93(2013) 431-469
BERNSTEIN THEOREM AND REGULARITY FOR A CLASS OF MONGE-AMP`ERE EQUATIONS
Huaiyu Jian & Xu-Jia Wang
Abstract
In this paper we first introduce a transform for convex func- tions and use it to prove a Bernstein theorem for a Monge-Amp`ere equation in half space. We then prove the optimal global regularity for a class of Monge-Amp`ere type equations arising in a number of geometric problems such as Poincar´e metrics, hyperbolic affine spheres, and Minkowski type problems.
1. Introduction
In this paper we introduce a transform for convex functions, prove a Bernstein theorem for a Monge-Amp`ere equation in half space, and establish the global regularity of solutions to a class of Monge-Amp`ere equations by new techniques. A well-known transform for convex func- tions is the Legendre transform, given by
(1.1) y = Du(x),
u∗(y) = x·y−u(x).
The Legendre transform is very useful in the study of Monge-Amp`ere type equations and optimal transportation, and in the theory of convex bodies.
In this paper we introduce the following new transform for convex functions:
y=x/u(x), u#(y) = 1/u(x), (1.2)
whereuis defined in a domain Ω⊂Rn. This transform is closely related to the concept of polar set in the theory of convex bodies; a geometric interpretation of it will be given in Section 2. See in particular Lemma 2.1 for its relation to the Legendre transform (1.1). Some properties of the transform are summarized in the following theorem.
Received 12/22/2011.
431
Theorem 1.1. Let u be a convex function. Then (a) u# is convex if u∗/u >0, and concave if u∗/u <0.
(b) (u∗)#= (u#)∗.
(c) Denote Q[u] =|u|n+2detD2u, and w= (u#)∗. Then
(1.3) Q[u] = 1/Q[w]
if u# is convex, and Q[u] = 1/|Q[w]|if u# is concave.
The transform u → u# has some other properties that should be useful in studying Monge-Amp`ere type equations. For example, if u is a convex function satisfying u(0) = |Du(0)| = 0 and |Du(x)| → ∞ as
|x| → ∞, then the transform (1.2) sendsx= 0 toy=∞ andx=∞to y= 0. This property is shared by the Kelvin transform
(1.4) y=ψ(x) :=x/|x|2,
v(y) =J(n−2)/2nu(x),
where J is the Jacobian of the mapping ψ. Another property shared by the transforms (1.2) and (1.4) is the invariance of certain quantities, that is,Q[u] for (1.2) and|u|−n+2n−2∆ufor (1.4). Moreover,Q[u] is closely related to the Blaschke-Santalo inequality, just like|u|−n+2n−2∆uis related to the Sobolev inequality.
The Kelvin transform is very useful in the study of semilinear elliptic equations involving critical Sobolev exponents. It is natural to use the transform (1.2) to study the equation
(1.5) detD2u= η(x)
|u|n+2 in Ω,
where Ω is a convex domain inRn, andη∈C(Ω) is a positive function.
Equation (1.5) arises in several geometric problems such as the Hilbert metric (Poincar´e metric) in convex domains [24], affine spheres [6,7], the p-Minkowski problem [25], and the Minkowski problem in centro- affine geometry [10]. A parabolic version of equation (1.5) has been used in image processing. In Section 2 we will see that equation (1.5) is related to the Euler equation of the well-known Blaschke-Santalo inequality [26,30].
A special case of (1.5) is whenη≡1, which is the equation for affine hyperbolic spheres. By the rotation of coordinates,
(1.6)
y1=−xn+1,
yk=xk, k= 2, . . . , n, yn+1=x1,
equation (1.5) (with η≡1) can be rewritten as (1.7) detD2u= ux1
x1 n+2
inRn,+:=Rn∩ {x1 >0}.
Equation (1.7) and the more general equation (4.2) also arise in opti- mal transportation; namely u is the potential function of the optimal transportation from Rn,+ to Rn,+.
Equation (1.7) is invariant under both transforms (1.1) and (1.2). As an application of the transform (1.2), we prove the following Bernstein property.
Theorem 1.2. Let u be a smooth convex solution to (1.7) in Rn,+. If u(0,x) =˜ |˜x|2/2, where x˜= (x2, . . . , xn), then
either u(x) =|x|2/2, or u(x) =|˜x|2/2.
(1.8)
Bernstein theorems usually concern the classification of entire solu- tions and have been a key issue in PDEs. But the Bernstein theorem in half-space is also important. For example, an interesting problem is the Bernstein theorem for the equation
(1.9) detD2u = 1 inRn,+, u = 12|˜x|2 on{x1 = 0}
(namely, a smooth convex solution to (1.9) must be u(x) = 12|x|2 up to a linear function). If the Bernstein property for (1.9) is true, then one can recover the global C2,α regularity of convex solutions to the Dirichlet problem of the Monge-Amp`ere equation [4, 18, 31], and it can also be used to establish the global W2,p estimate, extending the interior estimate of Caffarelli [2].
As another application of the transform (1.2), in this paper we also prove the regularity up to boundary for the graphs of solutions to the Dirichlet problem
detD2u=η(x)/|u|n+2 in Ω, u= 0 on∂Ω,
(1.10)
where Ω is a bounded, uniformly convex domain in Rn, and η is a positive, sufficiently smooth function. There are a number of deep works on boundary regularity for related problems such as (1.12), and (1.14) below [7,12,19,32,33]. Letu∈C∞(Ω)∩C0(Ω) be a solution to (1.10).
If η ≡1, then (−u)−1P
uxixjdxidxj is a Hilbert metric in Ω [24], and u∗, the Legendre transform ofu, defines an affine hyperbolic sphere [7].
