Electronic Journal of Differential Equations, Vol. 2021 (2021), No. 86, pp. 1–18.

ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu

SINGULAR MONGE-AMPÈRE EQUATIONS OVER

CONVEX DOMAINS

MENGNI LI

Abstract. In this article we are interested in the Dirichlet problem for a

class of singular Monge-Ampère equations over convex domains being either
bounded or unbounded. By constructing a family of sub-solutions, we prove

the existence and global Hölder estimates of convex solutions to the problem

over convex domains. The global regularity provided essentially depends on
the convexity of the domain.

1. Introduction

Let us consider the Dirichlet problem of Monge-Ampère equation

detD2u = |u|−α in Ω,

u = 0 on ∂Ω,
(1.1)

where α > 0 is a constant, Ω ⊆ Rn(n > 2) is a convex domain and u : Ω → R is a
convex function. We note that the problem (1.1) is invariant under translation and
rotation transformations. In this paper, our main purpose is to settle the issue of
the existence and global regularity of the solution u to problem (1.1) over convex
domains, including both bounded convex domains and unbounded convex domains.

This type of equations is one of the most important fully nonlinear partial differ-
ential equations and plays a fundamental role in a profusion of geometric applica-
tions. It is known that the equation in (1.1) arises from the Lp-Minkowski problem
when we denote α = 1− p. As a generalization of the classical Minkowski problem
[21], the Lp-Minkowski problem was first proposed to explore convex bodies in Rn+1

with given p-area measures by Lutwak [20], and was furthermore associated with
self-similar solutions to Gauss curvature flows by Andrews [1] and Urbas [23]. In
particular, the Lp-Minkowski problem with p = −n− 1 (namely α = n+ 2), corre-
sponding to the critical exponent case, is interpreted as the centroaffine Minkowski
problem [6, 9]. To elaborate a little bit on a solution u to problem (1.1) with
α = n + 2, (−1/u)

∑
uxixjdxidxj gives the Hilbert metric in convex domain [19],

and the Legendre transform of u defines a complete hyperbolic affine sphere [4, 5].
The readers may consult [3, 7, 10, 15, 22] and the references therein for more related
topics.

2010 Mathematics Subject Classification. 35J96, 52A20, 35B65.
Key words and phrases. Dirichlet problem; Hölder estimate; bounded convex domain;

unbounded convex domain.
c©2021. This work is licensed under a CC BY 4.0 license.

Submitted November 26, 2020. Published October 18, 2021.

1



2 M. LI EJDE-2021/86

In the past four decades, a great deal of mathematical effort has been devoted to
developing the global regularity theory of the problem (1.1); see [2, 4, 8, 12, 13, 14,
16, 17, 18, 22] for example. A pivotal observation has been that the equation in (1.1)
becomes singular on the boundary ∂Ω by virtue of u = 0 there. This singularity
will inevitably lead to the phenomena that the gradient Du may blow up at the
boundary and hence the optimal global regularity of the solution u should be Hölder
continuous. Specifically, based on the (a, η) type domain introduced by Jian and Li
[12], the corresponding Hölder exponent for a class of Monge-Ampère type equations
can be independent of the smoothness of domain but only essentially depends on
the convexity of domain [12, 13, 18]. However, only the bounded domains were
addressed, except for [11] where the existence of solution to (1.1) with α = n + 2
was obtainable on a class of unbounded domains. The main motivation of this
paper is to extend such result to (1.1) with more general α, and moreover prove the
existence and global regularity results on convex domains being either bounded or
unbounded.

Before stating the main results, we first review the concept of (a, η) type domain
in [12] to describe the convexity of domain. Roughly speaking, the less is the
parameter a, the more convex is the domain. We also refer the readers to [18] for
a careful understanding of the geometry at one point as the parameter a varies.

Definition 1.1. Suppose that Ω is a bounded convex domain in Rn and x0 ∈ ∂Ω.
We say x0 is (a, η) type if there exist numbers a ∈ [1,+∞] and η > 0 such that
after translation and rotation transforms, we have

x0 = 0 and Ω ⊆ {x = (x′, xn) ∈ Rn : xn > η|x′|a}.

The domain Ω is called (a, η) type domain if its every boundary point is (a, η) type.

We note that for the case a ∈ [1, 2), there exists no (a, η) type domain though
some boundary points might be (a, η) type. Thus we need only consider the (a, η)
type domain with a ∈ [2,+∞] from now on.

Based on the existence of the solution to (1.1) over bounded convex domains
[4, 13], we can derive the following global regularity result for (1.1), which can be
regarded as a direct consequence of setting F (x, u,∇u) = |u|−α with α > 0 in [18].
We point out that we merely pay attention to the singular case α > 0 in this paper,
despite the fact that the assumption α > 0 can be trivially relaxed to α > 0 in
what follows.

Theorem 1.2. Suppose Ω ⊂ Rn is an (a, η) type domain with a ∈ [2,+∞]. If u is

a convex generalized solution to the problem (1.1), then u ∈ C
2(a+n−1)
a(n+α) (Ω) and

|u|
C

2(a+n−1)
a(n+α) (Ω)

6 C(a, η, α, n,diam(Ω)).

As our first goal is to show that this global regularity result for problem (1.1)
over bounded convex domains reflects the relation of the Hölder exponent with the
convexity of the domain. In particular, when we take a = 2, Ω corresponds to a
bounded convex domain satisfying exterior sphere condition (see [12, Definition 2.1
and Lemma 2.1] for details), and moreover, when we take a = +∞, Ω represents a
general bounded convex domain (see [18, Remark 2.3] for details). We would like
to individually present this global regularity result for the two extreme cases a = 2
and a = +∞ as the following two corollaries in light of their geometric significance.



