C^{2} A Priori Estimates for Degenerate Monge-Ampère Equations

Pengfei Guan^{∗} Department of Mathematics McMaster University Hamilton, Ontario L8S 4K1 Canada

1 Introduction

In this paper, we are concerned with the Dirichlet problem for degenerate Monge-Ampère equation:

\{ \begin{matrix}det(u_{ij}) = f  in Ω ⊂ R^{n} \\ u|_{∂Ω} = ϕ . \end{matrix}  (1)

The Dirichlet problem for elliptic Monge-Ampère equations has received considerable study. It is quite complete due to the important works of Caffarelli- Nirenberg-Spruck[CNS1] and Krylov[K1] (see also references in above papers for the history of the problem). When f is only assumed non-negative, the equation is degenerate. In this case, in general one can only expect the solution to be C^{1,1}. Thus, much attention has been directed to obtain C^{2} a priori estimates of the problem. The interior and global C^{2} estimates were obtained respectively by Trudinger-Urbas[TU2] and Caffarelli-Nirenberg-Spruck[CNS2] for the homogeneous equation:

\{ \begin{matrix}det(u_{ij}) = 0  in Ω ⊂ R^{n} \\ u|_{∂Ω} = ϕ . \end{matrix}  (2)

C^{1,1} regularity for inhomogeneous degenerate equation (1) was studied by Caffarelli-Kohn-Nirenberg-Spruck[CKNS] and Krylov[K2] (see also the references therein) under the assumption thatf = g^{n} for some C^{1,1} function g. Since F (u_{ij}) = det(u_{ij})^{1/n} is concave, the assumption f = g^{n} seems natural for obtaining C^{2} estimates for the equation (1). This assumption has been used in most of the works related to the regularity problem of the equation (1). Except in [C], Caffarelli obtained C^{1,α} regularity for convex solutions of equation (1) under a very general condition that the right hand side of (1) is a measure satisfying the doubling property. As we know, the assumption f = g^{n} is quite restrictive. Even in n = 2 case, not every smooth function is a square of a C^{1,1} function. In dealing with geometric problems, one would like to know C^{1,1} (and better) regularity of the solution when f is only assumed to be nonnegative and smooth (e.g., [GL1]). Therefore, we ask: is the solution of equation (1) in C^{1,1} if f is nonnegative and smooth?

∗Supported in part by NSERC grant OGP0046732 and a Sloan Fellowship.

In this paper, we will prove that for n = 2 and 3, when f is smooth enough, the solution of the equation with homogeneous Dirichlet boundary condition is indeed C^{1,1}. For general n, we introduce a sharp sufficient condition to obtain C^{2} estimates for the homogeneous Dirichlet boundary value problem of the degenerate Monge-Ampère equation (1). Our result for n = 2 and 3 is a consequence of our general result. Our sufficient condition is satisfied by any C^{1,1} nonnegative functions when n = 2 and by any C^{3,1} nonnegative functions when n = 3 (the smoothness assumptions for n = 2 and 3 are also sharp). As for higher regularity result in n = 2, we refer [G].

To state our results, we let Ω be a bounded strictly convex domain in R^{n}, let f be a non-negative function in R^{n}. We assume:

Condition (C): there is a positive constant A, such that, for any x in Ω,

(n − 1)f(x)Δf(x) − (n − 2)|∇f(x)|^{2} ≥ −Af(x)^{2−1/(n−1)} ;  (3)

and, there is δ > 0, for any x in Ω, with dist(x, ∂Ω) ≤ δ, then,

|∇f(x)| ≤ Af(x)^{1−1/(n−1)} .  (4)

Theorem 1 Suppose Ω be a bounded strictly convex domain in R^{n} with C^{2,1} boundary. If f ≥ 0, and f satisfied Condition (C), then, the Monge-Ampère equation (1) with the homogeneous boundary condition ϕ ≡ 0, has an unique C^{1,1} convex solution u, with

∥u∥_{C^{2}(Ω̄)} ≤ C,  (5)

where the constant C depends only on Ω, A, δ, sup(f), and modulus continuity of f in Ω.

To characterize the condition, we introduce some notation. We say a function g is PSEUDO-SUBHARMONIC, if there is a constant A, such that the function g + A|x|^{2} is subharmonic. Then the first inequality in Condition (C) is equivalent to say f^{\frac{1}{n−1}} pseudosubharmonic in Ω, that is,

inf\{Δf^{\frac{1}{n−1}} (x)\} ≥ −(n − 1)^{−1}A ,  (6)

for x in Ω. And the second inequality is to say f^{\frac{1}{n−1}} is Lipschitz continuous near ∂Ω, that is,

|∇f^{\frac{1}{n−1}} | ≤ \frac{A}{n − 1}  (7)

for x in Ω with dist(x, ∂Ω) ≤ δ. It’s easy to check that when f = |g|^{m} for m ≥ n − 1, with C^{1,1} function g, f will satisfy the condition. In fact, Condition (C) is satisfied by a much wider class of functions as we see from following lemmas (see also Lemma 16 for further discussions).

Lemma 2 Suppose that

f = ∑_{j=1}^{N} |g_{j}|^{m_{j}},  (8)

for some C^{1,1} functions g_{j} in Ω̄, and for some real constants m_{j} ≥ n − 1, then f satisfies Condition (C) for any δ.

Moreover, in the cases n = 2 and 3, Condition (C) is quite general. We have

Lemma 3 When n = 2, if f = |g| for some C^{1,1} function g in Ω̄, then f satisfies Condition (C) for any δ; when n = 3, if f is a C^{3,1} nonnegative function on R^{3}, then f satisfies Condition (C) for any δ.

For n = 2, Lemma 3 is trivial. For n = 3, Lemma 3 is a consequence of Lemma 2 and following important sum-of-squares lemma of Fefferman about nonnegative functions in R^{n}.

Lemma 4 If f is a C^{3,1} nonnegative function on R^{n}, with ∥f∥_{C^{4}(R^{n})} ≤ A, then there is N (only depends on n), and there are functions g_{1}, g_{2}, …, g_{N}, with g_{j} in C^{1,1}(R^{n}), with ∥g_{j}∥_{C^{2}(R^{n})} ≤ C, such that f = ∑_{j=1}^{N} g_{j}^{2}, where constant C depends only on n and A.

Therefore, we deduce

Corollary 5 Suppose that

f = ∑_{j=1}^{N} |g|_{j}^{m_{j}},  (9)

for some m_{j} ≥ n−1, g_{j} ∈ C^{1,1}(Ω̄), then the degenerate Monge-Ampère equation (1) with the boundary condition ϕ ≡ 0 has an unique C^{1,1} convex solution. In particular, when n = 2, f = |g|, for some g ∈ C^{1,1}(Ω̄), or when n = 3, 0 ≤ f ∈ C^{3,1}(R^{3}), then equation (1) with the boundary condition ϕ ≡ 0 has an unique C^{1,1} convex solution.

Remark 6 (i) The Condition (C) is sharp in the sense the (n − 1)-power growth can’t be reduced. An example given by Wang (Example 3 in [W]) shows that, for any ε > 0, there is a C^{1,1} nonnegative function g, such that f = g^{n−1−ε}. But, the equation (1) with the boundary condition ϕ ≡ 0 has no C^{1,1} convex solution. By the same example, the regularity assumptions in Corollary 5 for n = 2 and 3 are also sharp. Similar results were obtained in [GL2] for Monge-Ampère type equations related to Alexandrov problem on S^{n} (no boundary).

(ii) For the smoothness assumption (on f), we only need f^{\frac{1}{n−1}} to be pseudo-subharmonic in Ω, and Lipschitz near the boundary. We note that, for any α ≥ 0, the function f(x) = |x|^{α} satisfies (C) in any dimension if o ∉ ∂Ω.