The existence and uniqueness of smooth solutions u∈C∞(Ω)∩C0(Ω) to (1.10) was proved in [7]. Due to the singularity on the right hand side of (1.10), the gradient Du necessarily blows up at the boundary. One cannot expect the regularity of the solution u up to the boundary. But we want to know if its graph Mu, as a hypersurface inRn+1, is smooth up to the boundary. In this paper we prove
Theorem 1.3. Suppose Ω is a bounded, uniformly convex domain in Rn with C∞ boundary. Suppose η > 0 and η ∈ C∞(Ω). Let u be a convex solution to (1.10). Then the graph Mu is Cn+2,α smooth up to its boundary, for any α∈(0,1).
In§5.5 we will introduce a compatibility condition, which implies that the Cn+2,α regularity is probably optimal, even if ∂Ω and η are C∞ smooth. A similar phenomenon occurs for Fefferman’s equation (1.12) below. We point out that the global C1,1 estimate in Theorem 1.3 (in the two-dimensional case) was first observed by Loewner and Nirenberg [24], and theC2,αregularity was obtained by Lin and Wang [22] by con- structing proper sub- and super-solutions (see Remark 4.1 for details).
In this paper we will present a different proof for the C2,α regularity.
Here our main interest is the optimal regularity, stated in Theorem 1.3.
It is also a question raised by S.T. Yau in his lecture at the Chinese Academy of Sciences in 2006.
Due to the singularity on the right hand side of (1.7), our proofs of Theorems 1.2 and 1.3 involve new techniques. Our proof of Theorem 1.2 and the C2,α regularity in Theorem 1.3 is based on the transform (1.2) and the method of moving planes. The method of moving planes and the Kelvin transform have been used in semilinear elliptic equations to prove the rotational symmetry of solutions and Liouville theorems.
But for Monge-Amp`ere type equations it is new to use it to establish the regularity of solutions, and this should be of interest itself. We ex- pect more applications of the transform (1.2) in Monge-Amp`ere type equations.
For the Ck,α (k ≥ 3) regularity of (1.7), we introduce an iteration for an ordinary differential equation, which is the linearized equation of (1.7), by regarding ˜xas parameters. The iteration improves the regular- ity of solutions step by step until theCn+3−ǫregularity. It also applies to the linear elliptic equation with singular coefficients near the boundary,
(1.11) Xn
i,j=1aij(x)uxixj+Xn i=1
bi(x)
d(x)uxi+c(x)u=f,
and yields the global Cβ+1 regularity if β >0 is not an integer, where β = ab·γ
ijγiγj, γ is the unit outer normal of ∂Ω, and where d(x) is the distance fromxto the boundary∂Ω. Ifβis an integer, our iteration gives a necessary and sufficient compatibility condition for regularity higher than Cβ+1. This compatibility condition implies that the regularity in Theorem 1.3 is optimal.
An interesting and related problem is the boundary value problem for the equation of Fefferman [12]
(−1)ndet
v, vz¯j vzi, vziz¯j
= 1 in Ω, v= 0 on∂Ω.
(1.12)
The equation is obtained by the change v = e−u from the complex Monge-Amp`ere equation [12]
detuziz¯j =e(n+1)u in Ω, u=∞ on ∂Ω, (1.13)
where Ω is a strictly pseudoconvex domain in Cn. If u is a solution to (1.13), then P ∂2u
∂zi∂¯zjdzid¯zj is a complete K¨ahler-Einstein metric on Ω [12], in a way similar to the real case (1.10) considered in [24], where
−1 u
Puxixjdxidxj is a Hilbert metric in Ω.
By computing the formal power series expansion, Fefferman observed that the solution hypersurface to (1.12) should beCn+2−δ smooth up to the boundary, and Cheng and Yau [8] were able to prove the Cn+3/2−δ regularity for any smallδ >0. Lee and Melrose [19] finally obtained the optimal Cn+2−δ regularity result observed in [12].
Another boundary value problem for the Monge-Amp`ere equation with singularity at the boundary was studied by Urbas [32, 33]. He proved that there exists a positive constantKsuch that when∂Ω∈C∞ is uniformly convex, there is a unique (up to a constant) convex solution to
detD2u
(1 +|Du|2)(n+2)/2 =K in Ω,
|Du|=∞ on ∂Ω.
(1.14)
The solution itself is not smooth at the boundary, but its graph is aC∞ smooth hypersurface up to the boundary.
We would like to point out that equations (1.7) or (1.10), (1.12), and (1.14) contain different singularities and the regularity of solutions to these equations is quite different. For equation (1.14), by the elliptic regularity theory and using the support function of the convex hyper- surface, higher regularity follows readily from the C2,α regularity. The higher regularity for (1.7) and (1.12) is more complicated. For equation (1.12), which becomes v∆v = |Dv|2 −1 in C1, Lee and Melrose [19]
proved the following interesting singularity profile. There exist func- tions ψj = φ(n+1)j0 aj, j ≥ 1, aj ∈ C∞(Ω), such that for all integers N ≥1,
(1.15) u−XN
j=0ψj[log(−φ0)]j ∈C(n+1)N−1(Ω),
where φ0 is a defining function for the domain Ω, namely−φ0(x)≈dx when x is close to∂Ω, wheredx is the distance fromx to ∂Ω.
By a formal computation in dimension 2, a solution to equation (1.7) does not seem to have the singularity profile (1.15). Our argument im- plies that there exists a smooth function a(˜x) such that
(1.16) u(x) =p(x) +a(˜x)xn+31 logx1+q(x)
where p is a polynomial of order n+ 3 and q is a higher order term, namely q(x) =o(|x|n+3) near x= 0. By our compatibility condition, if a≡0, thenq∈C∞(Ω).