EJDE-2021/86 SINGULAR MONGE-AMPÈRE EQUATIONS 3

Corollary 1.3. Suppose Ω ⊂ Rn is a bounded convex domain satisfying exterior
sphere condition. If u is a convex generalized solution to the problem (1.1), then

u ∈ C
n+1
n+α (Ω) and

|u|
C
n+1
n+α (Ω)

6 C(a, η, α, n,diam(Ω)).

Corollary 1.4. Suppose Ω ⊂ Rn is a bounded convex domain. If u is a convex

generalized solution to the problem (1.1), then u ∈ C
2

n+α (Ω) and

|u|
C

2
n+α (Ω)

6 C(α, n, diam(Ω)).

We remark here that 2(a+n−1)
a(n+α) ∈ [ 2

n+α ,
n+1
n+α ] for any a ∈ [2,+∞], and it equals

n+1
n+α when a = 2 and equals 2

n+α when a = +∞. Similar to [18], the proof of
Theorem 1.2 relies on carefully constructing sub-solutions and can be divided into
two parts 2 6 a < +∞ and a = +∞. In addition, the proof of Corollary 1.4, i.e.,
the case a = +∞, will provide great convenience for the subsequent study on the
unbounded domains.

The second goal of this paper concerns the existence and global regularity of the
solution to (1.1) over unbounded convex domains. On the one hand, we are inspired
by the corresponding result over bounded convex domains [4, 13], for which we refer
the readers to Theorem 5.1 in Section 5 with its proof based on Corollary 1.4. On
the other hand, it is not surprising to generalize the existence result in [11] to the
problem (1.1) over unbounded domains. The key ingredient still lies in a delicate
construction of sub-solutions. Precisely, our second main result is stated as follows.

Theorem 1.5. Suppose Ω ⊂ Rn is an unbounded convex domain such that ∂Ω is
strictly convex at some point x0 ∈ ∂Ω. Then problem (1.1) admits a convex solution

u ∈ C∞(Ω) ∩ C(Ω). Moreover, for any r > 0, u ∈ C
2

n+α (Ω ∩Br(0)) and

|u|
C

2
n+α (Ω∩Br(0))

6 C(α, n, diam(Ω ∩Br(0))).

Here Br(0) denotes the ball in Rn centered at the origin with radius r.

An outline of this paper is as follows. In Section 2, we revisit necessary results on
the convexity and establish the general setup for choosing sub-solutions. To better
understand the geometry of (a, η) domain, we deal with two particular cases a = 2
and a = +∞ of Theorem 1.2 in Section 3 and Section 5 respectively. Moreover,
Section 4 and Section 5 provide a complete proof for Theorem 1.2. Finally, Section
6 is devoted to the construction of sub-solutions and the existence of solutions to
(1.1) on unbounded convex domains, which completes the proof of Theorem 1.5.

2. Preliminaries

2.1. Useful observations on convexity. Firstly, we give a brief review on convex
bodies in Rn as follows.

Remark 2.1. The definitions of convex bodies and strictly convex bodies in Rn are
well known:

(i) A subset Ω ⊂ Rn is said to be convex if for every two points x, y ∈ Ω, the
line segment joining x to y is contained in Ω, that is, for any t ∈ [0, 1], we
have tx+ (1− t)y ∈ Ω.



4 M. LI EJDE-2021/86

(ii) A subset Ω ⊂ Rn is said to be strictly convex if for every two points x, y ∈ Ω,
the line segment joining x to y is strictly contained in Ω, that is, for any
t ∈ (0, 1), we have tx+ (1− t)y ∈ Ω◦, where Ω◦ denotes the interior of Ω.

From the geometric intuition, the curvature at every boundary point of smooth con-
vex domains is nonnegative, while the curvature at every boundary point of smooth
strictly convex domains is positive. For instance, balls are not only convex but also
strictly convex, while cubes are merely convex rather than strictly convex. In addi-
tion, the concept of (a, η) type as in Definition 1.1 gives a more precise description
of the convexity of domains, where we refer the readers to [18, Remarks 2.2 and
2.3] for more details.

Later on, we present the following fact based on rotation and translation trans-
forms of convex domains.

Lemma 2.2. Given a general convex domain Ω, an invertible n-order matrix A ∈
Mn and a point x0 ∈ Rn, we denote

Ω̃ := AΩ + x0 = {y : there exists x ∈ Ω such that y = Ax+ x0}.

If u(x) is a convex solution to (1.1) on Ω, then

ũ(Ax+ x0) := |detA|
2

n+αu(x)

is a convex solution to (1.1) on Ω̃.

Proof. Let x̃ = Ax+ x0. Then there holds

detD2
xu = (detA)

2
detD2

x̃u = |detA|2 detD2
x̃u.

According to the equation in (1.1), i.e., |u|α detD2
xu = 1, we infer

|u|α|detA|2 detD2
x̃u = 1.

Let ũ = |detA|
2

n+αu. Then we have

|u|α = |detA|−
2α
n+α |ũ|α

and

detD2
x̃u = detD2

x̃

(
|detA|−

2
n+α ũ

)
= |detA|−

2n
n+α detD2

x̃ũ.

Combining the previous three formulas, we derive

|ũ|α detD2
x̃ũ = 1.

The proof of the lemma is now complete. �

We turn to review an interesting lemma concerning the boundary Hölder regu-
larity of convex functions over convex domains, for which we refer the readers to
[12, Lemma 2.3] for the proof.

