Weak Solutions to Monge–Ampère Type Equations on Compact Hermitian Manifold with Boundary

Abstract

We prove the bounded subsolution theorem for the complex Monge–Ampère type equation, with the right-hand side being a positive Radon measure, on a compact Hermitian manifold with boundary.

1 Introduction

Let $(\overline{M},\omega )$ be a smooth compact n-dimensional Hermitian manifold with the non-empty boundary $\partial M$. In [24], we studied weak quasi-plurisubharmonic solutions of the Dirichlet problem for the complex Monge–Ampère equation. In this paper, we extend those results to the complex Monge–Ampère type equation where the right-hand side depends also on the solution.

Let $\mu$ be a positive Radon measure on $M = \overline{M} \setminus \partial M$. Suppose that $F(u,z): \mathbb {R}\times M \rightarrow \mathbb {R}^+$ is a non-negative function. Let $\varphi \in C^0(\partial M)$. We consider the Dirichlet problem

$$\begin{aligned} {\left\{ \begin{array}{ll} u\in PSH(M,\omega ) \cap L^\infty (\overline{M}), \\ (\omega + \mathrm{{d}}d^c u)^n = F(u,z) \mu ,\\ u = \varphi \quad \text {on } \partial M. \end{array}\right. } \end{aligned}$$

(1.1)

When M is a bounded strictly pseudoconvex domain in $\mathbb {C}^n$, the problem for plurisubharmonic functions ($\omega$ is just the zero form) was studied by Bedford and Taylor [2] for $\mu = dV_{2n}$ the Lebesgue measure, $F \in C^0(\mathbb {R}\times \overline{M})$ and $F^{1/n}$ convex and non-decreasing in u. Generalizations for weak solutions were done by Cegrell and Kołodziej [6, 8, 22] with more general function F(u, z) and a positive Radon measures $\mu$. Classical smooth solutions were obtained in [5]. On the product of a compact Kähler manifold and an annulus this problem, with $F \equiv 0$, is the geodesic equation on the space of Kähler potentials of the manifold. For $F(u,z) = e^{\lambda u}$ with $\lambda \in \mathbb {R}$, one obtains the Kähler–Einstein metric equations as in Cheng and Yau [9], where the authors obtained complete Kähler–Einstein metrics, but then the potentials tend to infinity when the argument approaches the boundary. Berman [4] discovered that one may use the solution of a family of Monge–Ampère type equation to approximate various envelopes of quasi-plurisubharmonic functions. This approximation process is very useful in studying the regularity of global envelopes (see [10, 14, 28]).

On a compact Hermitian manifold with boundary, if $F(u,z) = 1$ and $\mu$ is a positive Radon measure, we obtained weak solutions of the problem under the hypothesis that a subsolution exists, [24]. To deal with a general (non-Kähler) Hermitian metric $\omega$, we needed to adapt a suitable comparison principle from [23]. Furthermore, we can no longer rely on assumption that the boundary is pseudoconvex. We bypassed this by employing the Perron envelope and the existence of a bounded subsolution. Here, we use similar strategy to obtain the following result.

Theorem (c.f. Theorem 3.2) Assume that F(u, z) is a bounded non-negative function which is continuous and non-decreasing in the first variable and $\mu$-measurable in the other one. Let $\mu$ be a positive Radon measure which is locally dominated by Monge–Ampère measures of bounded plurisubharmonic functions. Then, the Dirichlet problem (1.1) has a solution if and only if there is a bounded subsolution $\underline{u} \in PSH(M,\omega ) \cap L^\infty (\overline{M})$ satisfying $\lim _{z\rightarrow x} u (z) = \varphi (x)$ for every $x\in \partial M$ and

$$\begin{aligned} (\omega +\mathrm{{d}}d^c \underline{u})^n \ge F(\underline{u}, z) \mu \quad \text {on } M. \end{aligned}$$

In many cases of interest, the extra assumption on $\mu$ (the local domination by Monge–Ampère measures of bounded plurisubharmonic functions) is always satisfied once the subsolution $\underline{u}$ exists, for example, $F(t,z) = e^{\lambda t}$ with $\lambda \in \mathbb {R}$ or $\mu = \omega ^n$. If $F(t,z) = e^{\lambda t}$ with $\lambda >0$, then we also have the uniqueness of the solution (Corollary 3.3).