We would like to point out that the transform (1.2) and its proper- ties in Section 2 below were found by the authors in 2005. In the last few years, the authors asked several researchers in the area whether the transform (1.2) is new. In early 2009, John Loftin told us that the transform coincides with the conormal map in affine geometry, which transforms a hyperbolic affine sphere to its dual sphere [23]; see Exam- ple 4 below. The authors would like to thank John Loftin for this.
This paper is arranged as follows. In Section 2 we discuss in some detail the transform (1.2) and prove Theorem 1.1. In Section 3 we prove Theorem 1.2. In Sections 4 and 5 we prove Theorem 1.3 for the cases k= 2 and k≥3, respectively.
Acknowledgments.The first author was supported by NSFC (Grant No. 11131005) and the Doctoral Programme Foundation of Institution of Higher Education of China. The second author was supported by Australian Research Council DP1094303 and DP120102718.
2. A transform for convex functions
To make the notation simpler, we write h = u# and the transform (1.2) as
y= x u(x), h(y) = 1
u(x). (2.1)
Obviously the inverse transform is given by x= y
h(y), u(x) = 1
h(y). (2.2)
Hence (u#)#=u; namely, the transform of u# isu itself.
Geometric interpretation. The transform (2.1) has a clear geometric interpretation. Let Mu ={(x, u(x), x∈Ω} be the graph of the convex function u. For any point X = (x, u(x))∈ Mu, consider a ray Rfrom the origin O, given byR ={tX : t >0}. Ifu(x) >0, then the ray R intersects with the hyperplane P ={xn+1= 1} at a pointY = (y,1)∈ P, where y = x/u(x). We have two similar triangles OxX and OyY, where x, y are regarded as points on the plane {xn+1 = 0}. Then the value of h at y is equal to the ratio
|y|
|x| = 1
u(x) = |Y|
|X|.
The following properties of the transform are easy to verify.
(i) If u(0) = 0, Du(0) = 0, and u is uniformly convex, then h is asymptotically quadratic.
(ii) If Ω is a convex domain containing the origin and u < 0 is a convex function in Ω, vanishing on∂Ω, then−his an entire convex function inRn, asymptotic to a convex cone determined by Ω.
(iii) For any vector ξ 6= 0, like the transform x →x/|x|2 in (1.4), the transform (2.1) maps the ray Rξ={tξ: t >0}to the ray itself.
(iv) From the above geometric interpretation, we have the following monotonicity of the transform. Letu, P,R, X, Y be as above. If ˆu is another positive convex function such that ˆX = (ˆx,u(ˆˆ x)) is a point onRbetweenXandY, then ˆh(y)< h(y). As a consequence, for any positive constants C1 and C2, we have
if u(x)≥C1|x|2, then h(y)≤C1|y|2; if 0≤u(x)≤C2|x|2, then h(y)≥C2|y|2.
The transform has some other interesting properties. For example, it transforms linear functions to linear functions (Example 1), and trans- forms the quadratic function u(x) = P
aijxixj to itself (Example 2).
For relation with polar set of a convex body, see Example 4 below. Now we compute
∂yk
∂xi = δki
u − xkuxi
u2 and
uxi = −hyk
h2
∂yk
∂xi
= −hyk h2
δki
u −xkuxi u2
= −hyk
h +ykhykuxi
h .
Hence
(2.3) uxi = hyi
ykhyk−h = hyi h∗ and
u∗ = xiuxi−u (2.4)
= yi h
hyi h∗ − 1
h
= 1
h∗. Lemma 2.1. We have
(2.5) (u∗)#= (u#)∗.
Proof. By (2.4), we have
h∗ = 1 u∗. By definition (2.1), we have
(u∗)#= 1 u∗.
Hence (u∗)#=h∗= (u#)∗. q.e.d.
Next we compute the second derivatives.
uxixj = hyi h∗
yl
∂yl
∂xj (2.6)
= hyiyl
h∗ −ykhyihykyl h∗2
δlj
u −xluxj u2
= h
h∗ hil− yk
h∗hyihykyl
δlj− yl h∗hyj
= h
h∗ hij −yk
h∗hihkj− yl
h∗hjhli+ykyl
h∗2hihjhkl
= h
h∗ δik−aibk
{hkl} δlj−blaj , where
ai =hyi, bk = yk h∗. We have
det(δij−aibj) = 1−X akbk
= 1−ykhyk
h∗
= −h h∗. Hence
detD2u = hn
h∗ndetD2h|det(δij −aibj)|2 (2.7)
= hn+2
h∗n+2 detD2h.
In particular, we have (see Remark 2.2 below)
(2.8) un+2detD2u= 1
h∗n+2detD2h∗.
Proof of Theorem 1.1. Part (b) of Theorem 1.1 was proved in Lemma 2.1. Part (a) was verified in (2.6). Part (c) follows from (2.8).
Remark 2.1. In order to keeph convex, we may let y = x
u(x), h(y) =− 1
u(x) if u∗/u <0.
Remark 2.2. In comparison to (2.8), we recall the corresponding for- mula for the Legendre transform (1.1). For the Legendre transform (1.1), by differentiating either x=Du∗(y) or y=Du(x) we have
{D2u∗(y)} · {D2u(x)}=I, namely D2u∗ is the inverse ofD2u. It follows that (2.9) detD2u= 1/detD2u∗.
Let us compute a few examples.
Example 1.
u(x) =a·x+b (b6= 0).
The graph of u is a linear function. Then byy = u(x)x = a·x+bx ,we have x= 1−a·yby and
h(y) = 1
b(1−a·y).
