v ˜ ∣ 2 .

And for 1 ≤ k < n , v is a solution of

σ k ( λ ( a i j ) ) = f ( x ) e ( n − k ) v ( 1 + ∣ ∇ v ∣ 2 ) k − 1 2 .

Then we consider function P = v − v ˜ .

At the maximum point y 0 of P , we have

0 = ∇ v − ∇ v ˜

and

0 ≥ { v i j − v ˜ i j } .

The Newton-MacLaurin inequality tells us if λ ( a i j ) ∈ Γ ¯ k

f ( x ) e ( n − k ) v ( 1 + ∣ ∇ v ∣ 2 ) k − 1 2 = σ k ( λ ( a i j ) ) ≤ C n k n k ( 1 + ∣ ∇ v ∣ 2 div ( ∇ v 1 + ∣ ∇ v ∣ 2 ) + n ) k ≤ C n k n k ( 1 + ∣ ∇ v ˜ ∣ 2 div ( ∇ v ˜ 1 + ∣ ∇ v ˜ ∣ 2 ) + n ) k = C n k n k λ k f ( x ) ( 1 + ∣ ∇ v ˜ ∣ 2 ) k .

So we have

max ( v − v ˜ ) ≤ C − v ˜ ( y 0 ) .

Let z 0 be the maximum point of v , then

v ( z 0 ) − v ˜ ( z 0 ) ≤ max ( v − v ˜ ) .

So we obtained

v ( z 0 ) ≤ C k , n , inf ∫ S n f 1 k , ∣ ∇ f 1 k ∣ , η ,

which in turn gives positive lower bound of ρ .□

5 C 1 , 1 estimate

Now we have C 1 estimate and positive lower bound of radial function ρ . Then C 1 , 1 estimate of ρ is equivalent to the curvature estimate of the following curvature equation:

(5.1) σ k ( κ 1 , … , κ n ) = f X ∣ X ∣ ∣ X ∣ − 1 − n u = ϕ ( X ) u ,

where X ∈ M , u ≔ ⟨ X ( x ) , x ⟩ , and ϕ ( X ) = f X ∣ X ∣ ∣ X ∣ − 1 − n . And there exists a positive constant a such that u − a > 0 .

Theorem 5.1

For 1 < k ≤ n , we suppose f ∈ C 2 ( S n ) satisfies conditions,

Δ S n f 1 k − 1 ≥ − A

and

∣ ∇ S n f 1 k − 1 ∣ ≤ A .

Then the smooth hypersurface which satisfies curvature equation (5.1) has the following curvature estimate:

H ≤ C ( A , k , n , ∣ ∣ ρ ∣ ∣ C 1 ( S n ) , max ∣ f ∣ , min ρ ) .

For curvature estimate, we consider the auxiliary function

(5.2) P = log H − log ( u − a ) + γ ( t ) ,

where t ≔ ∣ X ∣ 2 .

Remark 5.1

In order to deal with C 1 , 1 estimate for the degenerate case, we need to add term contained ∣ X ∣ 2 . And here a is introduced for the technical reason.

Proof

It is convenient to work on orthonormal frame on M directly. We choose an orthonormal frame such that e 1 , e 2 , … , e n are tangent to M and e n + 1 is normal. We denote F l m ≔ ∂ σ k ∂ h m l . At maximum point of P , we have

(5.3) 0 = P i = H i H − u i u − a + γ ′ t i

and

(5.4) F i j P i j = F i j H i j H − F i j H i H j H 2 − F i j u i j u − a + F i j u i u j ( u − a ) 2 + F i j γ ″ t i t j + F i j γ ′ t i j .

We list the following well-known formulas for hypersurfaces in R n + 1 .

(5.5) X i j = − h i j e n + 1 ( Gauss formula ) ,

(5.6) ( e n + 1 ) i = h i j e j ( Weingarten equation ) ,

(5.7) h i j k = h i k j ( Codazzi formula ) ,

(5.8) R i j k l = h i k h j l − h i l h j k ( Gauss equation ) ,

where R i j k l is the curvature tensor. By (5.5)–(5.8), we have the following commutation formula:

(5.9) H i i = ∑ l h i i l l + ∑ j , k h j k 2 h i i − ∑ l h i l h l i H .