Lemma 2.3. Let Ω be a bounded convex domain and u ∈ C(Ω) be a convex function
in Ω with u|∂Ω = 0. If there exist λ ∈ (0, 1] and M > 0 such that

|u(x)| 6Mdλx, ∀x ∈ Ω,

where dx = dist(x, ∂Ω), then u ∈ Cλ(Ω) and

|u|Cλ(Ω) 6M
(
(diam(Ω))λ + 1

)
.



EJDE-2021/86 SINGULAR MONGE-AMPÈRE EQUATIONS 5

2.2. Equivalent conditions of sub-solution. We present the comparison prin-
ciple and define sub-solutions of (1.1). For simplicity of expression, we denote

H[W ] := detD2W · |W |α.

We have an application of the comparison principle for fully nonlinear equations,
i.e. [8, Theorem 17.1].

Theorem 2.4 (comparison principle). Let u, v ∈ C(Ω)∩C2(Ω) satisfy F [u] > F [v]
in Ω and u 6 v on ∂Ω. It then follows that u 6 v in Ω.

Definition 2.5. A non-positive function W is called a sub-solution of (1.1) if

detD2W > |W |−α in Ω,

i.e. H[W ] > 1 in Ω.

For convenience of constructing sub-solutions in the next sections, we give two
equivalent conditions for which W is a sub-solution to the problem (1.1). In fact,
the only difference between these two equivalent conditions lies in what the variable
r represents.

Lemma 2.6. Consider W (x) = W (r) and write for i, j ∈ {1, 2, . . . , n},

Wr =
∂W

∂r
, Wi =

∂W

∂xi
, Wij =

∂2W

∂xi∂xj
.

Then a non-positive function W is a sub-solution to the problem (1.1) if and only
if

H[W ] =
(Wr

r

)n−1

Wrr|W |α > 1 in Ω.

Proof. For i, j ∈ {1, 2, . . . , n}, by direct computation, we have

Wi = Wr
xi
r
,

Wij =
Wr

r
δij +

(
Wrr −

Wr

r

)xi
r

xj
r
.

Consequently,

detD2W =
Wr

r
I +

(
Wrr −

Wr

r

)
θθT ,

where I is the unit matrix and θT =
(
x1

r , . . . ,
xn
r

)
. We notice that all the n eigen-

values of matrix θθT are 1, 0, . . . , 0 and thus all eigenvalues of matrix D2W are
Wrr,

Wr

r , . . . ,
Wr

r . Then we have the following explicit formula for detD2W :

detD2W =
(Wr

r

)n−1

Wrr.

As a result, we obtain

H[W ] = detD2W · |W |α =
(Wr

r

)n−1

Wrr|W |α.

With W 6 0 on ∂Ω, the lemma follows immediately. �

Lemma 2.7. Consider W (x) = W (r, xn) and write for i, j ∈ {1, 2, . . . , n},

Wr =
∂W

∂r
, Wi =

∂W

∂xi
, Wij =

∂2W

∂xi∂xj
.



6 M. LI EJDE-2021/86

Then a non-positive function W is a sub-solution to the problem (1.1) if and only
if

H[W ] =
(Wr

r

)n−2(
WrrWnn − |Wrn|2

)
|W |α > 1 in Ω.

Proof. Let

D2W :=

(
A α
αT Wnn

)
,

where αT = (Wn1, . . . ,Wn(n−1)) and A is the (n− 1)-order matrix. We infer that

detD2W = detA(Wnn − αTA−1α).

For k, l ∈ {1, 2, . . . , n− 1}, a direct computation gives

Wk = Wr
xk
r
,

Wkl =
Wr

r
δkl +

(
Wrr −

Wr

r

)xk
r

xl
r
,

Wkn = Wrn
xk
r
.

Thus we obtain

A =
Wr

r
I +

(
Wrr −

Wr

r

)
θθT ,

where I is the unit matrix and θT =
(
x1

r , . . . ,
xn−1

r

)
. We note that all the n − 1

eigenvalues of matrix θθT are 1, 0, . . . , 0 and hence all eigenvalues of matrix A
are Wrr,

Wr

r , . . . ,
Wr

r . In particular, α is an eigenvector of A with respect to the
eigenvalue Wrr. Then we obtain

detA =
(Wr

r

)n−2

Wrr,

αTA−1α = αT
1

Wrr
α =

|Wrn|2

Wrr
.

This implies the following explicit formula for detD2W :

detD2W =
(Wr

r

)n−2

Wrr

(
Wnn −

|Wrn|2

Wrr

)
=
(Wr

r

)n−2(
WrrWnn − |Wrn|2

)
.

Therefore,

H[W ] = detD2W · |W |α =
(Wr

r

)n−2(
WrrWnn − |Wrn|2

)
|W |α.

Combining this with W 6 0 on ∂Ω, we have proved the lemma. �

3. Bounded convex domains satisfying exterior sphere condition

In this section, we focus on a bounded convex domain Ω satisfying exterior sphere
condition, or in other words, Ω is a (2, η) type domain. We construct a sub-solution
to the problem (1.1) which exploits the strength of exterior sphere condition, and
then give an alternative proof of Corollary 1.3 for the case α > 1. In fact, a complete
proof of Corollary 1.3 for all α > 0 is given in the following Section 4 since a = 2
is a particular case of 2 6 a < +∞.

We begin with several simplifications of the problem:

(i) By Lemma 2.3, it suffices to show that

|u(y)| 6 C(η, α, n,diam(Ω))d
n+1
n+α
y , ∀y ∈ Ω. (3.1)



EJDE-2021/86 SINGULAR MONGE-AMPÈRE EQUATIONS 7

(ii) For any point y ∈ Ω, we can find z ∈ ∂Ω be the nearest boundary point
to y. Without loss of generality, we assume that the domain Ω satisfies
the exterior sphere condition with radius R. By some translations and
rotations, we can further assume z = 0, 0 ∈ ∂Ω ∩ ∂BR(y0), Ω ⊂ BR(y0),
and the line yz is the xn-axis.

