A remark on the continuous subsolution problem for the complex Monge-Amp\`ere 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-Amp\`ere equation has the continuous solution. The modulus of continuity of the solution is 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: (1.1) u ∈ P SH(Ω) ∩ C 0 (Ω), (dd c u) n = dµ, u = ψ on ∂Ω.
It was shown in [9] 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 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 [11,12] 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. Our 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 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).
Acknowledgement. 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.

Proof of Theorem 1.1
In this section we will prove Theorem 1.1. We need the following lemma. The proof of this lemma is based on a similar idea as the one in [10, 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 2.1. 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 (in the sense that its relative extremal function [4] is continuous) as µ is a Radon measure. Denote by ϕ ε the standard regularization of ϕ. We choose (up to a constant depending only on Ω, Ω ′ ) in what follows we will write cap(K) for either one of these capacities. We have Let u K the relative extremal function for K with respect to Ω ′ . Consider the set Hence, by the comparison principle [4], Note that The comparison principle, the bounds (2.4) and the volume-capacity inequality from [1] (in the last inequality below) give us that 1 n (we assume that ε is so small that it satisfies (2.2), otherwise the inequality (2.1) holds true by increasing the constant) and plug in the formula for δ we get that This combined with (2.3) gives the desired inequality.
We are ready to finish the proof of the theorem. It follows from Lemma 2.1 that a suitable function h for the measure µ which satisfies (1.2) is By changing the variable s = 1/x, and then t = e −τ /s , this is equivalent to The finiteness 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 [9] the Dirichlet problem (1.1) has a unique solution.
To deal with the general case consider the exhaustion of Ω by 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 which gives the continuity of u. The proof is completed.

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) (definition below) under the additional condition that a given measure is locally dominated by capacity.
Recall 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 Borel set E ⊂ Ω, The following result [8, Lemma 2] will be used in what follows.
Proof. See Appendix.
Thanks to the arguments in [11,Lemma 2.11] it is easy to see that there exists δ 0 > 0 such that 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 [7, Theorem 1.1] (see also [11,Proposition 2.10]) we get Lemma 3.2. There are uniform constants C, α 1 depending only on Ω, α, µ such that for every 0 < δ < δ 0 .
Thanks to this lemma we know that the right hand side tends to zero as δ decreases to zero. We will use the property "locally dominated by capacity" to obtain a quantitative bound via Lemma 3.1.
End of Proof of Theorem 1.3. Let us denote the support of µ by K. Since u ∞ is controlled by a contant C = C(α, Ω, µ), without loss of generality we may assume that − 1 ≤ u ≤ 0. Then for every 0 < ε < 1 We shall now estimate the second term on the right hand side. Let us fix the notation that will be used later on. We may assume that Ω ⊂⊂  where I s = I(z s , δ) and B Is = B(z s , 2δ) for some z s ∈ I 0 . Hence Using (3.3), (3.4), and then applying Lemma 3.1 for r = 2δ and R = 2R 0 , we have for B s := B(z s , 4δ) corresponding to each cube I s : where 2R 0 = dist(K, ∂Ω). Therefore, combining the above inequalities, we get that We conclude from this and Lemma 3.2 that If we choose ε = (log R 0 /δ) −1/2 then Theorem 1.3 follows.

Appendix
For the reader's convenience we give the details of the proof of Lemma 3.1. The following inequality is due to Alexander and Taylor [2, Lemma 3.3]. In particular, if R/r = 3 then the constant C depends only on n.
If R = 3r, then C is also independent of r.
We are ready to prove Lemma 3.1. We shall reformulate it as in [8, Lemma 2] and follow the proof given there.
Proof. From the logarithmic convexity of the function r → sup |z−z0|<r v(z) it follows that for z ∈ B M \ B 0 and a 0 :