And

(5.10) u i = X , ∑ l h i l e l ,

(5.11) u i j = h i j + ∑ l h i l j ⟨ X , e l ⟩ − ∑ l h i l h j l u .

(5.12) ∣ X ∣ i 2 = 2 ⟨ X , e i ⟩ , ∣ X ∣ i j 2 = 2 δ i j − 2 h i j u .

Take first and second derivatives to our equation (5.1).

(5.13) F i j h i j l = ϕ l u + ϕ u l

and

(5.14) F i j , p q h p q l h i j l + F i j h i j l l = ϕ l l u + 2 ϕ l u l + ϕ u l l .

Now we assume that at the maximum point h i j is diagonal. Inserting (5.3) and (5.9)–(5.14) to (5.4), we have

(5.15) 0 ≥ Δ ϕ u + 2 ∇ ϕ ∇ u + ϕ H + H l ⟨ X , e l ⟩ ϕ − ϕ ∣ A ∣ 2 u H − F i j , p q h p q l h i j l H + k ϕ u ∣ A ∣ 2 H − F i i h i i 2 − F i i u i u − a − γ ′ t i 2 − k ϕ u u − a − ( ϕ l u + ϕ u l ) ⟨ X , e l ⟩ u − a + F i i h i i 2 u u − a + F i i u i 2 ( u − a ) 2 + 4 γ ″ F i i ⟨ X , e i ⟩ 2 + 2 γ ′ ∑ i F i i − 2 γ ′ k ϕ u 2 .

Now we recall Lemma 3.2 from [3], which in fact use the concavity of σ k σ 1 1 k − 1 ( h i j ) . If ( h i j ) ∈ Γ ¯ k , then

(5.16) − F i j , p q h i j l h p q l H ≥ σ k H ( σ k ) l σ k − H l H 1 k − 1 − 1 ( σ k ) l σ k − 1 + 1 k − 1 H l H .

Then from equations (5.1), (5.13), and (5.3), we further estimate

(5.17) − F i j , p q h i j l h p q l H ≥ ϕ u H ϕ l ϕ + 2 γ ′ t l − a u l u ( u − a ) ( 2 − k ) ϕ l ( k − 1 ) ϕ + γ ′ k k − 1 t l − u l ( 2 u − k − 2 k − 1 a ) u ( u − a ) ≥ − ( k − 2 ) u ∣ ∇ ϕ ∣ 2 ( k − 1 ) H ϕ − C ( ϕ + ∣ ∇ ϕ ∣ ) ,

where C is a constant depending on n , k , min ρ , max ρ , and sup ∣ ∇ ρ ∣ . Using (5.17), we obtain from (5.15)

(5.18) 0 ≥ − ( k − 2 ) u ∣ ∇ ϕ ∣ 2 ( k − 1 ) H ϕ − C ( ϕ + ∣ ∇ ϕ ∣ ) + Δ ϕ u H + 2 ∑ l ϕ l h l l ⟨ X , e l ⟩ H + ϕ − k ϕ u u − a − 2 γ ′ ⟨ X , e l ⟩ 2 ϕ + ( k − 1 ) ϕ u ∣ A ∣ 2 H + 4 γ ′ F i i h i i ⟨ X , e i ⟩ 2 u − a + 4 F i i ⟨ X , e i ⟩ 2 ( γ ″ − ( γ ′ ) 2 ) + a F i i h i i 2 u − a − ∑ i ϕ i u ⟨ X , e i ⟩ u − a + 2 γ ′ ∑ i F i i − 2 γ ′ k ϕ u 2 .

First if we choose H sufficiently large,

(5.19) − ( C + 1 ) ϕ − k ϕ u u − a − 2 γ ′ ⟨ X , e l ⟩ 2 ϕ + ( k − 1 ) ϕ u ∣ A ∣ 2 H − 2 γ ′ k ϕ u 2 > 0 .

Then we may choose γ ( t ) = − c log ( 2 max M ∣ X ∣ 2 − t ) , where c = min min M ( 2 max M ∣ X ∣ 2 − ∣ X ∣ 2 ) ( u − a ) a 4 ∣ X ∣ 4 , 1 , such that

(5.20) γ ′ ∑ i F i i + 4 γ ′ F i i h i i ⟨ X , e i ⟩ 2 u − a + 4 F i i ⟨ X , e i ⟩ 2 ( γ ″ − ( γ ′ ) 2 ) + a F i i h i i 2 u − a ≥ γ ′ ∑ i F i i − 4 ( γ ′ ) 2 ∑ i F i i ∣ X ∣ 4 a ( u − a ) ≥ 0 .