(iii) The assumption of the homogeneous boundary condition together with the second inequality (6) in Condition (C) are only needed in obtaining C^{2} bounds near the boundary. If one can get the C^{2} a priori estimates on the boundary by other means, then one only needs the first inequality (3) in Condition (C) to get global C^{2} bounds (see lemma 11 in section 3). For example, we have

Theorem 7 Let Ω be a bounded strictly convex domain in R^{n} with C^{3,1} boundary. Let f be a nonnegative function in Ω. Suppose following condition is satisfied:

Condition (C’): there is a constant A, such that, for any x in Ω,

(n − 1)f(x)Δf(x) − (n − 2)|∇f(x)|^{2} ≥ −Af(x)^{2−1/(n−1)} ,  (10)

and assume that for some positive constant δ, when dist(x, ∂Ω) ≤ δ, f(x) ≥ γ > 0 and is Lipschitz. Then for any function ϕ ∈ C^{3,1}(∂Ω) the Dirichlet problem has an unique C^{1,1} convex solution u, with

∥u∥_{C^{2}(Ω̄)} ≤ C,  (11)

where C depends only on Ω, γ, δ, A, sup(f) and ∥ϕ∥_{C^{4}(∂Ω)}.

The plan of this paper is as follow. In the next section, we derive C^{2} bounds on the boundary. Interior C^{2} bounds together with the proof of theorems are given in section 3. In section 4, we extend our techniques to treat the equation of prescribed Gauss curvature. In the last section, we prove lemmas in this introduction.

We wish to express our thanks to C. Fefferman for helpful comments and for permission to include his unpublished result (Lemma 4) in this paper, which fits into the case n = 3 perfectly. We would like to thank J.J. Kohn for helpful discussions and encouragements. We want also to thank Y.Y. Li, our interest for the degenerate Monge-Ampère equations is in large stemed from joint works with him [GL1] and [GL2]. The research was conducted while the author was visiting Princeton University. We want to thank Princeton University for the hospitality.

2 C^{2} Bounds on the Boundary

Now we set the stage for the estimates. We assume f > 0 in the process of establishing a priori estimates.

C^{0} and C^{1} estimates can be done in the same way as in [CNS1] and [CKNS]. We won’t repeat them here. Therefore, we assume:

∥u∥_{C^{1}(Ω̄)} ≤ C.  (12)

where C depends only on Ω, Sup(f) and ∥ϕ∥_{C^{2}(Ω̄)}.

We turn to the estimates for the second derivatives of u. First, we shall get the estimation on the boundary. Let p ∈ ∂Ω be a boundary point. Choose coordinates x_{1}, x_{2}, …, x_{n} with origin at p and such that r_{i}(0) = 0 for i ≤ n − 1, and r_{n}(0) = −1, where r is the defining function of Ω. Near the origin p = 0, we may represent u as

u = ϕ + σr  (13)

for some function σ. We have

u_{n}(0) = ϕ_{n}(0) − σ(0) .  (14)

So that by (12), |σ(0)| ≤ C. Hence

u_{ij}(0) = ϕ_{ij}(0) + σ(0)r_{ij}(0)  for i, j ≤ n − 1 .  (15)

This gives

|u_{ij}(0)| ≤ C  for i, j ≤ n − 1 ,  (16)

where C depends only on Ω, ∥u∥_{C^{1}(Ω̄)}, ∥ϕ∥_{C^{2}(Ω̄)}.

Now, we want to estimate |u_{in}| at the origin for 1 ≤ i ≤ n − 1. This will be done by a careful construction of a barrier function following [CKNS]. But, our arguments here are little subtler, since we only assume now |∇f| ≤ Af^{1− \frac{1}{n−1}} near the boundary.

For ε > 0, set S_{ε} = {x ∈ U | r(x) ≤ 0, x_{n} ≤ ε}, where U is a neighborhood of the origin. We note that, near the origin, r can be expressed as

r(x) = −x_{n} + ∑_{i,j}^{n−1} a_{ij}x_{i}x_{j} + O(|x|^{3}) .

Let

T_{α} = \frac{∂}{∂x_{α}} − \frac{r_{α}}{r_{n}} \frac{∂}{∂x_{n}} .  (17)

T_{α} are tangential vector fields on ∂Ω, T_{α}r = T_{α}(u − ϕ) = 0 on the boundary, ∀ α = 1, 2, …, n − 1.

Define

w = ±T_{α}(u − ϕ) − B̃x_{n} + B|x|^{2} .  (18)

Claim 8 Assume that for some δ > 0, |∇f(x)| ≤ Af^{1− \frac{1}{n−1}} (x), ∀ x ∈ \{x ∈ Ω | dist(x, ∂Ω) ≤ δ}. There are positive constants ε, B, B̃ which depend only on A, Ω, sup_{Ω} f, δ and ∥u∥_{C^{1}(Ω̄)}, such that max_{x∈S_{ε}} w = w(0) = 0.

Remark 9 This is the only place we use the second inequality (4) in Condition (C).

Proof of the Claim. Let

L = u^{ij}(x) \frac{∂^{2}}{∂x_{i}∂x_{j}} ,  (21′)

where (u^{ij}) = (u_{ij})^{−1}. Applying T_{α} on the equation log det(u_{ij}) = log f

L(T_{α}u) = \frac{T_{α}f}{f} − u^{ij} (\frac{r_{α}}{r_{n}} )_{i} u_{nj} − u^{ij} (\frac{r_{α}}{r_{n}} )_{j} u_{ni} + u^{ij} (\frac{r_{α}}{r_{n}} )_{ij} u_{n} .  (19)

Observe that u^{ij}u_{ni} = δ_{nj}, we have

L(T_{α}u) = \frac{T_{α}f}{f} − 2 (\frac{r_{α}}{r_{n}} )_{n} + u^{ij} (\frac{r_{α}}{r_{n}} )_{ij} u_{n} .  (20)

By (12) and (18), and the assumption |∇f| ≤ Af^{1− \frac{1}{n−1}} in \{x ∈ Ω | dist(x, ∂Ω) ≤ δ\}

Lw ≥ B ∑_{i=1}^{n} u^{ii} − Af^{− \frac{1}{n−1}} − C (∑_{i=1}^{n} u^{ii} + 1)  (21)

in {x ∈ Ω | dist(x, ∂Ω) ≤ δ}, where C depends only on ∥r∥_{C^{3}}, ∥u∥_{C^{1}} and ∥ϕ∥_{C^{3}}. Now, let

B = 10(n + 1)^{2}(A + C + sup_{Ω}f^{\frac{1}{n−1}} + 1) .  (22)

We pick 0 < ε ≤ δ small, such that 2x_{n} − ∑_{i,j}^{n−1} a_{ij}x_{i}x_{j} ≥ 0 and |\frac{r_{α}}{r_{n}} | ≤ 1 in S_{ε}. We choose

B̃ = 10 (\frac{∥u − ϕ∥_{C^{1}}}{ε} + B + max_{i,j≤n−1} |a_{ij}| + (1 + ∥\frac{r_{α}}{r_{n}} ∥_{C^{2}})(∥u∥_{C^{1}} + ∥ϕ∥_{C^{3}})) .  (23)

We shall show with the choice of above B, ε, B̃, max_{S_{ε}} w = w(0) = 0.

We decompose S_{ε} = S_{ε}^{1} ∪ S_{ε}^{2}, where S_{ε}^{1} = S_{ε} ∩ {Δu ≥ 1}, S_{ε}^{2} = S_{ε} ∩ {Δu < 1}. Step 1: Lw > 0 in S_{ε}^{1}.

