# A Remark on the Continuous Subsolution Problem for the Complex Monge-Ampère Equation

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## Abstract

We prove that if the modulus of continuity of a plurisubharmonic subsolution satisfies a Dini-type condition then the Dirichlet problem for the complex Monge-Ampère equation has the continuous solution. The modulus of continuity of the solution also given if the right hand side is locally dominated by capacity.

## Keywords

Dirichlet problem Complex Monge-Ampère equation Weak solutions Subsolution problem## Mathematics Subject Classification (2010)

32W20 32U40## 1 Introduction

*ψ*be a continuous function on the boundary of

*Ω*. We look for the solution to the equation:

*PSH*stands for plurisubharmonic functions, and \(d^{c} = i (\overline {\partial } -\partial )\). It was shown in [9] and [10] that for the measures satisfying certain bound in terms of the Bedford-Taylor capacity [4], the Dirichlet problem has a (unique) solution. The precise statement is as follows.

*h*is admissible, then so is

*A*

*h*for any number

*A*> 0. Define

*F*

_{h}(

*x*) a Borel measure

*μ*satisfies

*E*⊂

*Ω*. Then, by [9] the Dirichlet problem (1.1) has a solution.

*μ*has density with respect to the Lebesgue measure in

*L*

^{p},

*p*> 1 then this bound is satisfied [9]. By the recent results in [12, 13] if

*μ*is bounded by the Monge-Ampère measure of a Hölder continuous plurisubharmonic function

*φ*

*h*, and consequently, the Dirichlet problem (1.1) is solvable with Hölder continuous solution. The main result in this paper says that we can considerably weaken the assumption on

*φ*and still get a continuous solution of the equation.

*φ*on \(\bar {\varOmega }\), i.e.,

*φ*(

*z*) −

*φ*(

*w*)|≤

*ϖ*(|

*z*−

*w*|) for every \( z,w\in \bar {\varOmega }\). Let us state the first result.

### **Theorem 1.1**

*Let*\(\varphi \in PSH({\varOmega }) \cap C^{0}(\bar {\varOmega }), \)

*φ*= 0

*on*

*∂*

*Ω*

*.*

*Assume that its modulus of continuity satisfies the Dini type condition*

*μ*satisfies

*μ*≤ (

*d*

*d*

^{c}

*φ*)

^{n}in

*Ω*, then the Dirichlet problem (1.1) admits a unique solution.

Let us mention in this context that it is still an open problem if a continuous subsolution *φ* implies the solvability of (1.1).

The modulus of continuity of the solution to the Dirichlet problem (1.1) was obtained in [3] for *μ* = *f**d**V*_{2n} with *f*(*x*) being continuous on \(\bar {\varOmega }\). We also wish to study this problem for the measures which satisfy the inequality (1.2). For simplicity, we restrict ourselves to measures belonging to \(\mathcal H(\alpha ,{\varOmega })\). In other words, we take the function *h*(*x*) = *C**x*^{nα} for positive constants *C*,*α* > 0 in the inequality (1.2).

We introduce the following notion, which generalizes the one in [8]. Consider a continuous increasing function \(F_{0}:[0,\infty ) \to [0,\infty )\) with *F*(0) = 0.

### **Definition 1.2**

*μ*is called uniformly locally dominated by capacity with respect to

*F*

_{0}if for every cube

*I*(

*z*,

*r*) =:

*I*⊂

*B*

_{I}:=

*B*(

*z*,2

*r*) ⊂⊂

*Ω*and for every set

*E*⊂

*I*,

According to [1], the Lebesgue measure *d**V*_{2n} satisfies this property with \(F_{0} = C_{\alpha } \exp (-\alpha / x^{-1/n})\) for every 0 < *α* < 2*n*. The case *F*_{0}(*x*) = *C**x* was considered in [8]. We refer the reader to [5] for more examples of measures satisfying this property. Here is our second result.

### **Theorem 1.3**

*Assume*\(\mu \in \mathcal H(\alpha ,{\varOmega })\)

*with*

*compact support and satisfying the condition*(1.4)

*for some*

*F*

_{0}

*. Then,*

*the modulus of continuity of the solution*u

*of the Dirichlet problem*(1.1)

*satisfies for*0 <

*δ*<

*R*

_{0}

*and*2

*R*

_{0}= dist(supp

*μ*,

*∂*

*Ω*) > 0

*,*

*C*,

*α*

_{1}depend only on

*α*,

*μ*,

*Ω*.

