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

## 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.

## Introduction

In this note, we consider the Dirichlet problem for the complex Monge-Ampère equation in a strictly pseudoconvex domain $${\varOmega }\subset \mathbb {C}^{n}$$. Let ψ be a continuous function on the boundary of Ω. We look for the solution to the equation:

$$\begin{array}{@{}rcl@{}} && u\in PSH({\varOmega}) \cap C^{0}(\bar{\varOmega}), \\ && (dd^{c} u)^{n} = d\mu, \\ && u = \psi \quad \text{on }\partial\varOmega. \end{array}$$
(1.1)

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

Let $$h :\mathbb R_{+} \rightarrow (0, \infty )$$ be an increasing function such that

$${\int}_{1}^{\infty} \frac{1}{x [h(x) ]^{\frac{1}{n}} } dx < +\infty.$$

We call such a function admissible. If h is admissible, then so is Ah for any number A > 0. Define

$$F_{h}(x) = \frac{x}{h(x^{-\frac{1}{n}})}.$$

Suppose that for such a function Fh(x) a Borel measure μ satisfies

$${\int}_{E} d\mu \leq F_{h}(\text{cap}(E))$$
(1.2)

for any Borel set EΩ. Then, by  the Dirichlet problem (1.1) has a solution.

This statement is useful as long as we can verify the condition (1.2). In particular, if μ has density with respect to the Lebesgue measure in Lp, p > 1 then this bound is satisfied . By the recent results in [12, 13] if μ is bounded by the Monge-Ampère measure of a Hölder continuous plurisubharmonic function φ

$$\mu \leq (dd^{c}\varphi)^{n} \quad \text{in }{\varOmega} ,$$

then (1.2) holds for a specific 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.

Let $$\varpi (t):= \varpi (t;\varphi ,\bar {\varOmega })$$ denote the modulus of continuity of φ on $$\bar {\varOmega }$$, i.e.,

$$\varpi(t)= \sup \left\{|\varphi(z) -\varphi(w)| : z,w \in \bar{\varOmega}, |z-w| \leq t\right\}.$$

Thus |φ(z) − φ(w)|≤ ϖ(|zw|) 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 onΩ.Assume that its modulus of continuity satisfies the Dini type condition

$${{\int}_{0}^{1}} \frac{[\varpi(t)]^{\frac{1}{n}}}{t |\log t|} dt <+\infty.$$
(1.3)

If the measure μ satisfies μ ≤ (ddcφ)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  for μ = fdV2n 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) = Cxnα for positive constants C,α > 0 in the inequality (1.2).

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

### Definition 1.2

The measure μ is called uniformly locally dominated by capacity with respect to F0 if for every cube I(z,r) =: IBI := B(z,2r) ⊂⊂Ω and for every set EI,

$$\mu(E) \leq \mu(I) F_{0}\left( \text{cap} (E, B_{I}) \right).$$
(1.4)

According to , the Lebesgue measure dV2n satisfies this property with $$F_{0} = C_{\alpha } \exp (-\alpha / x^{-1/n})$$ for every 0 < α < 2n. The case F0(x) = Cx was considered in . We refer the reader to  for more examples of measures satisfying this property. Here is our second result.

### Theorem 1.3

Assume$$\mu \in \mathcal H(\alpha ,{\varOmega })$$withcompact support and satisfying the condition (1.4) for someF0. Then,the modulus of continuity of the solution u of the Dirichlet problem(1.1) satisfies for0 < δ < R0and2R0 = dist(suppμ,Ω) > 0,

$$\varpi(\delta;u,{\varOmega}) \leq \varpi(\delta;\psi,\partial{\varOmega}) + C \left[ \left( \log \frac{R_{0}}{\delta}\right)^{-\frac{1}{2}} + F_{0} \left( \frac{C_{0}}{[\log (R_{0}/\delta)]^{\frac{1}{2}}}\right)\right]^{\alpha_{1}},$$

where the constants C,α1 depend only on α,μ,Ω.