∀ q ∈ S_{ε}^{1}, we pick new coordinates y_{1}, y_{2}, …, y_{n} near q, which is an orthonormal linear transformation of x_{1}, x_{2}, …, x_{n}, such that the matrix \{\frac{∂^{2}u}{∂y_{β}∂y_{γ}} \} is diagonal at q. Since ∑_{β=1}^{n} \frac{∂^{2}u}{∂y_{β}^{2}}(q) = Δu(q) ≥ 1, there is 1 ≤ σ ≤ n, such that u_{σσ}(q) = \frac{∂^{2}u}{∂y_{σ}^{2}} (q) ≥ \frac{1}{n}. At q,

∑_{i=1}^{n} u^{ii}(q) = tr{u^{ij}(q)} = tr{u_{ij}(q)}^{−1} = tr{u_{βγ}(q)}^{−1} = tr{u^{γβ}(q)}
= ∑_{β=1}^{n} u^{ββ}(q) ≥ ∑_{β≠σ} u^{ββ}(q) ≥ (n − 1) ( ∏_{β≠σ} u^{ββ}(q))^{\frac{1}{n−1}}
= (n − 1)^{(∏_{β=1}^{n} u^{ββ}(q))^{\frac{1}{n−1}}} /(u^{σσ}(q))^{\frac{1}{n−1}}
= (n − 1)(det(u^{βγ}(q)))^{\frac{1}{n−1}} · uσσ^{\frac{1}{n−1}} (q)
≥ (n − 1) (\frac{1}{n})^{\frac{1}{n−1}} f^{− \frac{1}{n−1}} (q) .  (24)

With our choice of B, from (21), we get Lw(q) > 0.

Step 2: If max_{S_{ε}} w = w(x_{0}) for some x_{0} ∈ S̄_{ε}, then x_{0} ∈ ∂S_{ε}.

By Step 1, x_{0} ∉ S_{ε}^{1}. If x_{0} ∈ S_{ε}^{2}, since x_{0} is an interior point, w_{n}(x_{0}) = 0. From the definition (18),

0 = w_{n}(x_{0}) = ±u_{αn} ∓ \frac{r_{α}}{r_{n}} u_{nn} ∓ (\frac{r_{α}}{r_{n}} )_{n} u_{n} − B̃ + 2x_{n}^{0}B

∓ϕ_{αn} ± \frac{r_{α}}{r_{n}} ϕ_{nn} ± (\frac{r_{α}}{r_{n}} )_{n} ϕ_{n} .

By our choice of B, ε, B̃, at x_{0}, by the convexity of u

6 ≤ |u_{αn} − \frac{r_{α}}{r_{n}} u_{nn}| ≤ \frac{u_{αα}}{2} + \frac{3}{2} u_{nn} ≤ \frac{3}{2}Δu.

We arrive at Δu(x_{0}) ≥ 4, contradiction.

Finally, it is easy to see w|_{∂S_{ε}} ≤ 0 with our choice of B, ε, B̃. This completes the proof of the claim.

The claim implies

|(T_{α}u)_{n}(0)| ≤ C .  (25)

In turn, we get

|u_{in}(0)| ≤ C  for 1 ≤ i ≤ n − 1  (26)

where C depends only on A, Ω, ∥ϕ∥_{C^{3}}, sup_{Ω} f and δ.

To estimate u_{nn}(0), as in [TU1], we pick a principal coordinate system at 0 (i.e., leave x_{n} fixed, make an orthonormal linear transformation in x_{1}, …, x_{n−1}). In that principle

system (r_{ij}(0))_{i,j≤n−1} is a diagonal matrix [ \begin{matrix}k_{1} & & & & 0 \\ & . & . & . \\ 0 & & & & k_{n−1} \end{matrix} ] where k_{1}, …, k_{n−1} are the

principal curvatures of ∂Ω at 0. In view of (15), at 0

∇^{2}u − ∇^{2}ϕ = [ \begin{matrix}(u − ϕ)_{n}k_{1} & 0 & ⋯ & 0 & (u − ϕ)_{n1} \\ 0 \\ ⋮ & & & & ⋮ \\ 0 & ⋯ & & (u − ϕ)_{n}k_{n−1} \\ (u − ϕ)_{n1} & & ⋯ & & (u − ϕ)_{nn} \end{matrix} ]  (27)

We may assume that −ϕ is convex in Ω, so that we have, by the convexity, at 0,

|(u − ϕ)_{n}| ≥ (sup_{Ω} |u − ϕ|) /(diamΩ) .  (28)

Now, we may solve equation (1) for u_{nn}(0), if ϕ ≡ 0. (Here is the only place we make use of the homogeneous boundary condition.) We get

0 ≤ u_{nn}(0) ≤ C  (29)

where C depends only on A, Ω, sup_{Ω} f, δ, and \frac{1}{max_{Ω} |u|}.

In conclusion, we have proved,

Lemma 10 If ϕ ≡ 0 on ∂Ω, we have |∇^{2}u| ≤ C on ∂Ω, where C depends only on A, Ω, sup_{Ω} f, δ and \frac{1}{max |u|}.

3 C^{2}-Estimates of u in Ω and the Proofs of Theorems 1 and 7

Now we turn to estimate |∇^{2}u| in the interior. We prove

Lemma 11 Suppose f > 0, f satisfies Condition (C′) in Ω. Suppose u ∈ C^{4}(Ω) ∩ C^{2}(Ω̄) is convex, u satisfies Monge-Ampère equation (1) in Ω. Then,

sup_{Ω} |∇^{2}u| ≤ C(1 + sup_{∂Ω} |∇^{2}u|)  (30)

where C depends only on A and diam(Ω).

Proof. Let H(x) = Δu(x), define

G(x) = H(x)e^{α \frac{|x|^{2}}{2}}  (31)

where

α = \frac{1}{2(2n^{2} + 1)diam(Ω)} .  (32)

By the convexity of u, it is sufficient to obtain an upper bound for G. Let p ∈ Ω̄, G(p) = max_{Ω} G(x). If p ∈ ∂Ω, (32) holds automatically. We may assume p ∈ Ω and 0 ∈ Ω. Then at p, ∇G = 0, and ∇^{2}G ≤ 0. That is, at p

H_{β} = −αx_{β}H,  β = 1, 2, …, n  (33)

and

(H_{ij}) ≤ (α^{2}x_{i}x_{j} − αδ_{ij})H .  (34)

At p, making an orthonormal linear transformation if necessary, we may assume (u_{ij}) is diagonal, and u_{11} ≤ u_{22} ≤ ⋯ ≤ u_{nn}. Since (u_{ij}) is positive definite at p, by (34)

L(H) ≡ u^{ij}H_{ij} ≤ (α^{2}u^{ij}x_{i}x_{j} − α ∑_{i=1}^{n} u^{ii}) H

≤ −α(1 − α diam^{2}(Ω)) ∑_{i=1}^{n} u^{ii}H .  (35)

By the chain rule,

\frac{∂u^{ij}}{∂u_{kℓ}} = −u^{ik}u^{ℓj} .  (36)

We have

D^{2} (det(u_{ij})^{\frac{1}{n−1}} )
= \frac{1}{n − 1} det(u_{ij})^{\frac{1}{n−1}} \{u^{ij}D^{2}u_{ij} + \frac{1}{n − 1} u^{ℓk}Du_{ℓk}u^{ij}Du_{ij} − u^{ik}u^{ℓj}Du_{ij}Du_{ℓk}\} (37)

Since det(u_{ij})^{\frac{1}{n−1}} = f^{\frac{1}{n−1}} , and

D^{2} (f^{\frac{1}{n−1}} ) = \frac{1}{n − 1} f^{−2+ \frac{1}{n−1}} \{fD^{2}f − (1 − \frac{1}{n − 1}) |Df|^{2}\} .  (38)