Organization In Sect. 2, we study the problem (1.1) in a bounded strictly pseudoconvex domain in $\mathbb {C}^n$. We give a partial generalization of a result in [8] and study the convergence in capacity in dimension $n=2$. In Sect. 3, we prove the main theorem. We also give a sketch of the proof of Hölder continuity of solutions of the Laplace equation on compact Hermitian manifold with boundary in the appendix.

2 Monge–Ampère type equations and stability of solutions

In this section, we first study the problem (1.1) in the special case $M \equiv \Omega$ a bounded strictly pseudoconvex domain in $\mathbb {C}^n$. This will provide the construction of the lift of quasi-plurisubharmonic function in the Perron envelope method for the manifold case.

Let us recall the following version of the comparison principle for a background Hermitian metric, see [23, Theorem 3.1].

Lemma 2.1

Let $\Omega$ be a bounded open set in $\mathbb {C}^n$. Fix $0<\theta <1$. Let $u,v \in PSH(\Omega , \omega ) \cap L^\infty (\Omega )$ be such that $\liminf _{z\rightarrow \partial \Omega } (u-v) \ge 0$. Suppose that $-s_0 = \sup _{\Omega }(v-u) >0$ and $\omega + \mathrm{{d}}d^c v \ge \theta \omega$ in $\Omega$. Then, for any $0< s < \theta _0:= \min \{\frac{\theta ^n}{16 \mathbf{B}}, |s_0| \}$,

$$\begin{aligned} \int _{\{u< v + s_0 +s\}} (\omega + \mathrm{{d}}d^c v)^n \le \left( 1+ \frac{s\mathbf{B}}{\theta ^n} C_n \right) \int _{\{u<v+s_0 +s\}} (\omega + \mathrm{{d}}d^c v)^n, \end{aligned}$$

where $C_n$ is a dimensional constant and $\mathbf{B}>0$ is a constant such that on $\overline{\Omega }$,

$$\begin{aligned} -\mathbf{B}\omega ^2 \le 2n \mathrm{{d}}d^c \omega \le \mathbf{B}\omega ^2, \quad -\mathbf{B}\omega ^3 \le 4n^2 \mathrm{{d}}\omega \wedge \mathrm{{d}}^c \omega \le \mathbf{B}\omega ^3. \end{aligned}$$

It gives a useful comparison principle for solutions of Monge–Ampère type equations. We shall use it frequently in the paper.

Proposition 2.2

Let $\Omega$ be a bounded open set in $\mathbb {C}^n$. Let $u, v \in PSH(\Omega , \omega ) \cap L^\infty (\Omega )$ be such that $\liminf _{z\rightarrow \partial \Omega } (u-v)(z) \ge 0$. Let $F(t,z): \mathbb {R}\times \Omega \rightarrow \mathbb {R}^+$ be a non-negative function which is non-decreasing in t and $d\mu$-measurable in z. Suppose $\mu \le \nu$ as measures and

$$\begin{aligned} (\omega + \mathrm{{d}}d^c u)^n = F(u,z) \mu , \quad (\omega + \mathrm{{d}}d^c v)^n = F(v, z) \nu \end{aligned}$$

in $\Omega$. Then, $u \ge v$ on $\Omega$.

Proof

We may additionally assume that $\liminf _{z\rightarrow \partial \Omega } (u-v) \ge 2a>0$. The general case follows after replacing u by $u+2a$ and letting $a \rightarrow 0$. Thus $\{u< v\} \subset \Omega ' \subset \subset \Omega$ for some open set $\Omega '$. By subtracting from u, v a constant C and replacing F(t, z) by $F(t+C, z)$, we may also assume that $u, v \le 0$. Arguing by contradiction, suppose that $\{u<v\}$ was non-empty. Since $\Omega$ is bounded, there exists a bounded, strictly plurisubharmonic function $\rho \in C^2(\overline{\Omega })$ such that $-C \le \rho \le 0$. Then one can first multiply $\rho$ by small positive constant and then fix small positive constants $\theta , \tau >0$ such that the set $\{u< (1+\tau )^\frac{1}{n} v + \rho \} \subset \subset \Omega$ is non-empty and

$$\begin{aligned} \mathrm{{d}}d^c \rho \ge 2\theta \omega , \quad 1+ \theta \ge (1+\tau )^\frac{1}{n} . \end{aligned}$$