## Preliminaries

Here, we gather some basic facts from pluripotential theory taken from , and used in the sequel. Given a compact set K in a domain $${\varOmega }\subset \mathbb {C}^{n},$$ its relative extremal function uK is given by

$$u_{K} = \sup \{ u\in PSH ({\varOmega} ) : u<0, u\leq -1 \text{ on } K\}.$$

Its upper semicontinuous regularization $$u_{K}^{\ast }$$ is plurisubharmonic. When uK is continuous, we call K a regular set. It is easy to see that the 𝜖-envelope

$$K_{\epsilon } =\{ z : \operatorname{dist} (z, K) \leq \epsilon \}$$

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

The relative capacity of a compact set K with respect to Ω (now usually called the Bedford-Taylor capacity) is defined by the formula

$$\text{cap} (K, {\varOmega} ) = \sup \left\{ {\int}_{K} (dd^{c} u )^{n} : u\in PSH ({\varOmega} ), -1\leq u\leq 0 \right\},$$

and by , can be expressed as

$$\text{cap} (E, {\varOmega} ) = {\int}_{K} (dd^{c} u_{K}^{\ast } )^{n} .$$

We say that a positive Borel measure μ belongs to $$\mathcal H(\alpha ,{\varOmega })$$, α > 0, if there exists a uniform constant C > 0 such that for every compact set EΩ,

$$\mu(E) \leq C \left[\text{cap} (E,{\varOmega}) \right]^{1+ \alpha}.$$

## 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 measureμiscompactly supported. Fix 0 < α < 2nandτ = α/(2n + 1).There exists a uniform constant C such that for every compact setKΩ,

$$\mu(K) \leq C \left\{ \varpi\left( \exp \left( \frac{-\tau}{2[\text{cap}(K)]^{\frac{1}{n}}}\right) \right) + \exp\left( \frac{2n\tau -\alpha}{2[\text{cap}(K)]^{\frac{1}{n}}} \right)\right\} \cdot \text{cap}(K),$$
(3.1)

where cap(K) := cap(K,Ω).

### Proof

Fix a compact subset K ⊂⊂Ω. Without loss of generality, we may assume that K is regular. Denote by φε the standard regularization of φ in the terminology of . We choose ε > 0 so small that

$$\operatorname{supp} \mu \subset {\varOmega}^{\prime\prime} \subset\subset {\varOmega}^{\prime}\subset {\varOmega}_{\varepsilon} \subset \varOmega,$$

where Ωε = {zΩ : dist(z,Ω) > ε}. Since for every $$K \subset {\varOmega }^{\prime \prime }$$ we have

$$C_{0} \text{cap}(K,{\varOmega}) \leq \text{cap}(K, {\varOmega}^{\prime}) \leq C_{0}^{-1} \text{cap}(K,{\varOmega})$$

(for a constant C0 depending only on $$\varOmega , {\varOmega }^{\prime }$$) in what follows we shall write cap(K) for either one of these capacities. We have

$$0\leq \varphi_{\varepsilon} - \varphi \leq \varpi(\varepsilon) := \delta \quad \text{on } {\varOmega}^{\prime}.$$

Let uK 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,

$$K \subset K^{\prime} \subset \left\{u_{K} < -\frac{1}{2} \right\} \subset {\varOmega}^{\prime}.$$
(3.2)

Hence, by the comparison principle ,

$$\text{cap}(K^{\prime}) \leq 2^{n} \text{cap}(K).$$
(3.3)

Note that

$$dd^{c} \varphi_{\varepsilon} \leq \frac{C}{\varepsilon^{2}} dd^{c} |z|^{2}, \quad \|\varphi_{\varepsilon} + u_{K}\|_{\infty}=:M \leq \|\varphi\|_{\infty} +1.$$
(3.4)

The comparison principle, the bounds (3.4), and the volume-capacity inequality from [1, Theorem A] (in the last inequality below) give the following:

$$\begin{array}{@{}rcl@{}} {\int}_{K^{\prime}} (dd^{c} \varphi )^{n} &\leq& {\int}_{K^{\prime}} (dd^{c} (3\delta u_{K} + \varphi_{\varepsilon}) )^{n} \\ &\leq& 3\delta{\int}_{K^{\prime}} \left[dd^{c} (u_{K} + \varphi_{\varepsilon})\right]^{n} + {\int}_{K^{\prime}} (dd^{c} \varphi_{\varepsilon} )^{n} \\ &\leq& 3\delta M^{n} \text{cap}(K^{\prime}) + C(\alpha) \varepsilon^{-2n} \exp\left( \frac{-\alpha}{[\text{cap}(K^{\prime})]^{\frac{1}{n}}} \right) \text{cap}(K^{\prime}). \end{array}$$