We get the identity

f^{−2} \{fD^{2}f − (1 − \frac{1}{n − 1}) |Df|^{2}\}

= \{u^{ij}D^{2}u_{ij} + \frac{1}{n − 1} u^{ℓk}u^{ij}Du_{ℓk}Du_{ij} − u^{ik}u^{ℓj}Du_{ij}Du_{ℓk}\} .  (39)

Summing up,

f^{−2} \{fΔf − (1 − \frac{1}{n − 1}) |∇f|^{2}\}

= \{L(H) + \frac{1}{n − 1} ∑_{β=1}^{n} u^{ℓk}u^{ij}u_{βℓk}u_{βij} − ∑_{β=1}^{n} u^{ik}u^{ℓj}u_{βij}u_{βℓk}\} .  (40)

Using the fact that (u^{ij}) is diagonal at p,

f^{−2} \{fΔf − (1 − \frac{1}{n − 1}) |∇f|^{2}\}

= \{L(H) + \frac{1}{n − 1} ∑_{β=1}^{n} u^{ii}u^{jj}u_{βii}u_{βjj} − ∑_{β=1}^{n} u^{ii}u^{jj}u_{βij}^{2}\} .  (41)

For each fixed β, we wish to compute \frac{1}{n−1} u^{ii}u^{jj}u_{βii}u_{βjj} − u^{ii}u^{jj}u_{βij}^{2}. Let I_{β} = \{1 ≤ i ≤ n | u_{βii} > 0}, J_{β} = {1 ≤ i ≤ n | u_{βii} < 0}. We divide into two cases:

(i) |I_{β}| ≤ n − 1, |J_{β}| ≤ n − 1 (ii) I_{β} ∪ J_{β} = {1, 2, …, }, one of I_{β}, J_{β} is empty.

In case (i)

\frac{1}{n − 1} u^{ii}u^{jj}u_{βii}u_{βjj} − u^{ii}u^{jj}u_{βij}^{2} ≤ \frac{1}{n − 1} (∑_{i=1}^{n} u^{ii}u_{βii})^{2} − ∑_{i=1}^{n}(u^{ii}u_{βii})^{2}

≤ \frac{1}{n − 1} ( ∑_{i∈I_{β}} u^{ii}u_{βii})^{2} + \frac{1}{n − 1} ( ∑_{i∈J_{β}} u^{ii}u_{βii})^{2} − ∑_{i∈I_{β}}(u^{ii}u_{βii})^{2} − ∑_{i∈J_{β}}(u^{ii}u_{βii})^{2}
≤ 0, since |I_{β}| ≤ n − 1, |J_{β}| ≤ n − 1 . (42)

In case (ii), we may assume J_{β} = ϕ. That is at p, u_{βii} > 0, i = 1, 2, …, n. By (33),

u_{β11} + u_{β22} + ⋯ + u_{βnn} = −αx_{β}H,

|u_{βii}| ≤ −αx_{β}H .  (43)

We deduce

\frac{1}{n − 1} u^{ii}u^{jj}u_{βii}u_{βjj} − u^{ii}u^{jj}u_{βij}^{2} ≤ \frac{1}{n − 1} (u^{nn}u_{βnn} + ∑_{i=1}^{n−1} u^{ii}u_{βii})^{2} − ∑_{i=1}^{n}(u^{ii}u_{βii})^{2}

= \frac{1}{n − 1}(u^{nn}u_{βnn})^{2} + \frac{1}{n − 1} (∑_{i=1}^{n} u^{ii}u_{βii})^{2} − ∑_{i=1}^{n}(u^{ii}u_{βii})^{2} + \frac{2}{n − 1} u^{nn}u_{βnn} ∑_{i=1}^{n−1} u^{ii}u_{βii}
≤ \frac{2}{n − 1} u^{nn}u_{βnn} ∑_{i=1}^{n−1} u^{ii}u_{βii}
≤ \frac{2α^{2}}{n − 1} u^{nn}|x|^{2}H^{2} ∑_{i=1}^{n−1} u^{ii}
≤ \frac{2α^{2}|x|^{2}}{(n − 1)u_{nn}} H ∑_{i=1}^{n} u^{ii}
≤ \frac{2nα^{2}|x|^{2}}{n − 1} H ∑_{i=1}^{n} u^{ii}
≤ \frac{2n}{n − 1} α^{2} diam^{2}(Ω)H ∑_{i=1}^{n} u^{ii}  (44)

(since H = ∑ u_{ii}, u_{nn} ≥ u_{n−1;n−1} ≥ ⋯ ≥ u_{11}). Put (42) and (44) into (41),

f^{−2} \{fΔf − \frac{n − 2}{n − 1} |∇f|^{2}\} ≤ \{L(H) + \frac{2n^{2}}{n − 1} α^{2} diam^{2}(Ω) ∑_{i=1}^{n} u^{ii}H\} .  (45)

In view of (35), we have at p

f^{−2} \{fΔf − \frac{n − 2}{n − 1} |∇f|^{2}\} ≤ −α\{1 − (2n^{2} + 1)α diam^{2}(Ω)\} (∑_{i=1}^{n} u^{ii}) H .  (46)

While at p

∑_{i=1}^{n} u^{ii} ≥ ∑_{i=1}^{n−1} u^{ii}

≥ (n − 1) ( ∏_{i=1}^{n−1} u^{ii})^{\frac{1}{n−1}}

= \frac{(∏_{i=1}^{n−1} u^{ii})^{\frac{1}{n−1}}}{(u^{nn})^{\frac{1}{n−1}}}
= (n − 1)f^{− \frac{1}{n−1}} unn^{\frac{1}{n−1}}
≥ \frac{n − 1}{n^{\frac{1}{n−1}}} f^{− \frac{1}{n−1}} H^{\frac{1}{n−1}} .  (47)

Insert (47) into (46), we get at p, by Condition (C′),

\frac{n − 1}{n^{\frac{1}{n−1}}} α\{1 − (2n^{2} + 1)α diam^{2}(Ω)\}H^{1+ \frac{1}{n−1}} ≤ f^{−2+ \frac{1}{n−1}} \{fΔf − \frac{n − 2}{n − 1} |∇f|^{2}\}
≤ \frac{A}{n − 1} .  (48)

If we pick α = \frac{1}{2(2n^{2}+1)diam^{2}(Ω)}, we get the desired estimate (31). The proof of the lemma is complete.

Proposition 12 If Ω ⊂ R^{n} is a bounded strictly convex domain, with ∂Ω ∈ C^{3}. If f > 0, f ∈ C^{3}(Ω̄), f satisfies Condition (C) in Ω, then the equation (1) with homogeneous boundary condition has an unique convex solution u, such that

∥u∥_{C^{2}(Ω̄)} ≤ C ,  (49)

where C depends only on A, δ, Ω, sup(f) and modulus continuity of f in Ω.

Proof. Since f > 0, by the result [CNS], there is an unique convex solution u ∈ C^{4}(Ω). By Lemmas 10 and 11,

∥u∥_{C^{2}(Ω̄)} ≤ C

with C depending only on A, δ, Ω, sup(f) and \frac{1}{max_{Ω} |u|}. If we can show max_{Ω} |u| > B > 0, with B depending only on Ω, sup(f) and modulus continuity of f in Ω. The proposition will follow.