Put $\widehat{v} = (1+\tau )^\frac{1}{n}v+ \rho$. Then,

$$\begin{aligned} \begin{aligned} \omega _{\widehat{v}}^n&= \left( \omega + \mathrm{{d}}d^c \rho + \mathrm{{d}}d^c (1+\tau )^\frac{1}{n} v \right) ^n \\&\ge \left[ (1+2\theta )\omega + \mathrm{{d}}d^c (1+\tau )^\frac{1}{n} v \right] ^n\\&\ge (1+\tau ) \omega _v^n. \end{aligned} \end{aligned}$$

(2.1)

Denote by U(s) the set $\{u< \widehat{v} + s_0 +s\}$ where $-s_0 = \sup _{\Omega } (\widehat{v} -u) >0$. Then, for $0<s<|s_0|$,

$$\begin{aligned} U(s) \subset \subset \Omega , \quad \sup _{U(s)} \{\widehat{v} +s_0 +s - u\} =s. \end{aligned}$$

It follows from Lemma 2.1 that for every $0< s < \min \{\frac{\theta ^n}{16 \mathbf{B}}, |s_0|\}$,

$$\begin{aligned} 0<\int _{U(s)} \omega _{\widehat{v}}^n \le \left( 1+ \frac{s \mathbf{B}}{\theta ^n} C_n\right) \int _{U(s)} \omega _u^n, \end{aligned}$$

where the first inequality holds because $\omega + \mathrm{{d}}d^c \widehat{v} \ge \theta \omega$. Note that $u < \widehat{v}$ on U(s). Hence, the monotonicity of F implies

$$\begin{aligned} \omega _u^n = F(u, z) \mu \le F(u, z) \nu \le F(\widehat{v},z) \nu \le F(v, z) \nu = \omega _v^n. \end{aligned}$$

Combining this and (2.1), we get

$$\begin{aligned} (1+\tau ) \int _{U(s)} \omega _v^n \le \left( 1+ \frac{s \mathbf{B}}{\theta ^n} C_n\right) \int _{U(s)} \omega _v^n. \end{aligned}$$

Therefore, $0< \tau \le s\mathbf{B} C_n/\theta ^n$, which is impossible for $s>0$ small enough. Hence, $u \ge v$ on $\Omega$ and the proof is completed. $\square$

Let $\Omega$ be a bounded strictly pseudoconvex domain in $\mathbb {C}^n$. Let $\varphi \in C^0(\partial \Omega )$ and $\mu$ be a positive Radon measure in $\Omega$. Suppose that $F(u,z): \mathbb {R}\times \Omega \rightarrow \mathbb {R}^+$ is non-negative function which is continuous and non-decreasing in the first variable and $\mu$-measurable in the other one. Consider the following Dirichlet problem

$$\begin{aligned} {\left\{ \begin{array}{ll} u \in PSH(\Omega , \omega ) \cap L^\infty (\overline{\Omega }), \\ (\omega + \mathrm{{d}}d^c u)^n = F(u,z) \mu , \\ \lim _{z\rightarrow x} u(z) = \varphi (x)\quad \text {for } x\in \Omega . \end{array}\right. } \end{aligned}$$

(2.2)

By Proposition 2.2, there is at most one solution to the problem. Furthermore, in the special case $\mu \equiv 0$, by [23, Theorem 4.2], there is a unique continuous (maximal) $\omega$-plurisubharmonic ($\omega$-psh for short) function h solving

$$\begin{aligned} (\omega + \mathrm{{d}}d^c h)^n \equiv 0, \quad h = \varphi \quad \text {on } \partial \Omega . \end{aligned}$$

(2.3)

We now study the existence of the solution. The Cegrell class $\mathcal {E}_0(\Omega )$ is the set of all functions $v \in PSH(\Omega ) \cap L^\infty (\Omega )$ satisfying

$$\begin{aligned} \lim _{z\rightarrow \partial \Omega } v(z)= 0 \quad \text {and}\quad \int _\Omega (\mathrm{{d}}d^cv)^n <+\infty . \end{aligned}$$

The following generalizes [22, Theorem 1.1], where the case $\omega =0$ was treated.