By (5.19) and (5.20), inequality (5.18) becomes

(5.21) 0 ≥ − ( k − 2 ) u ∣ ∇ ϕ ∣ 2 ( k − 1 ) H ϕ + Δ ϕ u H − C ∣ ∇ ϕ ∣ + 2 ∑ l ϕ l h l l ⟨ X , e l ⟩ H − ∑ i ϕ i u ⟨ X , e i ⟩ u − a + γ ′ ∑ i F i i ≥ − ( k − 2 ) u ∣ ∇ ϕ ∣ 2 ( k − 1 ) H ϕ − C ∣ ∇ ϕ ∣ + Δ ϕ u H + γ ′ ∑ i F i i .

We compute ∣ ∇ ϕ ∣ 2 and Δ ϕ explicitly,

(5.22) ϕ i = f i ∣ X ∣ − 1 − n + ( − 1 − n ) f ∣ X ∣ − 3 − n ⟨ X , e i ⟩ ∣ ∇ ϕ ∣ 2 ≤ ∣ ∇ f ∣ 2 ∣ X ∣ − 2 − 2 n + C ( f 2 + ∣ ∇ f ∣ f )

and

(5.23) Δ ϕ = Δ f ∣ X ∣ − 1 − n + 2 ( − 1 − n ) ∣ X ∣ − 3 − n f i ⟨ X , e i ⟩ + ( 1 + n ) ( 3 + n ) f ∣ X ∣ − 5 − n ⟨ X , e i ⟩ 2 − ( 1 + n ) f ∣ X ∣ − 3 − n ( n − H u ) ≥ Δ f ∣ X ∣ − 1 − n − C ( ∣ ∇ f ∣ + H f ) .

If on sphere we suppose

Δ S n f 1 k − 1 ≥ − A

and

∣ ∇ S n f 1 k − 1 ∣ ≤ A .

Because we already have estimates of ρ and ∇ ρ , the above conditions are equivalent to

(5.24) Δ f 1 k − 1 ≥ − A ( 1 + H )

and

(5.25) ∣ ∇ f 1 k − 1 ∣ ≤ A .

Due to the Newton-MacLaurin inequality, we have

(5.26) ∑ F i i ≥ c 1 ( n , k ) σ k k − 2 k − 1 H 1 k − 1 .

Finally, combining (5.22), (5.23), (5.24), (5.25), and (5.26), we obtain estimate of H from (5.21),

□ H 1 k − 1 ≤ Δ f ∣ X ∣ − 1 − n u f k − 2 k − 1 H − ( k − 2 ) ∣ ∇ f ∣ 2 ∣ X ∣ − 1 − n u ( k − 1 ) f k − 2 k − 1 f H − C ( ∣ ∇ f 1 k − 1 ∣ + f 1 k − 1 ) c 1 ∣ X ∣ − ( n + 1 ) ( k − 2 ) k − 1 γ ′ ≤ C ( A , k , n , ∣ ∣ ρ ∣ ∣ C 1 ( S n ) , max ∣ f ∣ , min ρ ) .

6 Proof of Theorems 1.1 and 2.1

Proof

First let f ε = f + ε , there exists a C 3 hypersurface which satisfies the prescribed curvature equation (2.1) by [3]. Then combining Theorems 4.1 and 5.1, we obtained a C 1 , 1 estimate not depending on ε . Theorems 1.1 and 2.1 are completed by taking ε → 0 .□

Acknowledgement

Guohuan Qiu would like to express gratitude to Professor Pengfei Guan for suggesting the problem and helpful discussions.

Funding information: Guohuan Qiu is partially supported by a research grant from the Research Grants Council of the Hong Kong Special Administrative region, China [Project No: CUHK14304120] and CUHK Direct Grant [Project Code: 4053340].
Conflict of interest: The authors state no conflict of interest.
Data availability statement: The data used to support the findings of this study are included within the article.

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Received: 2022-08-27

Accepted: 2022-10-30

Published Online: 2023-01-10

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https://doi.org/10.1515/ans-2022-0035

Keywords for this article

prescribed curvature measure problems; degenerate equations

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