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.
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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:
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
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 Fh(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 Lp, 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
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.,
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
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 [3] 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 [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 ⊂ BI := B(z,2r) ⊂⊂Ω and for every set E ⊂ I,
According to [1], 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 [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,
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 uK is given by
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
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 \(\mathcal H(\alpha ,{\varOmega })\), 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 measureμiscompactly supported. Fix 0 < α < 2nandτ = α/(2n + 1).There exists a uniform constant C such that for every compact setK ⊂Ω,
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 [10]. We choose ε > 0 so small that
where Ωε = {z ∈Ω : dist(z,∂Ω) > ε}. Since for every \(K \subset {\varOmega }^{\prime \prime }\) we have
(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
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,
Hence, by the comparison principle [4],
Note that
The comparison principle, the bounds (3.4), and the volume-capacity inequality from [1, Theorem A] (in the last inequality below) give the following:
Choose
(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 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
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
and so the sequence uj 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
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 := ddcv = ddc|z|2, to get
By Jensen’s formula
where σ2n− 1 is the area of the unit sphere,
and
Since n(t)/t2n− 2 is increasing, we have
From u < 0, it follows that N(R) < −u(0). Hence,
Combining the above inequalities, we get the desired estimate with the constant
If R = 3r, then C is also independent of r. □
Lemma 4.2
Denote forρ ≥ 0,Bρ = {|z − z0| < eρR0}. Givenz0 ∈Ωand twonumbersR > 1,R0 > 0 such thatBM ⊂⊂Ω, and givenv ∈ PSH(Ω) such that− 1 < v < 0, denote by E theset
where δ ∈ (0,1). Then, there exists C0 depending only on n such that
Proof
From the logarithmic convexity of the function \(r \mapsto \sup _{|z-z_{0}|<r} v(z)\) it follows that for z ∈ BR ∖ B0 and \(a_{0}:= \sup _{B_{0}} v\) we have
Hence,
Let \(u = u_{E,B_{2}}\) the relative extremal function of E with respect to B2. One has
So, for some z1 ∈Ḅ0, we have
Note that E ⊂{|z − z1| < 2R0}⊂ B2. Therefore, Lemma 4.1 gives
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
and for z ∈Ωδ set
Thanks to the arguments in [12, Lemma 2.11] it is easy to see that there exists δ0 > 0 such that
for every z ∈ ∂Ωδ 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
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
Then for every 0 < ε < 1
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
Then, by the assumption, μ satisfies for every cube
and for every set E ⊂ I, the inequality
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 3−s = 2δ, where \( s \in \mathbb N\). Then
where Is = I(zs,δ) and \(B_{I_{s}} = B(z_{s}, 2\delta )\) for some zs ∈ I0. Hence,
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
where 2R0 = dist(K,∂Ω). Therefore, combining the above inequalities, we get that
We conclude from this and Lemma 4.3 that
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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On the occasion of Lê Văn Thiêm’s centenary
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Kołodziej, S., Nguyen, N.C. A Remark on the Continuous Subsolution Problem for the Complex Monge-Ampère Equation. Acta Math Vietnam 45, 83–91 (2020). https://doi.org/10.1007/s40306-019-00347-0
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DOI: https://doi.org/10.1007/s40306-019-00347-0