Let m = sup_{Ω} f. By the continuity of f, ∃ x_{0} ∈ Ω, R > 0, such that B_{R}(x_{0}) ⊂ Ω,

f(x) ≥ \frac{m}{2} ∀ x ∈ B_{R}(x_{0}). Let f̃(x) = \{ \begin{matrix}(\frac{m}{2})^{\frac{1}{n}} (1 − \frac{|x−x_{0}|}{R}) , & x ∈ B_{R}(x_{0}), \\ 0, & x ∈ Ω\B_{R}(x_{0}) \end{matrix} Let h̃ be the

solution of the equation

\{ \begin{matrix}Δh̃ = f̃ \\ h̃|_{∂Ω} = 0 . \end{matrix}  (50)

Since Δu ≥ nf^{\frac{1}{n}}. We have

Δ(u − h̃) ≥ 0 in Ω , (51)

and

u − h̃ = 0 on ∂Ω . (52)

By the maximum principle u ≤ h̃, u ≤ 0, h̃ ≤ 0. That is, |u| ≥ |h̃|. So that, max_{Ω} |u| ≥ max_{Ω} |h̃| ≥ B with B depends only on Ω, m, and R.

Now, we are ready to prove Theorem 1. Proof of Theorem 1. Suppose f ≥ 0, f satisfies Condition (C). For any ε > 0, let g_{ε} = f^{\frac{1}{n−1}} + ε. Let f_{ε} = g_{ε}^{n−1}. We may assume f_{ε} ∈ C^{3}(Ω̄). Othewise, we can using following smoothing process:

∀ ρ > 0, define g̃_{ε,ρ} = g_{ε} ⋆ ϕ_{ρ}, with ϕ_{ρ}(x) = ϕ (\frac{|x|}{ρ}), ϕ(A) ∈ C_{0}^{∞}(R), ϕ(A) ≥ 0, ∫_{R} ϕ(A) = 1. Let f_{ε,ρ} = g̃_{ε,ρ}^{n−1}. We claim: (i)

(n − 1)f_{ε,ρ}Δf_{ε,ρ} − (n − 2)|∇f_{ε,ρ}| ≥ −Af_{ε,ρ}^{2− \frac{1}{n−1}}  (53)

(the same constant A in Condition (C) for f.) (ii) for ρ small,

|∇f_{ε,ρ}(x)| ≤ Af_{ε,ρ}^{1− \frac{1}{n−1}} (x)  (54)

for x ∈ \{x ∈ Ω | dist(x, ∂Ω) ≤ \frac{δ}{2}\}.

First, (i) is equivalent to Δg̃_{ε,ρ} = Δfε,ρ^{\frac{1}{n−1}} ≥ −\frac{A}{n−1}. Since Δg_{ε} = Δf^{\frac{1}{n−1}} ≥ −\frac{A}{n−1}, Δg̃_{ε,ρ} = Δ(g_{ε} ⋆ ϕ_{ρ}) = (Δg_{ϵ}) ⋆ ϕ_{ρ} ≥ −\frac{A}{n−1} ⋆ ϕ_{ρ} ≥ −\frac{A}{n−1}.

Secondly, (ii) is equivalent to |∇fε,ρ^{\frac{1}{n−1}} | ≤ A for x ∈ \{x ∈ Ω | dist(x, Ω) ≤ \frac{δ}{2}\}. Again,

|∇fε,ρ^{\frac{1}{n−1}} (x)| = |∇g_{ε,ρ}(x)| = |∇g_{ε} ⋆ ϕ_{ρ}(x)| ≤ A ⋆ ϕ_{ρ} = A,

for ρ ≤ \frac{δ}{2}, x ∈ \{x ∈ Ω | dist(x, Ω) ≤ \frac{δ}{2}\}.

Now, we have f_{ε,ρ} ∈ C^{3}(Ω̄), f_{ε,ρ} satisfies Condition (C) uniformly ∀ε > 0, 0 < ρ ≤ \frac{δ}{2}. By the result in [CNS] and proposition 15, there are u_{ε,ρ} ∈ C^{4}(Ω), with

∥u_{ε,ρ}∥_{C^{2}(Ω̄)} ≤ C

where C depends only on A, δ, Ω, sup(f) and modulus continuity of f in Ω, i.e., C independent of ε, ρ. First, we let ρ → 0, by convexity of u_{ε,ρ}, passing to a subsequence if necessary, we obtain a convex solution u_{ε}, with

\{ \begin{matrix}det(u_{ε}) = f_{ε}  in Ω \\ u_{ε}|_{∂Ω} = 0 \end{matrix}  (55)

and ∥u_{ε}∥_{C^{2}(Ω̄)} ≤ C. Now, we let ε → 0, the same argument gives a convex solution u, with ∥u∥_{C^{2}(Ω̄)} ≤ C. Finally, the uniqueness follows from the maximum principle (e.g. see [CKNS]).

Proof of Theorem 7. In [CKNS], they obtained that, if f > 0 near ∂Ω, then |∇^{2}u| ≤ C on the boundary for convex solution u satisfying

\{ \begin{matrix}det(u_{ij}) = f  in Ω \\ u|_{∂Ω} = ϕ , \end{matrix}  (56)

where C depends only on Ω, ∥ϕ∥_{C^{4}(∂Ω)}, sup_{Ω} f, inf(f) near the boundary. Combining this estimate with lemma 14, we get a priori estimate ∥u∥_{C^{2}(Ω̄)} ≤ C for f satisfying Condition (C′). Then, Theorem 7 can be proved using the same argument as in the proof of Theorem 1.

4 The Equation of Prescribed Gauss Curvature

We would like to extend our estimates in the previous section to the equation of prescribed Gauss curvature. Let Ω be a domain in R^{n} and u a function in C^{2}(Ω) with graph S ⊂ R^{n+1}. The Gauss curvature of S at a point (x, z) = (x, u(x)) ∈ S is given by the formula

K(x, z) = \frac{det(u_{ij}(x))}{(1 + |∇u|^{2})^{\frac{n+2}{2}}} .  (57)

The prescribing Gauss curvature problem is: given K in Ω, find u, such that the hypersurface S = (x, u(x)) has the prescribed Gauss curvature K(x) at (x, u(s)). If we prescribe the boundary of S, then the problem is equivalent to the Dirichlet problem for equation (57).

In [TU1], Trudinger-Urbas obtained a necessary and sufficient condition for the existence of a solution u ∈ C^{2}(Ω) ∩ C^{0,1}(Ω̄) for equation (57) with arbitrary u = ϕ on ∂Ω. Assume that Ω is a bounded strictly convex C^{1,1} domain in R^{n}, and K ∈ C^{1,1}(Ω) ∩ C^{0,1}(Ω̄), K(x) > 0, ∀x ∈ Ω. Their necessary and sufficient conditions are:

\{ \begin{matrix}∫_{Ω} K < w_{n} \\ K = 0 on ∂Ω \end{matrix}  (58)

where w_{n} is the volume of the unit ball in R^{n}.

In the same paper [TU1], they proved C^{1,1}(Ω̄) regularity of solution u of the equation

\frac{det(u_{ij})}{(1 + |∇u|^{2})^{\frac{n+2}{2}}} = K(x)  in Ω,  (59)

with homogeneous Dirichlet boundary condition, under the assumptions (in addition to (58)): ∂Ω ∈ C^{2,1}, K = g^{n} for some positive g ∈ W^{2,n}(Ω).

On the other hand, Ivochkina proved in [I] that there is a smooth convex solution u of (59) with homogeneous boundary condition if K positive, and

K(x) < λ,  (60)

for x in Ω, where λ is the minimal principal curvature of the boundary.

By modifying the arguments in the previous section, we shall show

Theorem 13 Suppose Ω is a bounded strictly convex domain in R^{n} with ∂Ω ∈ C^{2,1}. Let K be a nonnegative function in Ω, suppose K satisfies Condition (C) with δ = diam(Ω). If in additon, K satisfies either (58) or (60). Then the Dirichlet problem of equation (59) with u|_{∂Ω} = 0 has an unique convex C^{1,1}(Ω̄) solution u, with

∥u∥_{C^{2}(Ω)} ≤ C

where C depends only on Ω, A and sup(f).

