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.

1 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 [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.

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 F_h(x) a Borel measure μ satisfies

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

(1.2)

for any Borel set E ⊂Ω. Then, by [9] 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 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

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

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(|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

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

(1.3)

If the measure μ satisfies μ ≤ (dd^cφ)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 = fdV_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) = Cx^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

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

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

(1.4)

According to [1], the Lebesgue measure dV_2n 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 [8]. We refer the reader to [5] for more examples of measures satisfying this property. Here is our second result.

Theorem 1.3

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

2 Preliminaries

Here, we gather some basic facts from pluripotential theory taken from [4], and used in the sequel. Given a compact set K in a domain ${\varOmega }\subset \mathbb {C}^{n},$ its relative extremal function u_K 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 u_K 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 [4], 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 E ⊂Ω,

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

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

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

(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 φ_ε the standard regularization of φ in the terminology of [10]. We choose ε > 0 so small that

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

where Ω_ε = {z ∈Ω : 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 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,

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

(3.2)

Hence, by the comparison principle [4],

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

(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 = e^−τ/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 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

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

and so the sequence 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 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) = {|z − z₀| < ρ}. We use the Błocki inequality [6] for v(z) = |z|² − ρ² and β := dd^cv = dd^c|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 σ_2n− 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)/t^2n− 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,B_ρ = {|z − z₀| < e^ρR₀}. Givenz₀ ∈Ωand twonumbersR > 1,R₀ > 0 such thatB_M ⊂⊂Ω, and givenv ∈ PSH(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 z ∈ B_R ∖ B₀ 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 z₁ ∈Ḅ₀, 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 ⊂{|z − z₁| < 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 [9] 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 z ∈Ω_δ 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 z ∈ ∂Ω_δ 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 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 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 E ⊂ I, 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 3^2ns congruent cubes of diameter 3^−s = 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 I_s = I(z_s,δ) and $B_{I_{s}} = B(z_{s}, 2\delta )$ for some z_s ∈ I₀. 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 B_s := B(z_s,4δ) corresponding to each cube I_s

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

(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.