Example 2.
u(x) =X
aijxixj,
where (aij) is positive definite. To compute h = u#, let ξ be a unit vector inRn. By the geometric interpretation, it suffices to compute the transform of u restricted to the 2-plane spanned by the vectors ξ and en+1 = (0, . . . ,0,1). Therefore it suffices to compute the transform for
˜
u(t) =ct2, wheret∈R1 and c is a constant. By direct computation we have ˜h(s) =cs2. Hence
h(y) =X
aijyiyj. Namely, u#=u, the transform of u is itself.
Example 3.
u(x) = 1
2(a+|x|2).
By y= xu we have |y|= a+|x|2|x|2. Hence|x|= |y|1 1±p
1−a|y|2
.There- fore
h(y) = 1
u(x) = |y|
|x|
= |y|2
1±p
1−a|y|2
= 1
a 1∓p
1−a|y|2
=:h± if a6= 0.
We obtain
if a >0, thenh± are two parts of the same sphere;
if a= 0, thenh≡u;
if a <0, thenh± are two hyperboloids.
Note that in the above example, if |x|2 is replaced by P
aijxixj in u, then|y|2 inh can simply be replaced by P
aijyiyj.
Example 4.Letube an affine hyperbolic sphere asymptotic to a convex cone with vertex at the origin. Then u∗ (the Legendre transform ofu) is a convex function that satisfies
(2.10) |u∗|n+2detD2u∗ = c0 in Ω, u∗ = 0 on ∂Ω, for some constant c0 >0. By (2.8), h=u# satisfies (2.11) |h|n+2 detD2h = 1/c0 in Ω#,
h = 0 on∂Ω#, where Ω# is the polar body of Ω, given by [30]
Ω#={x∈Rn: hx, yi<1 ∀y∈Ω}.
Hence h∗ also defines an affine hyperbolic sphere. It is the dual affine hyperbolic sphere defined by the conormal map of u, given in [14]. For affine hyperbolic spheres, the explicit formula (2.1) for the dual affine sphere was first given by J. Loftin in [23].
There are a number of geometric problems that involve the following equation:
(2.12) det(∇2H+HI) = η(X)
Hn+2 onSn.
For example, ifη≡1, thenHis the support function of an elliptic affine sphere. For generalη, (2.12) is the equation for thep-Minkowski problem [25] or for the Minkowski problem in the centro-affine geometry [10].
The equation is also closely related to the Blaschke-Santalo inequality [26,30],
(2.13) inf
ξ∈KV(K) Z
Sn
1
(H−ξ·x)n+1 ≤ ω2n n+ 1, where K is a convex body,H is its support function, and
V(K) = 1 n+ 1
Z
Sn
Hdet(∇2H+HI)dσ
is the volume ofK. The infimum is taken over allξ, satisfyingH−ξ·x >
0. For symmetric convex hypersurfaces, (2.12) is the Euler equation of the Blaschke-Santalo inequality.
Let ube the projection of H on{xn+1 =−1}, given by (2.14) u(x) =p
1 +|x|2H(X), X= x
p1 +|x|2, −1 p1 +|x|2
. By direct computation,
detD2u= (1 +|x|2)−n+22 det(∇2H+HI).
Hence u satisfies (1.5).
3. A Bernstein theorem
In this section we employ transform (1.2) to prove a Bernstein theo- rem (Theorem 1.2) for the problem
detD2u = ux1 x1
n+2
inRn,+:=Rn∩ {x1 >0}, (3.1)
u = 1
2|˜x|2 on {x1 = 0}.
(3.2)
That is, a smooth convex solution to (3.1)–(3.2) is either u(x) ≡ 12|x|2 or u(x)≡ 12|˜x|2. For the equation
(3.3) detD2u= 1 in Rn,
it is well known that an entire convex solution must be a quadratic function, proved by J¨orgens [17] for n = 2, Calabi [5] for n ≤ 5, and for all n ≥ 2 by Pogorelov [27]. See [9] for a different proof. A more general result was proved by Caffarelli and Li in [3]. For semilinear elliptic equations in half space, a Liouville type theorem was proved by Li and Zhu in [20].
Here we use transform (1.2) and a method of moving planes to prove the Bernstein theorem for (3.1)–(3.2). As the reader will see below, technically our proof for the Monge-Amp`ere equation is quite different from that for the semilinear elliptic equation [20]. First we point out a nice property of equation (3.1), namely its invariance under transforms (1.1) and (1.2).
Lemma 3.1. Let u be a smooth convex solution to (3.1). Then both u∗ andu# satisfy (3.1).
Proof. By (1.1) and (2.9), it is apparent that the Legendre transform u∗ satisfies (3.1).
Leth:=u#be the transform ofu, given in (1.2). By (2.2) and (2.3) we have
ux1
x1 n+2
= hy1 h∗
y1
h(y) n+2
= hy1 h∗
h(y) y1
n+2
, where h∗ is the Legendre transform ofh. By (2.7),
detD2u= hn+2
h∗n+2detD2h.
Hence h satisfies (3.1). q.e.d.
Lemma 3.2. Letube a smooth convex solution to (3.1). Thenux1 = 0 on {x1 = 0}.
Proof. If|ux1(0)|>0, then|ux1(x)| ≥ 12|ux1(0)|forx∈Br+(0), where Br+(0) =Br(0)∩ {x1 >0}andr >0 is small, depending onu. Consider the integral
Z
Br+
detD2u= Z
B+r
ux1 x1
n+2
.
The left hand side is equal to Du(Br+(0)), but the right hand side is equal to∞. The contradiction implies Lemma 3.2. q.e.d.
Lemma 3.3. Let ube a smooth convex solution to (3.1)–(3.2). Then either u≡ 12|˜x|2 or ux1 >0 when x1 >0.