We notice that the tangent plane of Ω at z = 0 is unique since z = 0 is the nearest
boundary point to y. Moreover, y is on the line determined by 0 and y0 (with the
order 0, y, y0), and hence

dy = dist(y, ∂Ω) = |y − 0| = |y0 − 0| − |y0 − y| = R− |y0 − y|.
We are now ready to construct a sub-solution to (1.1) by using Lemma 2.6. Let

U(x) = −K(R2 − |y0 − x|2)
n+1
n+α = −K(R2 − r2)

n+1
n+α ,

where r = |y0 − x| and K is a positive constant to be determined such that U is a
sub-solution to (1.1) in Ω. It is trivial to see that U 6 0 on Ω, thus

U 6 u on ∂Ω. (3.2)

A routine computation leads us to

Ur = 2K
n+ 1

n+ α
(R2 − r2)

n+1
n+α−1r,

Urr = 2K
n+ 1

n+ α
(R2 − r2)

n+1
n+α−2

(
R2 +

α− n− 2

n+ α
r2
)
,

which gives

H[U ] =
(Ur
r

)n−1

Urr|U |α = 2nKn+α (n+ 1)n

(n+ α)n
(
R2 +

α− n− 2

n+ α
r2
)
.

We split α > 1 into the following two cases:

(i) When α > n+ 2, we derive that

H[U ] > 2nKn+α (n+ 1)n

(n+ α)n
(R2 + 0r2) = 2nKn+α (n+ 1)n

(n+ α)n
R2.

Then we can take M sufficiently large such that H[U ] > 1.
(ii) When 1 < α < n+ 2, we infer that

H[U ] > 2nKn+α (n+ 1)n

(n+ α)n
(
R2 +

α− n− 2

n+ α
R2
)

= 2n+1Kn+α (α− 1)(n+ 1)n

(n+ α)n+1
R2.

Then we can take M sufficiently large such that H[U ] > 1.

To sum up, for any α > 1, we always have

H[U ] > 1. (3.3)

According to (3.2) and (3.3), we obtain from Lemma 2.6 that U is a sub-solution
to the problem (1.1).

By Theorem 2.4 (comparison principle), we obtain

0 > u(y) > U(y).

Taking this inequality on yn-axis, we arrive at the conclusion that

|u(y)| 6 |U(y)|

= K(R2 − |y0 − y|2)
n+1
n+α



8 M. LI EJDE-2021/86

= K(R+ |y0 − y|)
n+1
n+α (R− |y0 − y|)

n+1
n+α

6 K(2R)
n+1
n+α d

n+1
n+α
y ,

which implies (3.1). Thus we have proved Corollary 1.3 for the case α > 1.

4. (a, η) type domains with 2 6 a < +∞

In this section, we prove Theorem 1.2 for the case 2 6 a < +∞, which can
imply Corollary 1.3 immediately. We present here only the main procedure of the
proof and refer the readers to [18] for a rigorous derivation since this section can
be regarded as its special case F (x, u,∇u) = |u|−α with α > 0.

Firstly, we will adopt the following simplifications:

(i) Thanks to Lemma 2.3, it suffices to show that

|u(y)| 6 C(a, η, α, n,diam(Ω))d
2(a+n−1)
a(n+α)
y , ∀y ∈ Ω. (4.1)

(ii) For any point y ∈ Ω, there exists z ∈ ∂Ω such that dist(y, z) = dy. Since
the domain Ω is (a, η) type, without loss of generality, we can assume z = 0
and take the line determined by y and z as the xn-axis such that

Ω ⊆ {x ∈ Rn : xn > η|x′|a}.

We construct a sub-solution to the problem (1.1) by using Lemma 2.7. From
now on, we let

U(r, xn) = −
((xn

ε

) 2
a − r2

)1/b

,

where b = n+α
a+n−1 and ε is positive constant to be determined such that U is a

sub-solution to (1.1) in Ω. It is obvious that U 6 0 on Ω and hence

U 6 u on ∂Ω. (4.2)

By straightforward calculation, we obtain

Ur =
2

b
|U |1−br,

Un = − 2

ab
|U |1−b

(xn
ε

) 2
a−1 1

ε
,

Urr =
2

b
|U |1−b − 4(1− b)

b2
|U |1−2br2,

Unn = −2(2− a)

a2b
|U |1−b

(xn
ε

) 2
a−2 1

ε2
− 4(1− b)

a2b2
|U |1−2b

(xn
ε

) 4
a−2 1

ε2
,

Urn =
4(1− b)
ab2

|U |1−2br
(xn
ε

) 2
a−1 1

ε
,

which yields

UrrUnn − |Urn|2 =
8(a− 2)(b− 1)

a2b3
|U |2−3b

(xn
ε

) 2
a−2

r2 1

ε2︸ ︷︷ ︸
I1

+
8(b− 1)

a2b3
|U |2−3b

(xn
ε

) 4
a−2 1

ε2︸ ︷︷ ︸
I2



EJDE-2021/86 SINGULAR MONGE-AMPÈRE EQUATIONS 9

+
4(a− 2)

a2b2
|U |2−2b

(xn
ε

) 2
a−2 1

ε2︸ ︷︷ ︸
I3

.