Theorem 2.3

Suppose that F(t, z) is a bounded non-negative function which is continuous and non-decreasing in the first variable and $\mu$-measurable in the second one. Suppose that $d\mu \le (\mathrm{{d}}d^c v)^n$ for a function $v\in \mathcal {E}_0(\Omega )$. Then, there exists a unique bounded $\omega$-psh function in $\Omega$ solving

$$\begin{aligned} \begin{aligned}&(\omega + \mathrm{{d}}d^c u)^n = F(u, z) \mathrm{{d}}\mu , \\&\lim _{z\rightarrow x} u(z) = \varphi (x) \quad \text {for } x \in \partial \Omega . \end{aligned} \end{aligned}$$

Proof

The proof follows the lines of [22, Theorem 1.1] (see also [6]), which applied Schauder’s fixed theorem. In the presence of the Hermitian background form, we need results from [24] to verify the hypothesis of that theorem.

We start with the following simple observation.

Lemma 2.4

We may assume that $\mu$ has compact support in $\Omega$.

Proof

Suppose that the problem is solvable for compactly supported measures. Let $\{\Omega _j\}_{j\ge 1}$ be an increasing exhausting sequence of open sets that are relatively compact in $\Omega$. Then we can find a sequence of functions: $u_j \in PSH(\Omega , \omega ) \cap L^\infty (\Omega )$ with $\lim _{z\rightarrow x} u_j(z) = \varphi (x)$ solving

$$\begin{aligned} (\omega + \mathrm{{d}}d^c u_j)^n = F(u_j, z) \mathbf{1}_{\Omega _j} \mathrm{{d}}\mu . \end{aligned}$$

By Proposition 2.2, this is a decreasing sequence and $v + h \le u_j \le h \quad \text {on } \overline{\Omega },$ where h is the maximal function defined in (2.3). Put $u = \lim u_j$. By monotone convergence theorems [13] (see also [3]), we have that $\omega _{u_j}^n$ converges weakly to $\omega _u^n$. Also $F(u_j, z) \mathbf{1}_{\Omega _j} \mathrm{{d}}\mu$ converges weakly to $F(u, z) \mathrm{{d}}\mu$ when $j\rightarrow \infty$. Thus, u satisfies the equation $\omega _u^n = F(u, z) \mathrm{{d}}\mu$. $\square$

By Lemma 2.4, we assume that $\text{ supp } \mu$ is compact in $\Omega$. Since F is bounded, without loss of generality, we may assume that $0 \le F \le 1$, as we can rescale $\mu$ and v by a positive constant. Let h be a maximal $\omega$-psh in $\Omega$ as in (2.3). Then, the set

$$\begin{aligned} \mathcal {A}= \{u \in PSH(\Omega , \omega ): v+ h \le u \le h\} \end{aligned}$$

is a convex and bounded set in $L^1(\Omega )$ with respect to $L^1-$topology. Thus, it is a compact set. We define the map $T : \mathcal {A}\rightarrow \mathcal {A}$, where $T(u) =w$ is the solution of

$$\begin{aligned} w \in \mathcal {A}, \quad (\omega + \mathrm{{d}}d^c w)^n = F(u, z) \mathrm{{d}}\mu , \end{aligned}$$

where this solution is uniquely determined by [24, Theorem 3.1].

Next, we shall verify that T is continuous. Let $\{u_j\}$ be a sequence in $\mathcal {A}$ such that $u_j \rightarrow u$ in $L^1(\Omega )$. Set $w = T(u)$ and $w_j = T(u_j)$. The continuity of T will follow if we have

$$\begin{aligned} \limsup w_j \le w \le \liminf w_j. \end{aligned}$$

(2.4)

Since $\mathcal {A}$ is compact in $L^1(\Omega )$, we may assume that $w_j$ converges to w in $L^1(\Omega ).$ In the arguments,we often pass to a subsequence which does not affect the proof. We also skip the renumbering of indices of those subsequences.

We need the following result which is essentially due to Cegrell [7].

Lemma 2.5

There is a subsequence of $\{F(u_j, z)\}_{j\ge 1}$ that converges to F(u, z) in $L^1(d\mu )$.

Proof of Lemma 2.5

Since $u_j$ is uniformly bounded, we may subtract a constant and assume that $u_j, u \le 0$ for all $j\ge 1$. Then it follows from [24, Lemma 2.1] that $\lim \int _\Omega u_j \mathrm{{d}}\mu = \int _\Omega u \mathrm{{d}}\mu$. Since $\mu$ is dominated by capacity, it follows from [24, Corollary 2.2] that $u_j \rightarrow u$ in $L^1(\mathrm{{d}}\mu )$. Passing to a subsequence, we have that $u_j$ converges to u almost everywhere in $\mathrm{{d}}\mu$. Since F(t, z) is continuous in t, the sequence $F(u_j, z)$ converges $\mu$-a.e to F(u, z). By the Lebesgue-dominated convergence theorem, we get the conclusion. $\square$