Proof. By Lemma 3.2 and the convexity ofu, we haveux1 ≥0. Hence
∀ x˜ ∈ Rn−1 and x1 > 0 we have u(x1,x)˜ ≥ 12|˜x|2. We claim that if u6≡ 12|˜x|2, then∀x˜∈Rn−1 and there existsx1>0 such thatu(x1,x)˜ >
1
2|˜x|2. Indeed, if there exists ˜x0∈Rn−1 such thatu(x1,x˜0) = 12|˜x0|2 for all x1 >0, by a translation of ˜xand subtraction of a linear function of
˜
x, we assume ˜x0 = 0. Then u(x1,0) = 0 for all x1 > 0. Extend u to Rn such that it is even in x1. Thenu is a convex function in Rn and u(x1,0)≡0. The convexity of u then implies that u is a function of ˜x, independent ofx1. By the boundary condition (3.2) we haveu≡ 12|˜x|2. The claim is proved.
Note that if there is a sequence ˜xk → x˜0 and x1,k → ∞ such that u(x1,k,x˜k) = 12|˜xk|2, by convexity it means u(x1,x˜0) = 12|˜x0|2 for all x1>0, which is ruled out by the above claim. Hence∀R >0,∃M >0 such that u(x1,x)˜ > 12|˜x|2 whenever x1> M and |˜x|< R.
To prove ux1(x1,x)˜ >0 whenx1>0, it suffices to prove it for ˜x= 0.
Let v(x) = 12ǫ|x|2. Then
detD2v > vx1
x1 n+2
.
By the above claim, we can choose ǫ > 0 small andR > 1 large such that v < u on ∂BR+(0). By the comparison principle, we obtain v ≤u inBR+(0). Hence ux1(0)≥0 andux1(x1,0)>0 when x1 >0. q.e.d.
In the following proof of Theorem 1.2, we always assumeu6≡ 12|˜x|2. Lemma 3.4. Let ube a smooth convex solution to (3.1)–(3.2). Then for any R >0, there exist constants C1, C2>0 such that
(3.4) C1x21 ≤u(x)−1
2|˜x|2≤C2x21 ∀ x∈BR+(0), where C1, C2 are allowed to depend on x˜ and R.
Proof. For any ˜x0 ∈ Rn−1, by a translation of the coordinates ˜x = (x2, . . . , xn), we may assume x0 = 0. By subtracting a linear function of ˜x, we may also assume thatu(0) = 0 andD˜xu(0) = 0.
In the previous lemma, we have shown that v(x) = 12ǫ|x|2 is a sub- barrier. Hence the first inequality was proved. Letw(x) = 12M|x|2. Then
detD2w < wx1 x1
n+2
if we choose M large such that w > u on ∂B+R(0). By the comparison principle, we obtain w≥u inBR+(0). Hence w is an upper barrier and we obtain the second inequality of (3.4). q.e.d.
Lemma 3.5. Let ube a smooth convex solution to (3.1)–(3.2). Then u11≡1 on {x1= 0}, where we denote uij =uxixj.
Proof. By Lemma 3.2, we have ux1 = 0 on {x1 = 0}. Hence by differentiating in ˜x we have
(3.5) u1k(x) = 0 on {x1= 0}.
From equation (3.1), it then follows:
(3.6) u11detDx2˜u=un+211 on {x1 = 0}.
Namely,
un+111 = detDx2˜u on{x1= 0}.
By (3.2), detD2˜xu≡1 on{x1 = 0}. q.e.d.
By Lemma 3.5, we have, fork≥1,
(3.7) Dxk˜u11= 0 on {x1 = 0}.
By Lemma 3.2, we also have
(3.8) Dkx˜ux1 = 0 on {x1= 0}.
The boundary condition also implies that if k≥3, (3.9) Dxk˜u= 0 on {x1= 0}.
Lemma 3.6. Let ube a smooth convex solution to (3.1)–(3.2). Then u111≡0 on {x1 = 0}.
Proof. Write equation (3.1) as (3.10) u11={(ux1
x1 )n+2+ ˜g}/detD2x˜u, where
(3.11) g˜=− X
i,j>1
r1irj1 ∂2detr
∂r1i∂rj1 at r =D2u.
Differentiating (3.10) in ˜x, we have
(3.12) u111=H(u11
x1
−ux1 x21 ) +G where
H = (n+ 2)[ux1
x1 ]n+1/detDx2˜u, (3.13)
G = ( ˜g
detD2˜xu)x1 + ( 1
detDx2˜u)x1(ux1 x1
)n+2. (3.14)
By (3.5), (3.7)–(3.9), and Lemma 3.5 we have G(0,x) = 0, H(0,˜ x) =˜ n+ 2.
Hence by Taylor expansion and (3.12) we have u111 = (n+ 2)u111 on {x1= 0},which implies the desired result. q.e.d.
Lemma 3.7. Letu∈C3,1(Br(0)) be a smooth convex function. Sup- pose at the origin 0, u has the expansion
(3.15) u(x) = 1
2|x|2+a(x) +b(x) where
a(x) =X
i,j,k
uijk(0)xixjxk,
|Dkb(x)|=O(|x|4−k) for 0≤k≤4.
Let h=u# be the transform of u. Then as y→ ∞,
(3.16) h(y) = 1
2|y|2+ 2a(y)
|y|2 +b∗(y), with
b∗(y) =O(1), Db∗(y) =O( 1
|y|) as |y| → ∞.