To estimate I1 + I2 + I3, we will choose δ ∈ (0, 1) and ε = ε(δ, a, η) > 0 such that

ε
(

1
δ

) a
2 6 η. As a consequence, we obtain

Ω ⊆ {x ∈ Rn|xn > η|x′|a} ⊆ {x ∈ Rn|δ
(xn
ε

) 2
a > r2},

and thus

|U |b =
(xn
ε

) 2
a − r2 ∈

[
(1− δ)

(xn
ε

) 2
a ,
(xn
ε

) 2
a

]
. (4.3)

Step 1: Consider the case α + 1 > 2, i.e. α > 1. We distinguish two cases:
2 6 a < α+ 1 and α+ 1 6 a < +∞.

Case 1: When 2 6 a < α + 1, we have b = n+α
a+n−1 > 1. In such a case, we obtain

I1, I2, I3 > 0 and hence

Urr · Unn − |Urn|2 > I2 =
8(b− 1)

a2b3
|U |2−3b

(xn
ε

) 4
a−2 1

ε2

(4.3)

>
8(b− 1)

a2b3
|U |2−3b

(
(1− δ)− a2 |U | ab2

) 4
a−2 1

ε2

=
8(b− 1)

a2b3
(1− δ)a−2|U |2−b−ab 1

ε2
.

This and the formula of H[·] in Lemma 2.7 imply

H[U ] =
(Ur
r

)n−2(
UrrUnn − |Urn|2

)
|U |α

>
(2

b
|U |1−b

)n−2 8(b− 1)

a2b3
(1− δ)a−2|U |2−b−ab 1

ε2
|U |α

=
(2

b

)n−2 8(b− 1)

a2b3
(1− δ)a−2 1

ε2
.

Since b > 1, we can take ε = C(a, α, n, δ) > 0 sufficiently small such that

H[U ] > 1 in Ω. (4.4)

By (4.2), (4.4) and Lemma 2.7, we conclude that U is a sub-solution to the problem
(1.1).

By Theorem 2.4 (comparison principle), we obtain

0 > u(y) > U(y).

Restricting this inequality onto yn-axis, we have

|u(y)| 6 |U(y)| 6
( yn
ε(a, α, n, δ)

) 2
ab

= C(a, η, α, n,diam(Ω))y
2
ab
n

= C(a, η, α, n,diam(Ω))d
2(a+n−1)
a(n+α)
y ,

which implies (4.1).



10 M. LI EJDE-2021/86

Case 2: When α+ 1 6 a < +∞, we have b = n+α
a+n−1 ∈ (0, 1]. Since a > α+ 1 > 2,

then I1 6 0, I2 6 0, I3 > 0. On the one hand, we observe that

I1 = (a− 2)r2
(xn
ε

)− 2
a I2

(4.3)

> δ(a− 2)I2.

In view of (4.3), we obtain xn > ε|U |
ab
2 and then(xn

ε

) 4
a−2
6
(
|U | ab2

) 4
a−2

= |U |2b−ab,

which gives rise to

I1 + I2 >
(
δ(a− 2) + 1

)
I2

=
(
δ(a− 2) + 1

)8(b− 1)

a2b3
|U |2−3b

(xn
ε

) 4
a−2 1

ε2

>
(
δ(a− 2) + 1

)8(b− 1)

a2b3
|U |2−3b|U |2b−ab 1

ε2

=
(
δ(a− 2) + 1

)8(b− 1)

a2b3
|U |2−b−ab 1

ε2
.

On the other hand, since (4.3) also yields

xn 6 ε(1− δ)−
a
2 |U | ab2

and moreover (xn
ε

) 2
a−2
>
(
(1− δ)− a2 |U | ab2

) 2
a−2

= (1− δ)a−1|U |b−ab,

we can infer that

I3 =
4(a− 2)

a2b2
|U |2−2b

(xn
ε

) 2
a−2 1

ε2

>
4(a− 2)

a2b2
|U |2−2b(1− δ)a−1|U |b−ab 1

ε2

=
4(a− 2)

a2b2
(1− δ)a−1|U |2−b−ab 1

ε2
.

It follows that

Urr · Unn − |Urn|2

> δ(a− 2) · I2 + I2 + I3

>
((
δ(a− 2) + 1

)8(b− 1)

a2b3
+

4(a− 2)

a2b2
(1− δ)a−1

)
|U |2−b−ab 1

ε2
.

We can proceed as in Case 1 and derive that

H[U ] >
(2

b

)n−2
((
δ(a− 2) + 1

)8(b− 1)

a2b3
+

4(a− 2)

a2b2
(1− δ)a−1

) 1

ε2
.

We remark here that we require(
δ(a− 2) + 1

)8(b− 1)

a2b3
+

4(a− 2)

a2b2
(1− δ)a−1 > 0, (4.5)

which is equivalent to

(a− 2)(1− δ)a−1 > (δ(a− 2) + 1)
(2

b
− 2
)
. (4.6)



EJDE-2021/86 SINGULAR MONGE-AMPÈRE EQUATIONS 11

In fact, we note that α > 1 and a > α+ 1 > 2 lead us to

a− 2 >
a(n+ 1)

n+ α
− 2 =

a(n− 1) + 2a

n+ α
− 2

>
2(n− 1) + 2a

n+ α
− 2 =

2(a+ n− 1)

n+ α
− 2

=
2

b
− 2.

Thus, we can take δ = C(a, α, n) > 0 small enough such that (4.6) holds and hence
(4.5) holds. As a result, we can take ε = C(a, α, n, δ) > 0 sufficiently small such
that

H[U ] > 1 in Ω. (4.7)

Using (4.2), (4.7) and Lemma 2.7, we derive that U is a sub-solution to problem
(1.1).