Corollary 14 Suppose K satisfies either (58) or (60). If K = ∑_{j=1}^{N} g_{j}^{m_{j}} with g_{j}^{m_{j}} ≥ 0 for some m_{j} ≥ n − 1, g_{j} ∈ C^{1,1}(Ω̄), then equation (59) with the homogeneous boundary condition has an unique C^{1,1} convex solution. In particular, when n = 2, K ∈ C^{1,1}(Ω̄), and when n = 3, K ∈ C^{3,1}(R^{3}), and if K satitsfies (58), then equation (59) with homogeneous Dirichlet boundary condition has an unique C^{1,1} convex solution.

Corollary 14 is an easy consequence of Theorem 13 and Lemmas 2 and 3. To prove Theorem 13, we only need to obtain C^{2} a priori estimates for u. C^{0} and C^{1} estimates have been established in [TU1] and [I] respectively. We now indicate how to obtain a bound for |∇^{2}u|.

First, we establish a bound of |∇^{2}u| on ∂Ω. As in Section 2, let p ∈ ∂Ω, we may assume p = 0. Choose coordinates x_{1}, …, x_{2} near the origin such that r_{i}(0) = 0 ∀ i ≤ n − 1, and r_{n}(0) = −1, where it is the defining function of Ω. We can get, as in Section 2

|u_{ij}(0)| ≤ C  for i, j ≤ n − 1 .  (61)

If we can bound |u_{in}(0)| for i ≤ n − 1, a bound for |u_{nn}(0)| can be obtained in the same way as in Section 2.

To bound |u_{in}(0)| for i ≤ n − 1, we let S_{ε}, S_{ε}^{1}, S_{ε}^{2} and w be the same as defined in Section 2. We define

L̃ = u^{ij} \frac{∂^{2}}{∂x_{i}∂x_{j}} − \frac{n + 2}{1 + |∇u|^{2}} ∑_{i=1}^{n} u_{i} \frac{∂}{∂x_{i}} .

Following the same argument as in Section 2, with a suitable choice of ε, B, B̃, we can show that w|_{∂S_{ε}} ≤ 0, and

L̃ w(x) > 0  in S_{ε}^{1} .  (62)

Similarly, we prove x cannot be a critical point of w if x ∈ S_{ε}^{2}. This will give

max_{x∈S_{ε}} w = w(0) = 0 .

¿From this, we conclude that |u_{in}(0)| ≤ C for i ≤ n − 1.

As for a bound of |∇^{2}u| at the interior, we have

Lemma 15 Suppose f > 0, f satisfies Condition (C) with δ = diam(Ω). Suppose u ∈ C^{4}(Ω) ∩ C^{2}(Ω̄) is convex, u satisfies equation (62) in Ω. Then

sup_{Ω} |∇^{2}u| ≤ C (1 + sup_{∂Ω} |∇^{2}u|)

where C depends only on A, sup_{Ω} |∇u| and diam(Ω).

Proof. We shall redo the proof of Lemma 11. Let H(x) = Δu(x), G(x) = H(x) e^{α|x|^{2}/2}, α = \frac{1}{2(2n^{2}+1)diam^{2}(Ω)}. We only need to bound G in Ω. We may assume 0 ∈ Ω, and

max_{Ω} G(x) = G(p) for some p ∈ Ω. Let f(x) = K(x)(1 + |∇u|^{2})^{\frac{n+2}{2}} . By the computations in Section 3, at p

\frac{n − 1}{n^{\frac{1}{n−1}}} α\{1 − (2n^{2} + 1)α diam^{2}(Ω)\}H^{1+ \frac{1}{n−1}}

≤ −f^{−2+ \frac{1}{n−1}} \{fΔf − \frac{n − 2}{n − 1} |∇f|^{2}\} .  (63)

We may assume (u_{ij}) is diagonal at p. At p, using the facts ∇G = 0, u_{αβ} = 0 for α ≠ β, we have

fΔf − \frac{n − 2}{n − 1} |∇f|^{2} = (1 + |∇u|^{2})^{n+2}\{KΔK − \frac{n − 2}{n − 1} |∇K|^{2}\}

+ \frac{2(n + 2)}{n − 1} (∑_{α=1}^{n} K_{α}u_{α}u_{αα}) K(1 + |∇u|^{2})^{n+1}

+(n + 2)K^{2}{(1 + |∇u|^{2})^{n+1} ∑_{α=1}^{n} u_{αα}^{2}

+(1 + |∇u|^{2})^{n} (n − \frac{(n + 2)(n − 2)}{n − 1}) ∑_{α=1}^{n} u_{α}^{2}u_{αα}^{2}\}

≥ (1 + |∇u|^{2})^{n+2} \{KΔK − \frac{n − 2}{n − 1} |∇K|^{2}\}

+\frac{2(n + 2)}{n − 1} (∑_{α=1}^{n} K_{α}u_{α}u_{αα}) K(1 + |∇u|^{2})^{n+1}

+(n + 2)K^{2}(1 + |∇u|^{2})^{n} (n + 1 − \frac{(n + 2)(n − 2)}{n − 1}) ∑_{α=1}^{n} u_{α}^{2}u_{αα}^{2}

≥ (1 + |∇u|^{2})^{n+2} \{KΔK − \frac{n − 2}{n − 1} |∇K|^{2}\}
+\frac{2(n + 2)}{n − 1} (∑_{α=1}^{n} K_{α}u_{α}u_{αα}) K(1 + |∇u|^{2})^{n+1} .  (64)

Put (64) into (63), using Condition (C) with δ = diam(Ω),

\frac{n − 1}{n^{\frac{1}{n−1}}} α\{1 − (2n^{2} + 1)α diam^{2}(Ω)\}H^{1+ \frac{1}{n−1}}

≤ C_{2} K^{−1+ \frac{1}{n−1}} |∇K|H − C_{1} K^{−2+ \frac{1}{n−1}} \{KΔK − \frac{n − 2}{n − 1}|∇K|^{2}\}
≤ C_{1}A + C_{2}H  (65)

where C_{1}, C_{2} depends only on sup_{Ω} |∇u| and n.

¿From (65), we get

\frac{n − 1}{n^{\frac{1}{n−1}}} α\{1 − (2n^{2} + 1)α diam^{2}(Ω)\}H (H^{\frac{1}{n−1}} − C_{2}) ≤ C_{1}A .  (66)

We conclude at p,

H ≤ (C_{2} + 1)^{n−1} + \frac{2C_{1}An^{\frac{1}{n−1}}}{(n − 1)α} .  (67)

So that,

max_{x∈Ω} G(x) ≤ C

where C depends only on A, sup_{Ω} |∇u| and diam(Ω).

5 Proof of the Lemmas in Section 1

As we already notice that Lemma 3 is a consequence of Lemma 4 by C. Fefferman for n = 3 (n = 2 case is trivial). In [GL2], we give a proof of Lemma 3 using a weaker version of Lemma 4. Since Lemma 4 is very useful, we produce a proof here. The proof follows the main steps suggested by C. Fefferman. We want to thank him for communicating the lemma together with the proof to us.

First, we prove Lemma 2.

Lemma 16 If g_{1} and g_{2} are pseudo-subharmonic in Ω, then the functions g_{1} + g_{2} and Max(g_{1}, g_{2}) are also pseudo-subharmonic. In particular, |g| is pseudo-subharmonic if g ∈ C^{1,1}(Ω̄). If f_{1} and f_{2} are two nonnegative functions in Ω, suppose f_{1} and f_{2} satisfies Condition (C) with constants A_{1} and A_{2} respectively. Then, the function f = f_{1} + f_{2} satisfies Condition (C) with the constant A = A_{1} + A_{2}.

