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

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 Ω ⊂ C n . Let ψ be a continuous function on the boundary of Ω. We look for the solution to the equation: (dd c u) n = dμ, u = ψ on ∂Ω. (1.1) Here, P SH stands for plurisubharmonic functions, and d c = i(∂ − ∂). 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 : R + → (0, ∞) be an increasing function such that We call such a function admissible. If h is admissible, then so is Ah for any number A > 0. Define Suppose that for such a function F h (x) a Borel measure μ satisfies 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 μ ≤ (dd c ϕ) n in Ω, 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.
Assume that its modulus of continuity satisfies the Dini type condition If the measure μ satisfies μ ≤ (dd 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 dV 2n with f (x) being continuous onΩ. 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 H(α, Ω). 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  (1.4) According to [1], the Lebesgue measure dV 2n satisfies this property with F 0 = C α exp(−α/x −1/n ) for every 0 < α < 2n. The case F 0 (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.

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 Ω ⊂ C n , its relative extremal function u K is given by Its upper semicontinuous regularization u * K is plurisubharmonic. When u K is continuous, we call K a regular set. It is easy to see that the -envelope 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 and by [4], can be expressed as We say that a positive Borel measure μ belongs to H(α, Ω), α > 0, if there exists a uniform constant C > 0 such that for every compact set E ⊂ Ω,

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 < 2n and τ = α/(2n + 1). There exists a uniform constant C such that for every compact set 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 in what follows we shall write cap(K) for either one of these capacities. We have Let u K be the relative extremal function of K with respect to Ω . Consider the set K = {3δu K + ϕ ε < ϕ − 2δ}. Then, Hence, by the comparison principle [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: 1 n (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 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 is a function which satisfies (1.2) for the measure μ once we have By changing the variable s = 1/x, and then t = e −τ/s , this is equivalent to The last inequality is guaranteed by (1.3). Thus, our assumption on the modulus of continuity (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 = {ϕ ≤ −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 and so the sequence u j 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 H(α, Ω) 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]. In particular, if R/r = 3 then the constant C depends only on n.
If R = 3r, then C is also independent of r.
Let us proceed with the proof of Theorem 1.3. Since μ ∈ H(α, Ω), 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 Thanks to the arguments in [12,Lemma 2.11] it is easy to see that there exists δ 0 > 0 such that u δ (z) ≤ u(z) + (δ; ψ, ∂Ω) (4.1) for every 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 onΩ 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 H(α, Ω) as in [ 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.