Choose

$$\varepsilon = \exp\left( \frac{-\tau}{[\text{cap}(K^{\prime})]^{\frac{1}{n}}} \right)$$

(we assume that 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

$$\begin{array}{@{}rcl@{}} \mu(K) &&\leq {\int}_{K^{\prime}} (dd^{c} (\varphi) )^{n} \\ &&\leq 3 M^{n} \varpi\left( \exp \left( \frac{-\tau}{[\text{cap}(K^{\prime})]^{\frac{1}{n}}}\right) \right) \cdot \text{cap}(K^{\prime}) + C \exp\left( \frac{2n\tau -\alpha}{[\text{cap}(K^{\prime})]^{\frac{1}{n}}} \right). \end{array}$$

This combined with (3.3) gives the desired inequality. □

We are ready to finish the proof of the theorem. It follows from Lemma 3.1 that

$$h (x)= \frac{1}{C \varpi(\exp (-\tau x))}$$

is a function which satisfies (1.2) for the measure μ once we have

$${\int}_{1}^{\infty} \frac{1}{x [h(x) ]^{\frac{1}{n}} } dx < +\infty.$$

By changing the variable s = 1/x, and then t = eτ/s, this is equivalent to

$${\int}_{0}^{e^{-\tau}} \frac{\left[\varpi(t) \right]^{\frac{1}{n}}}{t |\log t|} dt <+\infty.$$

The last inequality is guaranteed by (1.3). Thus, our assumption on the modulus of continuity 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.

To deal with the general case, consider the exhaustion of Ω by compact sets

$$E_{j} =\{ \varphi \leq -1/j \}$$

and define μj to be the restriction of μ to Ej. Denote by uj the solution of (1.1) with μ replaced by μj. By the comparison principle

$$u_{j} + \max (\varphi , -1/j ) \leq u \leq u_{j} ,$$

and so the sequence uj tends to $$u=\lim u_{j}$$ uniformly and the continuity of u follows. The proof is complete.

## 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 .

### Lemma 4.1

Let$$B^{\prime } = \{|z-z_{0}| <r \} \subset \subset B= \{|z-z_{0}| <R\}$$be two concentric balls centered atz0in$$\mathbb C^{n}$$.Let$$u \in PSH(B) \cap L^{\infty }(B)$$withu < 0. There is aconstant$$C = C(n, \frac {R}{r})$$independent of u suchthat

$${\int}_{B^{\prime}} (dd^{c} u)^{n} \leq C |u(z_{0})| \sup_{z\in B} |u(z)|^{n-1}.$$

In particular, if R/r = 3 then the constant C depends only on n.

### Proof

Without loss of generality, we may assume z0 = 0. Set ρ := (r + R)/2 and B(ρ) = {|zz0| < ρ}. We use the Błocki inequality  for v(z) = |z|2ρ2 and β := ddcv = ddc|z|2, to get

$$\begin{array}{@{}rcl@{}} {\int}_{B^{\prime}} (dd^{c} u)^{n} &&\leq \frac{1}{(\rho^{2} -r^{2})^{n-1}}{\int}_{B(\rho)} |v|^{n-1} (dd^{c} u)^{n} \\ &&\leq \frac{(n-1)! \|u\|_{B_{\rho}}^{n-1}}{(\rho^{2} -r^{2})^{n-1}} {\int}_{B(\rho)} dd^{c} u \wedge \beta^{n-1}. \end{array}$$