After passing to a subsequence, we may also assume that $u_j$ converges to u almost everywhere in $\mathrm{{d}}\mu$. Let us check the first inequality in (2.4). Define $\widehat{u}_k = \inf _{j\le k} u_j$ and denote by $\widehat{w}_k \in \mathcal {A}$ the sequence of solutions to

$$\begin{aligned} (\omega + \mathrm{{d}}d^c \widehat{w}_k)^n = F(\widehat{u}_k, z) \mathrm{{d}}\mu . \end{aligned}$$

Since $(\omega + \mathrm{{d}}d^c \widehat{w}_k)^n$ is increasing, by the comparison principle [23, Corollary 3.4], $\widehat{w}_k$ decreases to $\widehat{w} \in \mathcal {A}$. As $u_j$ converges to u, we have that $\widehat{u}_k$ increases to u (almost everywhere in $\mathrm{{d}}\mu$). Therefore,

$$\begin{aligned} (\omega + \mathrm{{d}}d^c \widehat{w})^n = \lim _{k\rightarrow +\infty }(\omega + \mathrm{{d}}d^c \widehat{w}_k)^n = \lim _{k\rightarrow +\infty } F (\widehat{u}_k, z) \mathrm{{d}}\mu = F(u, z) \mathrm{{d}}\mu , \end{aligned}$$

where the first equality holds by the convergence theorem in [13] (see also [3]) and the last equality follows from Lemma 2.5 and the Lebesgue monotone convergence theorem. By the uniqueness of the solution, $w = \widehat{w}$. Note that $\widehat{w}_k \ge w_k$ because $(\omega + \mathrm{{d}}d^c w_k)^n \ge (\omega + \mathrm{{d}}d^c \widehat{w}_k)^n$. Therefore, $w = \lim \widehat{w}_k \ge \limsup w_k.$

Next we prove the second inequality in (2.4). By Hartogs’ lemma for $\omega$-psh functions, $u = (\limsup _{j\rightarrow +\infty } u_j)^*$. Define $\widetilde{u}_k = (\sup _{j\ge k} u_j)^*$ and $\widetilde{w}_k = T (\widetilde{u}_k)$. Since $(\omega + \mathrm{{d}}d^c \widetilde{w}_k)^n = F(\widetilde{u}_k,z) \mathrm{{d}}\mu$ is decreasing, again by the comparison principle [23], one gets that the sequence $\widetilde{w}_k$ is increasing to some $\widetilde{w} \in \mathcal {A}$. Note also that

$$\begin{aligned} (\omega +\mathrm{{d}}d^c w_k)^n = F(u_k, z) \mathrm{{d}}\mu \le F (\widetilde{u}_k, z) \mathrm{{d}}\mu = (\omega + \mathrm{{d}}d^c \widetilde{w}_k)^n. \end{aligned}$$

Hence, $\widetilde{w}_k \le w_k$. Furthermore, $\widetilde{u}_k \downarrow u$ (almost everywhere in $\mathrm{{d}}\mu$). By the convergence theorem in [DK12] and Lemma 2.5, we infer

$$\begin{aligned} (\omega + \mathrm{{d}}d^c \widetilde{w})^n = \lim _{k\rightarrow +\infty } F(\widetilde{u}_k, z) \mathrm{{d}}\mu = F(u, z) \mathrm{{d}}\mu . \end{aligned}$$

As above by the uniqueness, $w = \widetilde{w}$. Then, $w = \lim \widetilde{w}_k \le \liminf w_k.$

Thus, we conclude the continuity of T. The Schauder theorem says that T has a fixed point $T(u) = u$. This gives the existence of a solution to the Dirichlet problem. $\square$

Remark 2.6

Thanks to the first lemma of the above proof, the theorem can be extended to the case of unbounded F, e.g., $F(u, z) = |u|^{-\alpha }$ with $\alpha >0$, in the Dirichlet problem:

$$\begin{aligned} {\left\{ \begin{array}{ll} u \in PSH(\Omega ) \cap L^\infty (\overline{\Omega }),\\ ( \mathrm{{d}}d^c u)^n = |u|^{-\alpha } \mathrm{{d}}\mu , \\ u= 0 \quad \text {on } \mathrm{{d}} \Omega . \end{array}\right. } \end{aligned}$$