Proof. The first assertion of the lemma follows easily from well-known properties of subharmonic functions. Since |g| = Max(g, 0) + Max(−g, 0), therefore, |g| is pseudo-subharmonic if g ∈ C^{1,1}(Ω̄). We now prove the last assertion of the lemma. Using smoothing process (as in the proof of Theorem 1) if necessary, without loss of generality, we may assume f_{1} and

f_{2} smooth and positive. First, if |∇f_{1}| ≤ A_{1}f_{1}^{1− \frac{1}{n−1}} , |∇f_{2}| ≤ A_{2}f_{2}^{1− \frac{1}{n−1}} , then

|∇f| ≤ |∇f_{1}| + |∇f_{2}| ≤ A_{1}f_{1}^{1− \frac{1}{n−1}} + A_{2}f_{2}^{1− \frac{1}{n−1}} ≤ (A_{1} + A_{2})f^{1− \frac{1}{n−1}} .

Suppose (n − 1)f_{j}(x)Δf_{j}(x) − (n − 2)|∇f_{j}(x)|^{2} ≥ −A_{j}f^{2− \frac{1}{n−1}} (x), ∀ x ∈ Ω, j = 1, 2. Fix x ∈ Ω, we have

(n − 1)fΔf − (n − 2)|∇f|^{2} = (n − 1) ∑_{i,j}^{2} f_{j}Δf_{i} − (n − 2) ∑_{i,j}^{2} ∇f_{i} · ∇f_{j}

= ∑_{i,j}{(n − 1)f_{j}Δf_{i} − (n − 2)∇f_{i} · ∇f_{j}}

= ∑_{i,j} \{(n − 1)\frac{f_{j}Δf_{i} + f_{i}Δf_{j}}{2} − (n − 2)∇f_{i} · ∇f_{j}\}

≥ ∑_{i,j} \{(n − 1)\frac{f_{j}Δf_{i} + f_{i}Δf_{j}}{2} − (n − 2)f_{i}^{−1}f_{j}^{−1} \frac{f_{j}^{2}|∇f_{i}|^{2} + f_{i}^{2}|∇f_{j}|^{2}}{2} \}
= \frac{1}{2} ∑_{i,j}(\{(n − 1)f_{j}Δf_{i} − (n − 2)f_{j}f_{i}^{−1}|∇f_{i}|^{2}\} + \{(n − 1)f_{i}Δf_{j} − (n − 2)f_{i}f_{j}^{−1}|∇f_{j}|^{2}\})
= \frac{1}{2} ∑_{i,j}(f_{i}^{−1}f_{j}\{(n − 1)f_{j}Δf_{i} − (n − 2)|∇f_{i}|^{2}\} + f_{j}^{−1}f_{i}\{(n − 1)f_{i}Δf_{j} − (n − 2)|∇f_{j}|^{2}\})
≥ − \frac{1}{2} ∑_{i,j} A_{i}f_{i}^{1− \frac{1}{n−1}} f_{j} − \frac{1}{2} ∑_{i,j} A_{j}f_{j}^{1− \frac{1}{n−1}} f_{i}

= −f ∑_{i=1}^{2} A_{i}f_{i}^{1− \frac{1}{n−1}}

≥ −f ∑_{i=1}^{2} A_{i}f^{1− \frac{1}{n−1}}
= −A f^{2− \frac{1}{n−1}}

Proof of Lemma 2. Suppose f = |h|^{m} with m ≥ n − 1, h ∈ C^{1,1}(Ω̄), first we note that g = |h| is Lipschitz continuous, and by Lemma 16, Δg ≥ −B for some constant B. Now,

(n − 1)fΔf − (n − 2)|∇f|^{2}
= (n − 1)g^{m}{mg^{m−1}Δg + m(m − 1)g^{m−2}|∇g|^{2}} − (n − 2)m^{2}g^{2m−2}|∇g|^{2}
= m(n − 1)g^{2m−1}Δg + m(m − n + 1)g^{2m−2}|∇g|^{2}
≥ m(n − 1) g^{2m−1}Δg
≥ −m(n − 1)Bg^{2m−1}
≥ −A f^{2− \frac{1}{n−1}} .

And since f^{\frac{1}{n−1}} = |h|^{\frac{m}{n−1}} , m ≥ n − 1, f^{\frac{1}{n−1}} is Lipschitz continuous.

Now, Lemma 2 follows directly from Lemma 16.

Finally, we prove Lemma 4.

Proof of Lemma 4. We will use induction on dimension n. We may assume ∥f∥_{C^{4}(R^{n})} ≤ 1. We make a Calderón-Zygmund decomposition of R^{n} into cubes {Q_{ν}} of diameters δ_{ν}. Stopping at Q_{ν} when

inf_{x∈Q_{ν}} (∑_{k=0}^{4} δ_{ν}^{k−4}|∇^{k}f(x)|) ≥ 4 .  (68)

We first note that δ_{ν} ≤ 1 by the assumption ∥f∥_{4} ≤ 1.

Observation. If denote 2Q_{ν} be the cube of diameter 2δ_{ν}, with the same center at Q_{ν}; and denote that Q_{ν}^{∗} be the cube of diameter (1 + \frac{1}{10^{4}(n+1)^{4}} )δ_{ν}, with the same center as Q_{ν}, then

(i)

∑_{k=0}^{4} δ_{ν}^{k−4}|∇^{k}f(x)| ≤ 5^{6}(n + 1)^{4}  ∀ x ∈ 2Q_{ν};  (69)

(ii)

inf_{x∈Q_{ν}^{∗}} (∑ δ_{ν}^{k−4}|∇^{k}f(x)|) ≥ \frac{1}{10^{5}(n + 1)^{4}}  (70)

Proof of Observation. Let Q̃_{ν} be the cube step before we get Q_{ν}. Then, Q̃_{ν} ⊃ Q_{ν}, and diameter of Q̃_{ν} = 2δ_{ν}. Since we didn’t stop at Q̃_{ν}, there is x_{0} ∈ \overline{Q̃_{ν}} such that ∑_{k=0}^{4}(2δ_{ν})^{k−4}|∇^{k}f(x_{0})| ≤ 4. That is

|∇^{k}f(x_{0})| ≤ 4(2δ_{ν})^{4−k},  k = 0, 1, 2, 3, 4 .  (71)

Use the fact |∇^{4}f(x)| ≤ 1, ∀ x ∈ R^{n}, by the Fundamental Theorem of Calculus, we get

|∇^{3}f(x)| ≤ 4nδ_{ν} + 8δ_{ν}  (72)

for x ∈ 2Q_{ν} (since dist(x, x_{0}) ≤ 4δ_{ν}).

Using (71) and (72), we get

|∇^{2}f(x)| ≤ 4nδ_{ν}(4nδ_{ν} + 8δ_{ν}) + 4(2δ_{ν})^{2}, ∀ x ∈ 2Q_{ν} .  (73)

So on, going backwards, we can easily get (69). While (70) is a simple consequence of (68) and (69), again by the Fundamental Theorem of Calculus.

Now, let Q_{ν}^{+} be the cube of diameter (1 + \frac{1}{10^{10}(n+1)^{4}} )δ_{ν}, with the same center as Q_{ν}. For x ∈ Q_{ν}^{+}, since f ≥ 0 in R^{n}, by Taylor expansion using the bounds in (69) and (70), we have, at x, either:

(i)

f(x) ≥ c̃_{n}δ_{ν}^{4}  (74)

or

(ii)

|∇^{2}f(x)| ≥ c̃_{n}δ_{ν}^{2} ,  (75)

where c̃_{n} > 0 is a universal constant depending only on n.