Proof. Byy= u(x)x we have yi = 2xi
|x|2
|x|2
|x|2+ 2a(x) + 2b(x)
= 2xi
|x|2
1− 2a(x) + 2b(x)
|x|2+ 2a(x) + 2b(x)
= 2xi
|x|2
1− 2a(x)
|x|2 +O(|x|2)
. Hence yi= |x|2xi2 1 +O(|x|)
andxi= |y|2yi2 1 +O(|y|1 )
. It follows that
|x| = 2
|y|
1
1 + 2a(x)/|x|2+ 2b(x)/|x|2
= 2|y|−1
1−2a(x)/|x|2−2b′(x)/|x|2
= 2|y|−1h
1−4a(y)/|y|4+ ˜b(y)i ,
where b′ satisfies|Dkb′(x)|=O(|x|4−k) for 0≤k≤4, and ˜bsatisfies
|˜b(y)| ≤C|y|−2, |D˜b(y)| ≤C|y|−3 asy → ∞.
Hence
u(x) = 1
2|x|2+a(x) +b(x)
= 2|y|−2
1−4a(y)/|y|4+ ˜b(y)2
+ 8a(y)/|y|6+ ˜b(y)/|y|2
= 2|y|−2
1−4a(y)/|y|4+ ˜b(y)2
+ 4a(y)/|y|4+ ˜b(y)
= 2|y|−2
1−4a(y)/|y|4 + ˜b(y) ,
where ˜b changes from line to line but all satisfy the above asymptotic behavior. We thus obtain
h(y) = 1
u(x) = |y|2/2
1−4a(y)/|y|4+ ˜b(y)
= 1
2|y|2 1 + 4a(y)/|y|4+ ˜b(y)
= 1
2|y|2+ 2a(y)/|y|2+b∗(y). q.e.d.
We are now ready to use the method of moving planes to prove The- orem 1.2. Denote
Σλ = {xn=λ},
Dλ,R = BR′ (0)× {λ < xn< λ+R}, Dλ,R∗ = BR′ (0)× {λ−R < xn< λ},
where Σλ is a plane, BR′ (0) = {x′ ∈ Rn−1 : |x′| < R, x1 > 0}, where x′ = (x1, . . . , xn−1). The domain Dλ,R is a cylinder in {x1 >0}, and Dλ,R∗ is the reflection of Dλ,R in the plane Σλ. For a point x =
(x1,· · · , xn) ∈ Dλ,R, we denote by xλ = (x1, . . . , xn−1,2λ−xn) the reflection of x in Σλ.
By the moving planes we want to prove that for any λ > 0 and all large R >0,
(3.17) h(x)> h(xλ) ∀x∈Dλ,R.
In the following we use x to denote the variables of h. If (3.17) holds, then by sendingλ→0 and R→ ∞, we obtain
h(x1, . . . , xn−1, xn)≥h(x1, . . . , xn−1,−xn).
As the axisxncan be chosen in any direction perpendicular tox1, hence h, and also u, are symmetric in ˜x. Note that we can choose any point
˜
x0 as a center. Hence the above proof implies thatu is symmetric with respect to ˜x (with center at ˜x0). We can then use the following lemma foru(x1,x), for any fixed˜ x1, to conclude thatuis a quadratic function.
Lemma 3.8. Let u be a convex function in Rn. Suppose that at any point x0 ∈Rn,u−Du(x0)(x−x0)is symmetric with respect tox0. Then up to a constant multiplication, u(x) = 12|x|2+ℓ(x), where ℓis a linear function.
Proof. By subtracting a linear function we assume thatu(0) = 0 and Du(0) = 0. By assumption, u is symmetric with respect to the origin 0. Hence u(x) = u(|x|). Therefore it suffices to consider u as a radial function defined on [0,∞). Hence the condition in the lemma implies that for any a >0,
u(a+t)−u(a−t) = [u(a+t)−u′(a)t]−[u(a−t)+u′(a)t]+2u′(a)t= 2u′(a)t.
If u ∈ C3, by the Taylor expansion at a, for sufficiently small t, we find that the third derivative of u must vanish everywhere. Hence u is a quadratic function.
If uis not C3 smooth, consider the case a= 0, and by subtracting a linear function we assume u′(0) = 0. Then the above formula meansu is even, so its mollification is also even. Hence its third derivative exists and vanishes ata= 0. Asais arbitrary,uis a quadratic function. q.e.d.
From Lemma 3.8 and by (3.17) (or the above discussions), we see that for any given x1,u(x1,x) is a quadratic function of ˜˜ x, i.e.
u(x1,x) =˜ 1
2|˜x|2+a(x1)˜x+b(x1), where a(x1) = (a2(x1), . . . , an(x1)). Hence
detD2u = b′′−Xn
i=2(a′i)2, ux1
x1 n+2
=
a′x˜+b′ x1
n+2,
where b′ and b′′ are the first and second derivatives of b. Notice that detD2u is independent of ˜x. Hence the right hand sides of the above two formulas are the same and we have a′ = 0,
b′′=
b′ x1
n+2.
Noting that b′(0) = 0, we have b′=x1 and sou= 12|x|2.
It remains to prove (3.17) for anyλ >0 and all largeR. This will be achieved in the following two steps.
(i) We show that (3.17) holds for all large R >0 whenλ > λ0, where λ0>0 depends only on the upper bound of|uijk(0)|. This step is for any h=u#, providedu satisfies the expression (3.15).
(ii) Whenuijk(0) = 0, by the maximum principle we show thatλ0 = 0.
Step (i) readily follows from (3.16). First note that by the boundary condition (3.2), we have
h(x) = 1
2|˜x|2 on {x1= 0}.
By (3.16) we have (3.18)
h(x)−h(xλ) = 1 2
x2n−(xn−2λ)2
+ 2a(x)
|x|2 −a(xλ)
|xλ|2
+
b∗(x)−b∗(xλ)
= 2λ[xn−λ] + 2a(x)
|x|2 −a(xλ)
|xλ|2 +
b∗(x)−b∗(xλ) . Note that
|Dxna∗(x)| ≤ CX
|uijk(0)| when |x|>>1, (3.19)
|Dxnb∗(x)| = O(|x|−1), (3.20)
where a∗(x) = a(x)|x|2. Hence ∃ λ0 such that if λ > λ0 and xn > λ, h(x)−h(xλ)>0.