## 2 Preliminaries

*K*in a domain \({\varOmega }\subset \mathbb {C}^{n}, \) its relative extremal function

*u*

_{K}is given by

*u*

_{K}is continuous, we call

*K*a regular set. It is easy to see that the

*𝜖*-envelope

*K*is regular, and thus any compact set can be approximated from above by regular compact sets.

*K*with respect to

*Ω*(now usually called the Bedford-Taylor capacity) is defined by the formula

*μ*belongs to \(\mathcal H(\alpha ,{\varOmega })\),

*α*> 0, if there exists a uniform constant

*C*> 0 such that for every compact set

*E*⊂

*Ω*,

## 3 Proof of Theorem 1.1

In this section, we shall prove Theorem 1.1. We need the following lemma. The proof of this lemma is based on a similar idea as the one in [11, Lemma 3.1] where the complex Hessian equation is considered. The difference is that we have much stronger volume-capacity inequality for the Monge-Ampère equation.

### **Lemma 3.1**

*Assume the measure*

*μ*

*is*

*compactly supported. Fix*0 <

*α*< 2

*n*

*and*

*τ*=

*α*/(2

*n*+ 1)

*.*

*There exists a uniform constant*C

*such that for every compact set*

*K*⊂

*Ω*

*,*

*K*) := cap(

*K*,

*Ω*).

### *Proof*

*K*⊂⊂

*Ω*. Without loss of generality, we may assume that

*K*is regular. Denote by

*φ*

_{ε}the standard regularization of

*φ*in the terminology of [10]. We choose

*ε*> 0 so small that

*Ω*

_{ε}= {

*z*∈

*Ω*: dist(

*z*,

*∂*

*Ω*) >

*ε*}. Since for every \(K \subset {\varOmega }^{\prime \prime }\) we have

*C*

_{0}depending only on \(\varOmega , {\varOmega }^{\prime }\)) in what follows we shall write cap(

*K*) for either one of these capacities. We have

*u*

_{K}be the relative extremal function of

*K*with respect to \({\varOmega }^{\prime }\). Consider the set \(K^{\prime } = \{ 3\delta u_{K} + \varphi _{\varepsilon } < \varphi - 2\delta \}\). Then,

*ε*is so small that it satisfies (3.2), otherwise the inequality (3.1) holds true by increasing the constant) and plug in the formula for

*δ*to get that

*μ*once we have

*s*= 1/

*x*, and then

*t*=

*e*

^{−τ/s}, this is equivalent to

*ϖ*(

*t*) implies that

*h*is admissible in the case of

*μ*with compact support. Then, by [10, Theorem 5.9] the Dirichlet problem (1.1) has a unique solution.

*Ω*by compact sets

*μ*

_{j}to be the restriction of

*μ*to

*E*

_{j}. Denote by

*u*

_{j}the solution of (1.1) with

*μ*replaced by

*μ*

_{j}. By the comparison principle

*u*

_{j}tends to \(u=\lim u_{j}\) uniformly and the continuity of

*u*follows. The proof is complete.

## 4 The Modulus of Continuity of Solutions

In this section, we study the modulus of continuity of the solution of the Dirichlet problem with the right hand side in the class \(\mathcal H(\alpha ,{\varOmega })\) under the additional condition that a given measure is locally dominated by capacity.

In what follows we need [8, Lemma 2] whose proof is based on the lemma due to Alexander and Taylor [2, Lemma 3.3]. For the reader’s convenience, we give the proofs. The latter can be simplified by using the Błocki inequality [6].

### **Lemma 4.1**

*Let*\(B^{\prime } = \{|z-z_{0}| <r \} \subset \subset B= \{|z-z_{0}| <R\}\)

*be two concentric balls centered at*

*z*

_{0}

*in*\(\mathbb C^{n}\)

*.*

*Let*\(u \in PSH(B) \cap L^{\infty }(B)\)

*with*

*u*< 0

*. There is a*

*constant*\(C = C(n, \frac {R}{r})\)

*independent of*u

*such*

*that*

*R*/

*r*= 3 then the constant

*C*depends only on

*n*.