In view of Theorem 2.4 (comparison principle), we obtain 0 > u(y) > U(y).
Restricting this inequality onto xn-axis, we have

|u(y)| 6 |U(y)| 6
( yn
ε(a, α, n, δ)

) 2
ab

= C(a, η, α, n,diam(Ω))y
2
ab
n

= C(a, η, α, n,diam(Ω))d
2(a+n−1)
a(n+α)
y ,

which leads us to (4.1).

Step 2: Consider the case α+ 1 6 2, i.e. α 6 1. In such a case, we always have

α+ 1 6 2 6 a < +∞.

We can adopt the same procedure as in Case 2 of Step 1 to obtain (4.1).
Up to now, we have proved Theorem 1.2 for the case 2 6 a < +∞. It remains

to prove Theorem 1.2 for the case a = +∞.

5. General bounded convex domains

In this section, we consider Ω as a general bounded convex domain, i.e., (+∞, η)
type domain. We first prove Corollary 1.4 and hence the a = +∞ limit case of
Theorem 1.2. We next apply Corollary 1.4 to provide a proof for the existence
result of solutions on bounded convex domains.

5.1. Global regularity of solution on bounded convex domain. First of all,
we adopt several simplifications:

(i) According to Lemma 2.3, we only need to show that

|u(y)| 6 C(α, n, diam(Ω))d
2

n+α
y , ∀y ∈ Ω.

(ii) For any point y ∈ Ω, letting z ∈ ∂Ω be the nearest boundary point to y, by
some translations and rotations, we can assume that z = 0, 0 ∈ Ω ⊂ Rn+,
and the line yz is the xn-axis with y over the plane z = 0. We remark here
that

dy = dist(y, ∂Ω) = |y − 0| = yn.



12 M. LI EJDE-2021/86

We now construct a sub-solution to (1.1) based on Lemma 2.7. Denote l =
diam(Ω) and let

V (r, xn) = −Mx
2

n+α
n (N2l2 − r2)

1
2 ,

where r = |x′| and M , N are positive constants to be determined such that V is a
sub-solution to the problem (1.1) in Ω. It is clear that V 6 0 on Ω and thus

V 6 u on ∂Ω. (5.1)

By straightforward calculation, we obtain

Vr = Mx
2

n+α
n (N2l2 − r2)−

1
2 r,

Vn = −M 2

n+ α
x

2
n+α−1
n (N2l2 − r2)

1
2 ,

Vrr = MN2l2x
2

n+α
n (N2l2 − r2)−

3
2 ,

Vnn = M
2

n+ α

(
1− 2

n+ α

)
x

2
n+α−2
n (N2l2 − r2)

1
2 ,

Vrn = M
2

n+ α
x

2
n+α−1
n (N2l2 − r2)−

1
2 r.

Therefore,

H[V ] =
(Vr
r

)n−2(
VrrVnn − |Vrn|2

)
|V |α

= Mn+αN2l2
2

n+ α

(
1− (1 + r2N−2l−2)

2

n+ α

)
(N2l2 − r2)

α−n
2 .

Observing that r = |x′| 6 diam(Ω) = l in Ω, we first take N = C(α, n, l) sufficiently
large such that

1− (1 + r2N−2l−2)
2

n+ α
> 0.

Because N2l2− r2 ∈ [(N2−1)l2, N2l2], we take M = C(α, n,N, l) sufficiently large
such that

Mn+αN2l2
2

n+ α

(
1− (1 + r2N−2l−2)

2

n+ α

)
(N2l2 − r2)

α−n
2 > 1.

It follows that

H[V ] > 1 in Ω. (5.2)

This with (5.1), (5.2), and Lemma 2.7 implies that V is a sub-solution to the
problem (1.1).

Using Theorem 2.4 (comparison principle), we obtain

0 > u(y) > V (y).

By taking this inequality on yn-axis, we can summarize that

|u(y)| 6 |V (y)| 6MNly
2

n+α
n = MNld

2
n+α
y .

This completes the proof of Corollary 1.4 as well as the a = +∞ case of Theorem
1.2. Combined with Section 4, we have thus completed the proof of Theorem 1.2.



EJDE-2021/86 SINGULAR MONGE-AMPÈRE EQUATIONS 13

5.2. Existence of solution on bounded convex domain. In fact, the existence
of solution to (1.1) on bounded convex domain can be seen as a particular case of [4,
Theorem 5] and [13, Theorem 1.1]. We state the result as the following theorem and
prove it by using Corollary 1.4. The proof below will serve as a heuristic argument
to the next section dealing with unbounded convex domains.

Theorem 5.1. Suppose Ω ⊂ Rn is a bounded convex domain. Then problem (1.1)

admits a convex solution u ∈ C∞(Ω) ∩ C(Ω). Moreover, u ∈ C
2

n+α (Ω) and

|u|
C

2
n+α (Ω)

6 C(α, n, diam(Ω)).

Proof. Let {Ωi} be a sequence of bounded C2 strictly convex domains such that
Ωi ⊂ Ωi+1 and

⋃∞
i=1 Ωi = Ω. In view of [4, Theorem 5], problem (1.1) admits a

convex solution ui ∈ C∞(Ωi) ∩ C(Ωi) for each Ωi. According to Corollary 1.4, we

have ui ∈ C
2

n+α (Ωi) and

|ui|
C

2
n+α (Ωi)

6 C(α, n, diam(Ωi)).

Let us define ui(x) = 0 for all x ∈ Ω \ Ωi. Then we obtain ui ∈ C
2

n+α (Ω) and the
uniform Hölder estimate

|ui|
C

2
n+α (Ω)

= |ui|
C

2
n+α (Ωi)

6 C(α, n, diam(Ω)).

By Theorem 2.4 (comparison principle) and the proof of Corollary 1.4, we also
obtain the decreasing property

0 > ui(x) > ui+1(x) > V (x), ∀x ∈ Ω.