By Jensen’s formula

$$u(0) + N(\rho) = \frac{1}{\sigma_{2n-1}} {\int}_{\{|\zeta|=1\}} u(\rho \zeta) d\sigma(\zeta),$$

where σ2n− 1 is the area of the unit sphere,

$$N(\rho) = {\int}_{0}^{\rho} \frac{n(t)}{t^{2n-1}} dt$$

and

$$n(t) = \frac{1}{\sigma_{2n-1}} {\int}_{\{|z| \leq t\}} {\varDelta} u(z) dV_{2n}(z) = a_{n} {\int}_{\{|z| \leq t\}} dd^{c} u \wedge \beta^{n-1}.$$

Since n(t)/t2n− 2 is increasing, we have

$$N(R) \geq {\int}_{\rho}^{R} \frac{n(t)}{t^{2n-1}} dt \geq \frac{n(\rho)}{\rho^{2n-2}} \log (R/\rho).$$

From u < 0, it follows that N(R) < −u(0). Hence,

$${\int}_{B_{\rho}} dd^{c}u \wedge \beta^{n-1} \leq \frac{n(\rho)}{a_{n}} \leq \frac{N(R)\rho^{n-2}}{\log(R/\rho)} \leq \frac{\rho^{2n-2} |u(0)|}{\log(R/\rho)}.$$

Combining the above inequalities, we get the desired estimate with the constant

$$C = \frac{(n-1)!\rho^{2n-2}}{(\rho^{2}-r^{2})^{n-1} \log (R/\rho)}.$$

If R = 3r, then C is also independent of r. □

### Lemma 4.2

Denote forρ ≥ 0,Bρ = {|zz0| < eρR0}. Givenz0Ωand twonumbersR > 1,R0 > 0 such thatBM ⊂⊂Ω, and givenvPSH(Ω) such that− 1 < v < 0, denote by E theset

$$E=E(\delta) = \left\{z \in B_{0} : (1-\delta)v \leq \sup_{B_{0}} v\right\},$$

where δ ∈ (0,1). Then, there exists C0 depending only on n such that

$$\text{cap}(E, B_{2}) \leq \frac{C_{0}}{R\delta}.$$

### Proof

From the logarithmic convexity of the function $$r \mapsto \sup _{|z-z_{0}|<r} v(z)$$ it follows that for zBRB0 and $$a_{0}:= \sup _{B_{0}} v$$ we have

$$v(z) \leq a_{0} \left( 1 - \frac{1}{R} \log\frac{|z-z_{0}|}{R_{0}}\right).$$

Hence,

$$a := \sup_{B_{2}} v \leq a_{0}\left( 1 - \frac{2}{R}\right).$$

Let $$u = u_{E,B_{2}}$$ the relative extremal function of E with respect to B2. One has

$$\frac{v-a}{a - a_{0}/(1-\delta)} \leq u.$$

So, for some z1Ḅ0, we have

$$u(z_{1}) \geq \frac{a_{0} - a}{a - a_{0}/(1-\delta)} \geq \frac{2(\delta -1)}{(M-2)\delta + 2}.$$

Note that E ⊂{|zz1| < 2R0}⊂ B2. Therefore, Lemma 4.1 gives

$$\text{cap}(E,B_{2}) = {\int}_{\{|z-z_{1}| < 6R_{0}\}} (dd^{c} u)^{n} \leq C_{0} \|u\|_{B_{2}}^{n-1} |u(z_{1})| \leq \frac{C_{0}}{R \delta}.$$

This is the desired inequality. □

Let us proceed with the proof of Theorem 1.3. Since $$\mu \in \mathcal H(\alpha ,{\varOmega })$$, according to  and [10, Theorem 5.9] we can solve the Dirichlet problem (1.1) to obtain a unique continuous solution u. Define for δ > 0 small

$${\varOmega}_{\delta} := \left\{z \in {\varOmega} : \operatorname{dist}(z, \partial {\varOmega}) > \delta\right\};$$

and for zΩδ set

$$u_{\delta}(z) := \sup_{|\zeta| \leq \delta} u(z+ \zeta).$$

Thanks to the arguments in [12, Lemma 2.11] it is easy to see that there exists δ0 > 0 such that

$$u_{\delta}(z) \leq u(z) + \varpi(\delta;\psi,\partial{\varOmega})$$
(4.1)

for every zΩδ and 0 < δ < δ0. 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 uδ 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 constantsC,α1dependingonly onΩ,α,μsuchthat

$$\sup_{{\varOmega}_{\delta}} (u_{\delta} - u) \leq \varpi(\delta; \psi, \partial{\varOmega}) + C \left( {\int}_{{\varOmega}_{\delta}} (u_{\delta} -u) d\mu \right)^{\alpha_{1}}$$

for every 0 < δ < δ0.