Definition: Let g be a function in a cube Q with diameter δ, we say ∥g∥_{C^{k}(Q)} is under control if ∑_{i=1}^{k} δ^{i−k}|∇^{i}g(x)| ≤ C for some constant C independent of δ.

In case (i), again by (69), there is a universal constant c′_{n} > 0, such that

\frac{c′_{n}}{2} δ_{ν}^{4} ≤ f(x) ≤ 5^{6}(n + 1)^{4}δ_{ν}^{4}  ∀ x ∈ Q′_{ν,x} .  (76)

where Q′_{ν,x} is cube centered at x, of diameter c′_{n}δ_{ν}. Let g(x) = \sqrt{f(x)}, then by (70) and (76), g(x) ∈ C^{1,1}(Q̄′_{ν,x}) with ∥g∥_{C^{2}} under control. So, f = g^{2} in Q′_{ν,x}.

In case (ii), we claim: there is a universal constant c′′_{n} > 0, such that

f(x) = g_{1}^{2}(x) + g̃_{2}(x′)  (77)

for x ∈ Q′′_{ν,x}, x′ is a (n − 1)-variable, where Q′′_{ν,x} is cube centered at x, of diameter c′′_{n}δ_{ν}, g_{1} ∈ C^{1,1}(Q̄′′_{ν,x}), g̃_{x}(x′) ≥ 0, g̃_{2} ∈ C^{3,1}(Q̄′′_{ν,x}), with ∥g_{1}∥_{C^{2}}, ∥g_{2}∥_{C^{4}} under control. The claim follows from

Assertion: Let f(x) = c + ∑ α_{k}x_{k} + ∑_{j,k} μ_{jk}x_{j}x_{k} + h(x) be n nonnegative C^{3,1} function in a unit cube A, satisfying (a) |μ_{11}| ≥ 1,

(b) |μ_{ik}| ≤ 10|μ_{11}|

(c) ∥g∥_{C^{4}(Q)} ≤ 4

(d) |∇^{2}g(0)| ≤ 10^{−6}n^{−2}|μ_{n}|

Then, in a universal dilate of A, either f ≥ c(n) or else

f(x, x′) = g_{1}^{2}(x) + g̃_{2}(x′)

where ∥f_{1}∥_{C^{2}}, ∥g̃_{2}∥_{C^{4}} are under control and g̃_{2} ≥ 0.

The proof of the assertion follows exact same lines of the proof of the assertion after Lemma 3.3 in [FP]. We won’t repeat it here.

Now, if n = 1, g̃_{2}(x′) ≡ constant ≥ 0, we can write

f = g_{1}^{2} + g_{2}^{2}

in Q_{ν,x}, for x ∈ Q_{ν}^{+}, where Q_{ν,x} is a cube centered at x, of diameter cδ, with universal constant c > 0 with ∥g_{1}∥_{C^{2}}, ∥g_{2}∥_{C^{2}} under control. If n ≥ 2, by induction, we can also write

f = ∑_{j=1}^{Ñ(n)} g_{j}^{2}  in Q_{ν,x}  (78)

with ∥g_{j}∥_{C^{2}(Q̄_{ν,x})} under control. If we pick a (finite) partition of unit for Q̄_{ν}^{∗}, we can write

f = ∑_{j=1}^{N(n)} g_{ν,j}^{2}  (79)

in Q_{ν}^{+}.

Now {f(x) > 0} ⊂ ⋃_{ν} Q_{ν}^{+}. For each Q_{ν}, with δ_{ν}, if Q̄_{ν′} ∩ Q̄_{ν} ≠ ϕ, by (70),

δ_{ν′} ≥ \frac{δ_{ν}}{(10^{5}(n + 1)^{4})^{4}} ,  (80)

and by (69)

δ_{ν} ≤ 10^{10}(n + 1)^{16}δ_{ν} .  (81)

Therefore, we conclude, for each Q_{ν}^{+}, there are only finite (universal) number of Q_{ν′}^{+} intersecting with it, and for each of them the diameter is comparable to δ_{ν}. Let ρ̃(x) ∈ C_{0}^{∞}(R^{n}), ρ̃(x) ≥ 0, ρ̃(x) = 1 in the unit cube Q, ρ̃(x) = 0 outside of the cube (1 + \frac{1}{10^{10}(n+1)^{4}} )Q.

Define ρ̃_{ν}(x) = ρ̃(\frac{x−x_{ν}}{δ_{ν}}). Let ρ_{ν}^{2} = \frac{ρ̃_{ν}^{2}}{(∑_{ν} ρ̃_{ν}^{2})}. We produce a partition of unity \{ρ_{ν}^{2}\} of ⋃_{ν} Q_{ν}^{∗}, such that

∑ ρ_{ν}^{2} = 1  (82)

and suppρ_{ν} ⊂ Q_{ν}^{+} ⊂ Q_{ν}^{∗}.

Now, we want to group Q_{ν}^{+}, Let A_{j} be the collection of cubes Q_{ν} with δ_{j} = \frac{1}{2^{j}}. By (69) and (70), ∃ m > 0 ( m = 10^{30}(n + 1)^{32} will be suffice), if Q_{ν_{1}} ∈ A_{j_{1}}, Q_{ν_{2}} ∈ A_{j_{2}}, with |j_{1} − j_{2}| ≥ m, then

Q_{ν_{1}}^{∗} ∩ Q_{ν_{2}}^{∗} = ϕ .  (83)

There is M > 0 ( M = 2^{n} will be suffice), we can write, ∀ j

A_{j} = A′_{j} ∪ A_{j}^{2} ∪ ⋯ ∪ A_{j}^{M}

such that if Q_{ν_{1}}, Q_{ν_{2}} ∈ A_{j}^{i_{1}}, if Q_{ν_{1}} ≠ Q_{ν_{2}}, then

Q_{ν_{1}}^{∗} ∩ Q_{ν_{2}}^{∗} = ϕ .  (84)

Now, let

B_{1}^{k} = ⋃_{j=1}^{∞} A_{1+(m+1)j}^{k}, …, B_{m}^{k} = ⋃_{j=1}^{∞} A_{m+(m+1)j}^{k}, k = 1, 2, …, M .

We have, by (87) and (88), ∀ Q_{ν_{1}}, Q_{ν_{2}} ∈ B_{i}^{k}, if Q_{v_{1}} ≠ Q_{ν_{2}}, then

Q_{ν_{1}}^{∗} ∩ Q_{ν_{2}}^{∗} = ϕ .  (85)

Therefore, we have by (79) and (85)

f(x) = ∑_{ν} ρ_{ν}^{2}f(x) = ∑_{i=1}^{m} ∑_{k=1}^{M} ( ∑_{Q_{ν}∈B_{i}^{k}} ρ_{ν}^{2}f(x))

= ∑_{i=1}^{m} ∑_{k=1}^{M} ( ∑_{Q_{ν}∈B_{i}^{k}} ρ_{ν}^{2} ∑_{j=1}^{N} g_{ν,j}^{2})

= ∑_{i=1}^{m} ∑_{k=1}^{M} ∑_{j=1}^{N} ( ∑_{Q_{ν}∈B_{i}^{k}}(ρ_{ν}g_{ν,j})^{2})

= ∑_{i=1}^{m} ∑_{k=1}^{M} ∑_{j=1}^{N} ( ∑_{Q_{ν}∈B_{i}^{k}} ρ_{ν}g_{ν,j})^{2} .

Clearly, ρ_{ν}g_{ν,j} ∈ C^{1,1}(R^{n}) ∀ ν, j. The proof of Lemma 4 is complete.

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