Using the diagonal technique of choosing subsequence, we obtain that {ui} is lo-
cally uniformly bounded. Due to the convexity of ui and [7, Corollary A.23], it
follows that ui is locally uniformly Lipschitz and thus {ui} is locally equicontinu-
ous. Thanks to Arzela-Ascoli theorem, a subsequence of {ui} (still denoted by {ui})
locally uniformly converges to a convex function u ∈ C(Ω), which also satisfies

|u|
C

2
n+α (Ω)

6 C(α,diam(Ω), n)

and hence u ∈ C
2

n+α (Ω). Moreover, u ∈ C(Ω) is a convex generalized solution to
(1.1) by [22, Lemma 2.2]. Based on Caffarelli’s interior C2,α regularity in [2, 8], we
can derive further regularity by bootstrapping from the equation in (1.1). Repeating
the bootstrap argument, we upgrade the regularity to u ∈ C∞(Ω). The theorem
follows immediately. �

6. Unbounded convex domains

It remains to concentrate on the existence and global regularity result over an
unbounded convex domain Ω. Similar to [11], we first construct sub-solutions to
problem (1.1) over unbounded convex domains, and then the next step is to prove
Theorem 1.5, which can be regarded as an application of Section 5 in spirit.



14 M. LI EJDE-2021/86

6.1. Construction of a sub-solution.

Lemma 6.1. Denote x = (x′, xn) and r = |x′|. If

Ω =
{

(x′, xn) ∈ Rn : xn >

√( 2n

n+ α

)nα− n
n+ α

r
}
,

then

W (x) = −
(( xn√(

2n
n+α

)n α−n
n+α

)2

− r2
) n
n+α

is a solution to (1.1) on Ω.

Proof. It is straightforward to compute that

Wr =
2n

n+ α

(( xn√(
2n
n+α

)n α−n
n+α

)2

− r2
) n
n+α−1

r,

Wn = −
2n
n+α(

2n
n+α

)n α−n
n+α

(( xn√(
2n
n+α

)n α−n
n+α

)2

− r2
) n
n+α−1

xn,

Wrr =
4nα

(n+ α)2

(( xn√(
2n
n+α

)n α−n
n+α

)2

− r2
) n
n+α−2

r2

+
2n

n+ α

(( xn√(
2n
n+α

)n α−n
n+α

)2

− r2
) n
n+α−1

,

Wnn =

4nα
(n+α)2((

2n
n+α

)n α−n
n+α

)2(( xn√(
2n
n+α

)n α−n
n+α

)2

− r2
) n
n+α−2

x2
n

−
2n
n+α(

2n
n+α

)n α−n
n+α

(( xn√(
2n
n+α

)n α−n
n+α

)2

− r2
) n
n+α−1

,

Wrn = −
4nα

(n+α)2(
2n
n+α

)n α−n
n+α

(( xn√(
2n
n+α

)n α−n
n+α

)2

− r2
) n
n+α−2

rxn,

which gives rise to

WrrWnn − |Wrn|2 =

4n2

(n+α)2(
2n
n+α

)n(( xn√(
2n
n+α

)n α−n
n+α

)2

− r2
) 2n
n+α−2

.

Putting this expression into the formula of H[·] in Lemma 2.7, we finally verify that

H[W ] =
(Wr

r

)n−2(
WrrWnn − |Wrn|2

)
|W |α = 1.

The proof is complete. �

Using Lemma 6.1, we can construct sub-solutions over unbounded convex do-
mains as follows.



EJDE-2021/86 SINGULAR MONGE-AMPÈRE EQUATIONS 15

Lemma 6.2. Suppose Ω is an unbounded convex domain in Rn such that ∂Ω is
strictly convex at some point x0 ∈ ∂Ω. Then

W (x) = −
(( xn√(

2n
n+α

)n α−n
n+α

)2

− r2
) n
n+α

is a sub-solution to (1.1) on Ω.

Proof. Since ∂Ω is strictly convex at some point x0 ∈ ∂Ω, then there exists a
tangent plane P of ∂Ω at x0 such that P ∩ ∂Ω = {x0}. By some translations and
rotations, we can assume x0 = 0 and P is given by the equation xn = 0. Without
loss of generality, we further assume that ∂Ω near the origin can be expressed as
xn = ϕ(x′) which is the graph of a function over the tangent plane xn = 0.

We use D to denote the largest domain such that ϕ is well-defined. Under these
assumptions, we have ϕ(0) = 0 and the origin is the lowest point of Ω, and thus
ϕ(x′) > 0 for all x′ ∈ ∂D. Moreover, for any small number δ > 0, we have

{x ∈ Rn−1 : |x′| < δ} b D,
h(δ) := min{ϕ(x′) : |x′| = δ} > 0.

We are now in a position to show that

Ω ⊂ Σ :=
{

(x′, xn) : xn >
h(δ)

δ
|x′| − h(δ)

}
. (6.1)

For any (x′, xn) ∈ ∂Ω, we can distinguish the following two cases to prove this
claim:

(i) When |x′| 6 δ, we can derive xn = ϕ(x′) > 0 > h(δ)
δ |x

′| − h(δ) and thus
(x′, xn) ∈ Σ.