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.

Let us denote the support of μ by K. Since $$\|u\|_{\infty }$$ is controlled by a constant C = C(α,Ω,μ), without loss of generality, we may assume that

$$-1 \leq u \leq 0.$$

Then for every 0 < ε < 1

$${\int}_{{\varOmega}_{\delta}} (u_{\delta} -u) d\mu \leq \varepsilon \mu({\varOmega}) + {\int}_{\{u < u_{\delta} - \varepsilon\} \cap K} d\mu.$$
(4.2)

We shall now estimate the second term on the right hand side. We may assume that 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 z0, of diameter 2r by

$$I(z_{0},r):= \{z = (x^{1}, \dots, x^{2n})\in \mathbb C^{n} : -r \leq x^{i} - {x_{0}^{i}}< r, \forall i = 1,\dots,2n\}.$$

Then, by the assumption, μ satisfies for every cube

$$I(z,r)=:I \subset B_{I} := B(z,2r) \subset\subset {\varOmega}$$

and for every set EI, the inequality

$$\mu (E) \leq \mu (I(z,r)) F_{0}\left( \text{cap} (E, B_{I}) \right),$$
(4.3)

where $$F_{0}: [0,\infty ] \to [0,\infty ]$$ is an increasing continuous function and F0(0) = 0.

Consider the semi-open cube decomposition of $${\varOmega } \subset \subset I_{0}:=[0,1)^{2n} \subset \mathbb R^{2n}$$ into 32ns congruent cubes of diameter 3s = 2δ, where $$s \in \mathbb N$$. Then

$$\{u < u_{\delta} -\varepsilon\} \cap I_{s} \subset \left\{z\in B_{I_{s}} : u< \sup_{B_{I_{s}}} u -\varepsilon\right\},$$
(4.4)

where Is = I(zs,δ) and $$B_{I_{s}} = B(z_{s}, 2\delta )$$ for some zsI0. Hence,

$${\int}_{\{u < u_{\delta} - \varepsilon\}} d\mu \leq {\sum}_{I_{s} \cap K \neq \emptyset }{\int}_{\{u< u_{\delta} -\varepsilon \} \cap I_{s}} d\mu.$$

Using (4.3), (4.4), and then applying Lemma 4.2 for r = 2δ and R = 2R0, we have for Bs := B(zs,4δ) corresponding to each cube Is

$$\begin{array}{@{}rcl@{}} {\int}_{\{u< u_{\delta} - \varepsilon\} \cap I_{s}} d\mu \leq \mu(I_{s}) F_{0}(\text{cap}(E(\varepsilon, u, B_{I_{s}}), B_{s})) \leq \mu(I_{s}) F_{0} \left( \frac{C_{0}}{\varepsilon \log (R_{0}/\delta)}\right), \end{array}$$
(4.5)

where 2R0 = dist(K,Ω). Therefore, combining the above inequalities, we get that

$${\int}_{\{u < u_{\delta} - \varepsilon\}} d\mu \leq \mu({\varOmega}) F_{0} \left( \frac{C_{0}}{\varepsilon \log (R_{0}/\delta)}\right).$$

We conclude from this and Lemma 4.3 that

$$\omega(\delta;u, \bar{\varOmega}) \leq \sup_{{\varOmega}_{\delta}} (u_{\delta} - u) \leq \varpi(\delta;\psi,\partial{\varOmega}) + C \left[\varepsilon + F_{0} \left( \frac{C_{0}}{\varepsilon \log (R_{0}/\delta)}\right)\right]^{\alpha_{1}}.$$

If we choose $$\varepsilon = (\log R_{0}/\delta )^{-1/2}$$, then Theorem 1.3 follows.

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## 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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