For step (ii), first we observe that equation (3.1) is invariant in re- flection, namely if h(x) satisfies (3.1), so does
hλ(x) :=h(x′,2λ−xn).
Therefore both h and hλ satisfy the equation (3.21) detD2h= hx1
x1 n+2
inDλ,R.
Hence by the strong maximum principle or the comparison principle, it follows that if
(3.22) h≥hλ on ∂Dλ,R,
then
either h > hλ in Dλ,R, (3.23)
or h≡hλ in Dλ,R. (3.24)
In the former case, (3.17) is proved. In the latter case we haveh=hλ in the whole{xn> λ}, and hencehis symmetric with respect to{xn=λ}.
By the boundary condition (3.2), λmust be zero.
Therefore it suffices to verify (3.22) for all largeR >0. The boundary
∂Dλ,R consists of the following parts, ∂Dλ,R=S4
i=1Γi, where Γ1 =∂Dλ,R∩ {xn=λ},
Γ2 =∂Dλ,R∩ {x1 = 0}, Γ3 =∂Dλ,R∩ {|x′|=R}, Γ4 =∂Dλ,R∩ {xn=λ+R}.
On Γ1, we obviously haveh=hλ.
On Γ2, by the boundary condition we obviously haveh > hλ, provided λ >0.
On Γ3, by (3.7)–(3.9) and Lemma 3.6, we haveuijk(0) = 0. Hence for any given λ >0, if R >1 is sufficiently large, by (3.19) and (3.20), we obtain from (3.18) that
h(x)−hλ(x) = h(x)−h(xλ)
≥ 2λ(xn−λ)−o(xn−λ)>0.
On Γ4, (3.22) follows from (3.18) and (3.20) asa∗≡0 in (3.19).
Therefore we have verified (3.22), and so also (3.17) by the maximum principle, for all λ >0. Hence by Lemma 3.8 and the discussion before it, we conclude that u is a quadratic function. Theorem 1.2 is proved.
Remark 3.1. For any α > 0, the above moving plane argument also implies that a smooth convex solution to
detD2u = ux1
x1
n+2+α
inRn,+, (3.25)
u = 1
2|˜x|2 on{x1= 0}
(3.26)
must be u(x) = 12|x|2 or u(x) = 12|˜x|2. Indeed, by the transform (2.1), equation (3.2) is changed to
(3.27) detD2h= h
h∗ α hx1
x1
n+2+α
.
When α ≥0, the comparison principle is applicable to the functions h and hλ, and from (3.22) we can still infer (3.23) or (3.24).
Remark 3.2. By the moving plane, we see that a smooth convex solu- tion to
detD2u = ux1 x1
n+2
inRn,+={x1 >0}, (3.28)
u = φ(˜x) on{x1 = 0}
(3.29)
is symmetric with respect to ˜xif φ is. One can easily verify that (3.7)–
(3.9) and Lemma 3.6 hold atx= 0, and hence uijk(0) = 0.
An interesting question is whether one can use the transform (1.2) to prove Bernstein theorems for other Monge-Amp`ere type equations, (3.30) detD2u=f(x, u, Du) inRn,
such as the case f ≡ 1. Let u be a solution to (3.30) and let h = u# be the transform of u. Another interesting question is the singularity removability forhnear the origin. Note that ifhis smooth at the origin, thenu is asymptotic to a quadratic function as infinity.
4. C2,α regularity
In this section we prove the global C2,α regularity for the problem detD2u= η(x)
|u|n+2 in Ω, u= 0 on∂Ω.
(4.1)
Namely, the graph of u is C2,α smooth up to the boundary. In this section we assume that Ω is a bounded, uniformly convex domain with Ck,α boundary, andηis positive andCk−2,α smooth,k≥3. By Cheng- Yau [7] and Caffarelli [2, 16], there is a unique convex solution u ∈ Ck,α(Ω)∩C0(Ω) to (4.1). By a translation of coordinates, we assume that the origin 0∈∂Ω, and e1 = (1,0, . . . ,0) is the inner normal of ∂Ω at 0. We make the rotation of the coordinates
y1 =−xn+1,
yk=xk, k= 2, . . . , n, yn+1=x1.
In the new coordinates, the graph ofunear the origin can be represented as
yn+1 =v(y).
Since ux1 <0 near the origin, we have vy1 >0 near the origin.
From equation (4.1), the Gauss curvature K = detD2u
[1 +|Du|2](n+2)/2
= η
|u|n+2[1 +|Du|2](n+2)/2.
Hence
detD2v = K[1 +|Dv|2](n+2)/2
= η
1 +|Dv|2 y12(1 +|Du|2)
n+22 . Claim:
1 +|Dv|2 1 +|Du|2
n+22
=vn+2y1 . Indeed, note that the tangent plane is given by
xn+1−ux1x1− · · · −uxnxn= 0.
After the above rotation of coordinates, it is given by yn+1+ 1
ux1y1+ux2
ux1y2+· · ·+uxn
ux1yn= 0.
Hence
vy1 =−1/ux1, vy2 =−ux2/ux1, . . .
vyn =−uxn/ux1, and the claim follows.
Therefore equation (4.1) can be written as (from now on we write v and y asu and x)
(4.2) detD2u=ηux1 x1
n+2
in BR+,
where η=η(u, x2, . . . , xn). By a rescaling, we assume that η(0) = 1
and near 0, the boundary ∂Ω is given by {x1 = φ(x2, . . . , xn)} in the original coordinates. Then the boundary condition for (4.2) is
(4.3) u=φ on {x1 = 0}.