(ii) When |x′| > δ, the convexity of Ω gives rise to

ϕ(x′)− ϕ( δ
|x′|x

′)

|x′ − δ
|x′|x

′|
>
ϕ( δ
|x′|x

′)− ϕ(0)

| δ|x′|x′ − 0|
,

which implies that

xn = ϕ(x′)

> ϕ(
δ

|x′|
x′) +

ϕ( δ
|x′|x

′)

δ
(|x′| − δ)

> h(δ) +
h(δ)

δ
(|x′| − δ)

=
h(δ)

δ
|x′|

>
h(δ)

δ
|x′| − h(δ).

Thus we obtain (x′, xn) ∈ Σ in such a case.

Summing up, for any (x′, xn) ∈ ∂Ω, we always have (x′, xn) ∈ Σ, namely ∂Ω ⊂ Σ.
We note that the lowest point of Ω, i.e., the origin, is in Σ. Thus by the convexity
of Ω, we obtain that Ω ⊂ Σ.

Now we take a linear transformation T : (x′, xn)→ (x̃′, x̃n) as follows:

x̃′ = x′,



16 M. LI EJDE-2021/86

x̃n =
δ

h(δ)

√( 2n

n+ α

)nα− n
n+ α

(xn + h(δ)) .

It follows that

TΣ =
{

(x̃′, x̃n) ∈ Rn : x̃n >

√( 2n

n+ α

)nα− n
n+ α

|x̃′|
}
.

We note that (6.1) gives TΩ ⊂ TΣ. Without loss of generality, by Lemma 2.2, we
can assume Ω ⊂ TΣ. According to Lemma 6.1, we obtain that

W (x) = −
(( xn√(

2n
n+α

)n α−n
n+α

)2

− r2
) n
n+α

is a solution to (1.1) on TΣ. Therefore it is a sub-solution to (1.1) on Ω as a result
of Ω ⊂ TΣ. The proof is complete. �

6.2. Proof of Theorem 1.5. Let {Ωi} be a sequence of bounded convex domains
such that Ωi ⊂ Ωi+1 and

⋃∞
i=1 Ωi = Ω. According to Theorem 5.1, the problem

(1.1) admits a convex solution ui ∈ C∞(Ωi)∩C(Ωi) for each Ωi. Corollary 1.4 also

gives rise to ui ∈ C
2

n+α (Ωi) as well as

|ui|
C

2
n+α (Ωi)

6 C(α, n, diam(Ωi)).

Define ui(x) = 0 for all x ∈ Rn \ Ωi. For any r > 0, we further obtain ui ∈
C

2
n+α (Ω ∩Br(0)) and the uniform Hölder estimate

|ui|
C

2
n+α (Ω∩Br(0))

= |ui|
C

2
n+α (Ωi∩Br(0))

6 C(α, n, diam(Ω ∩Br(0))).

Thanks to Theorem 2.4 (comparison principle) and Lemma 6.2, we also derive the
decreasing property

0 > ui(x) > ui+1(x) >W (x), ∀x ∈ Ω.

The diagonal technique of choosing subsequence leads us to the conclusion that {ui}
is locally uniformly bounded. By the convexity of ui and [7, Corollary A.23], we infer
that all ui are locally uniformly Lipschitz and thus {ui} is locally equicontinuous.
By Arzela-Ascoli theorem, a subsequence of {ui} (still denoted by {ui}) locally

uniformly converges to a convex function u ∈ C(Ω ∩Br(0)), which also satisfies

|u|
C

2
n+α (Ω∩Br(0))

6 C(α, n, diam(Ω ∩Br(0)))

and therefore u ∈ C
2

n+α (Ω ∩Br(0)). Since r > 0 is arbitrary, we can derive u ∈
C(Ω). In fact, if there exists a point y ∈ Ω such that u is not continuous at y,
then we will always find a sufficiently large r′ > 0 such that y ∈ Br′(0) and hence

y ∈ Ω ∩Br′(0), which contradicts u ∈ C(Ω ∩Br′(0)). Moreover, u ∈ C(Ω) is a
convex generalized solution to (1.1) by [22, Lemma 2.2]. Using Caffarelli’s interior
C2,α regularity in [2, 8], we can obtain further regularity by a bootstrap argument
from the equation in (1.1). Repeating the bootstrap argument, we can upgrade the
regularity to u ∈ C∞(Ω). This completes the proof.



EJDE-2021/86 SINGULAR MONGE-AMPÈRE EQUATIONS 17

Acknowledgments. The author would like to thank Profs. Huaiyu Jian and Pin
Yu for their helpful suggestions and encouragement along the way. Sincere thanks
also go to Drs. You Li and Haoyu Wang for providing useful comments and long-
term help. The author is also grateful to the anonymous referees for offering helpful
suggestions to improve the paper.

The major part of this paper was carried out while the author was participating
in summer social practice at Institute for Electronics and Information Technology
in Tianjin, Tsinghua University. The author acknowledges IEIT for the warm and
peaceful atmosphere that allows her to calm down in spare time and helps provoke
her thinking.

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Mengni Li
Department of Mathematical Sciences and Yau Mathematical Sciences Center, Ts-

inghua University, Beijing 100084, China
Email address: krisymengni@163.com


	1. Introduction
	2. Preliminaries
	2.1. Useful observations on convexity
	2.2. Equivalent conditions of sub-solution

	3. Bounded convex domains satisfying exterior sphere condition
	4. (a,) type domains with 2a<+
	5. General bounded convex domains
	5.1. Global regularity of solution on bounded convex domain
	5.2. Existence of solution on bounded convex domain

	6. Unbounded convex domains
	6.1. Construction of a sub-solution
	6.2. Proof of Theorem ??
	Acknowledgments

	References