### *Proof*

*z*

_{0}= 0. Set

*ρ*:= (

*r*+

*R*)/2 and

*B*(

*ρ*) = {|

*z*−

*z*

_{0}| <

*ρ*}. We use the Błocki inequality [6] for

*v*(

*z*) = |

*z*|

^{2}−

*ρ*

^{2}and

*β*:=

*d*

*d*

^{c}

*v*=

*d*

*d*

^{c}|

*z*|

^{2}, to get

*σ*

_{2n− 1}is the area of the unit sphere,

*n*(

*t*)/

*t*

^{2n− 2}is increasing, we have

*u*< 0, it follows that

*N*(

*R*) < −

*u*(0). Hence,

*R*= 3

*r*, then

*C*is also independent of

*r*. □

### **Lemma 4.2**

*Denote for*

*ρ*≥ 0

*,*

*B*

_{ρ}= {|

*z*−

*z*

_{0}| <

*e*

^{ρ}

*R*

_{0}}.

*Given*

*z*

_{0}∈

*Ω*

*and two*

*numbers*

*R*> 1

*,*

*R*

_{0}> 0

*such that*

*B*

_{M}⊂⊂

*Ω*,

*and given*

*v*∈

*P*

*S*

*H*(

*Ω*)

*such that*− 1 <

*v*< 0

*, denote by*E

*the*

*set*

*δ*∈ (0,1). Then, there exists

*C*

_{0}depending only on

*n*such that

### *Proof*

*z*∈

*B*

_{R}∖

*B*

_{0}and \(a_{0}:= \sup _{B_{0}} v\) we have

*E*with respect to

*B*

_{2}. One has

*z*

_{1}∈

*B*̣

_{0}, we have

*E*⊂{|

*z*−

*z*

_{1}| < 2

*R*

_{0}}⊂

*B*

_{2}. Therefore, Lemma 4.1 gives

*u*. Define for

*δ*> 0 small

*z*∈

*Ω*

_{δ}set

*δ*

_{0}> 0 such that

*z*∈

*∂*

*Ω*

_{δ}and 0 <

*δ*<

*δ*

_{0}. Here, we used the result of Bedford and Taylor [3, Theorem 6.2] (with minor modifications) to extend

*ψ*plurisubharmonically onto

*Ω*so that its modulus of continuity on \(\bar {\varOmega }\) is controlled by the one on the boundary. Therefore, for a suitable extension of

*u*

_{δ}to

*Ω*, using the stability estimate for measure in \(\mathcal H(\alpha ,{\varOmega })\) as in [7, Theorem 1.1] (see also [12, Proposition 2.10]), we get

### **Lemma 4.3**

*There are uniform constants*

*C*,

*α*

_{1}

*depending*

*only on*

*Ω*,

*α*,

*μ*

*such*

*that*

*δ*<

*δ*

_{0}.

Thanks to this lemma, we know that the right hand side tends to zero as *δ* decreases to zero. We shall use the property “locally dominated by capacity” to obtain a quantitative bound via Lemma 4.2.

*μ*by

*K*. Since \(\|u\|_{\infty }\) is controlled by a constant

*C*=

*C*(

*α*,

*Ω*,

*μ*), without loss of generality, we may assume that

*ε*< 1

*Ω*⊂⊂ [0,1]

^{2n}. Let us write \(z = (x^{1}, \dots ,x^{2n}) \in \mathbb R^{2n}\) and denote the semi-open cube centered at a point

*z*

_{0}, of diameter 2

*r*by

*μ*satisfies for every cube

*E*⊂

*I*, the inequality

*F*

_{0}(0) = 0.

^{2ns}congruent cubes of diameter 3

^{−s}= 2

*δ*, where \( s \in \mathbb N\). Then

*I*

_{s}=

*I*(

*z*

_{s},

*δ*) and \(B_{I_{s}} = B(z_{s}, 2\delta )\) for some

*z*

_{s}∈

*I*

_{0}. Hence,

*r*= 2

*δ*and

*R*= 2

*R*

_{0}, we have for

*B*

_{s}:=

*B*(

*z*

_{s},4

*δ*) corresponding to each cube

*I*

_{s}

*R*

_{0}= dist(

*K*,

*∂*

*Ω*). Therefore, combining the above inequalities, we get that

## Notes

### Acknowledgements

The first author was partially supported by NCN grant 2017/27/B/ST1/01145. The second author was supported by the NRF Grant 2011-0030044 (SRC-GAIA) of The Republic of Korea. He also would like to thank Kang-Tae Kim for encouragement and support.

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