Moreover, we have
u(0) = 0, u≥0 andux1 >0 near 0.
By Lemma 3.2,
ux1 = 0 on {x1 = 0}.
It follows that for any k≥0,
(4.4) Dkx˜ux1 = 0 on {x1= 0}, Dkx˜u=Dkx˜φ,
where ˜x = (x2, . . . , xn). In particular, we have u1k = 0 on x1 = 0, for any k≥2. Therefore at any boundary point, we can make a rotation of the axes x2, . . . , xn such that D2u is diagonal. By a unimodular linear
transform of ˜x, we may assume that uii(0) =ujj(0) for all 2≤i, j≤n.
Sinceφis uniformly convex, there is no loss in assuming thatuii(0) = 1 for all 2≤i≤n. By (3.6) we can then assume that
(4.5) D2u(0) =I
is the unit matrix.
To prove the regularity of the solution to (4.1), we employ the method of continuity, as was used in [8] for equation (1.12). It is worth noting that one may use the method of continuity in different ways. For ex- ample, we can write problems (4.2)–(4.3) as a Dirichlet problem for the support function of the solution on the south hemisphereSn,− and apply the method of continuity to the Dirichlet problem in the space C3,α(Sn,−). We can apply the method of continuity tov=u2inC3,α(Ω).
Instead of fixing the domain Ω and allowing η to vary, we can also fix an η and allow the domain Ω to vary. That is, letting Ωt be a family of uniformly convex domains such that Ω0 is the unit ball and Ω1= Ω, apply the method of continuity to Ωt.
Remark 4.1. (i) TheC2,αregularity at boundary was obtained in [22].
At a boundary point 0, one can construct a lower-barrier w1 and an upper-barrier w2 such that
(4.6) 0≤w2(x)−w1(x)≤C|x|2+α.
From (4.6), one can obtain the boundary C2,α estimate as follows. By rescaling, one can first prove the strict convexity of solutions and use the interior second derivative estimate of Pogorelov [28] to get theC1,1 estimate, such that equation (4.2) is uniformly elliptic. Then by (4.6) one can prove that for any ball Br(z), there is a quadratic polynomial Pz such that supBr(z)|u(x)−Pz(x)| ≤ Cr2+α. By using Campanato’s space, the C2,α regularity follows as in [1,29].
(ii) Unaware of the work [22], in early 2008 we found a similar proof.
We also constructed upper and lower barriers satisfying (4.6). Our barri- ers are as follows: Considering a boundary point 0, by a linear transform of ˜x, we may assume thatφ(˜x) = 12|˜x|2+O(|˜x|2+α) for someα∈(0,1).
Then it is straightforward to verify that w1(x) = 1
2|x|2+a|x|2+α, w2(x) = 1
2|x|2−a|x|2+α
are upper and lower barriers provided a is sufficiently large. Instead of using Camponato’s space, we observed that the C2,α estimate can be obtained from the interior regularity theory of Evans-Krylov [11,18] by a rescaling argument. That is, for any points x, y∈Rn,+ with|x−y|<
1
2min(dx, dy), where dx = dist(x, ∂Rn,+), by rescaling and the interior
regularity theory of Evans-Krylov we have |D2u(x)−D2u(y)| ≤C|x− y|α. For any two points x, y ∈ Rn,+, assuming dy ≤ dx, we choose a sequence of points (xk) on the line segment xy such that x0 = x and xN = y such that |xi −xi+1| < 12min(dxi, dxi+1) for i = 0,1, . . . , N.
Then
|D2u(x)−D2u(y)| ≤
N
X
i=1
|D2u(xi)−D2u(xi−1)|
≤X
C|xi−xi−1|α≤C|x−y|α.
Note thatN =∞if y is a boundary point. Note also that D2u isC2,α smooth on {x1 = 0}, which follows from ux1xk = 0 (Lemma 3.2) and (3.6), and so we may assume that the segment xy is parallel to the x1-axis.
(iii) Note that if one applies the regularity in [1,11,18,29] directly, one may get the C2,α regularity for a small α > 0 only. But for the Monge-Amp`ere equation (4.2), if η is sufficiently smooth, we have the C∞ interior regularity and the above argument implies the C2,α regu- larity, for any α >0.
In the following, we first use the method of moving planes to prove a continuity estimate for D2u at the boundary, then prove the C2,α regularity, for anyα∈(0,1). The proof is more complicated than the one in Remark 4.1, and also we need to assume∂Ω∈C3,α. We present the proof here because the technique is new and should be of some interest in the area. Moreover, some estimates will be needed in the next section for the proof of the higher regularity, which is more complicated. Our main interest is to obtain the optimal regularity, as with Fefferman’s equation (1.12), mentioned in the introduction.
4.1. Continuity estimate for D2u.Letu be a smooth convex solu- tion to (4.2) and (4.3). We assume ∂Ω ∈ C3,α and η ∈ C1,α. We will first prove for simplicity the regularity of solutions to (4.2)–(4.3) for the caseη ≡1. We then explain that our argument also applies to solutions with general smooth and positive η.
To start with, we first recall the estimate in Lemma 3.4, φ(˜x) +C1x21 ≤u(x)≤φ(˜x) +C2x21 inBR+(0),
obtained by proper construction of barriers, which also follows from (4.6). Hence by the uniform convexity of φwe have
(4.7) C1|x|2 ≤u(x)≤C2|x|2 inBR+(0),
for two positive constants C1, C2. From (4.7) and the convexity of uwe also have
(4.8) ∂x1u >0 in B+R(0).