Indeed, suppose that there exists a subsolution $\underline{u} \in PSH(\Omega ) \cap L^\infty (\overline{\Omega })$ such that $\underline{u} =0$ on $\partial \Omega$ and $(\mathrm{{d}}d^c \underline{u})^n \ge |\underline{u}|^{-\alpha } \mathrm{{d}}\mu .$ The proof above shows that for the sequence $\Omega _j \uparrow \Omega$ in Lemma 2.4, we can find $u_j \in PSH(\Omega ) \cap L^\infty (\overline{\Omega })$ such that

$$\begin{aligned} (\mathrm{{d}}d^c u_j)^n = |u_j|^{-\alpha } \mathbf{1}_{\Omega _j} \mathrm{{d}}\mu , \quad u_j =0 \text { on } \mathrm{{d}} \Omega . \end{aligned}$$

Since $|u_j|^\alpha (\mathrm{{d}}d^c u_j)^n \le |\underline{u}|^\alpha (\mathrm{{d}}d^c \underline{u})^n$, by the comparison principle $\underline{u} \le u_j \le 0$. Similarly, $u_j$ is a uniformly bounded and decreasing sequence. Therefore, $u = \lim _{j\rightarrow \infty } u_j$ is the unique solution to the Dirichlet problem.

Furthermore, one can obtain better regularity of the solution under stronger assumptions on $\mu$. Let us consider $\mu =f(z) \mathrm{{d}}V_{2n}$ with $f \in L^1(\Omega )$ and let $\rho$ be the strictly plurisubharmonic defining function for $\Omega$. Assume that $|\rho |^{-\alpha } f \in L^p(\Omega )$ for some $p>1$ (e.g., when $0<\alpha <1$ and $f\in L^p(\Omega )$ is bounded near the boundary $\mathrm{{d}}\Omega$.) Then, we have

$$\begin{aligned} |\rho |^{-\alpha } f \mathrm{{d}}V_{2n} = (\mathrm{{d}}d^c w)^n, \quad w = 0 \quad \text {on } \mathrm{{d}} \Omega \end{aligned}$$

for some Hölder continuous plurisubharmonic function w (see [16]). Therefore,

$$\begin{aligned} f dV_{2n} = |\rho |^\alpha (\mathrm{{d}}d^c w)^n \le |w + \rho |^\alpha [\mathrm{{d}}d^c (w+\rho )]^n. \end{aligned}$$

Therefore, $\rho + w$ is a Hölder continuous subsolution to the Dirichlet problem. So there is a unique bounded solution u. Furthermore,

$$\begin{aligned} | u|^{-\alpha } f \mathrm{{d}}V_{2n} = (|\rho |/|u|)^\alpha |\rho |^{-\alpha } f \mathrm{{d}}V_{2n}. \end{aligned}$$

A simple use of Hölder inequality shows that the measure on the right-hand side is well dominated by capacity on every compact set of $\Omega$. Thanks to [24, Remark 4.5], we get that u is continuous. Note that the Hölder continuity of u is proved in [18].

The above problem appeared in [2] and was more recently studied in [11] and [18, 19].

It is well known that the convergence of $\omega$-psh functions in $L^p(\Omega )$ does not imply the weak convergence of their associated Monge–Ampère operators. However, under the assumption of a uniform bound of the Monge–Amère measures, this is the case, see [8, Theorem 2.2]. We give a (partial) corresponding stability result for Hermitian background metrics.

Theorem 2.7

Suppose $\mathrm{{d}}\mu = (\mathrm{{d}}d^c v)^n$ for some $v \in \mathcal {E}_0(\Omega )$. Let $0\le f_j \le 1$ be a sequence of $\mathrm{{d}}\mu -$measurable functions such that $f_j \mathrm{{d}}\mu$ converges weakly to $f \mathrm{{d}}\mu$ as measures. Suppose that $u_j$ solves

$$\begin{aligned} {\left\{ \begin{array}{ll} u_j \in PSH(\Omega , \omega ) \cap L^\infty (\overline{\Omega }), \\ (\omega + \mathrm{{d}}d^c u_j)^n = f_j \mathrm{{d}}\mu , \\ \lim _{z\rightarrow x} u_j(z) = \varphi (x)\quad \text {for } x\in \mathrm{{d}} \Omega , \end{array}\right. } \end{aligned}$$

for a continuous function $\varphi$.

Then, for $u =\lim u_j$ in $L^1(\Omega )$, we have $(\omega + \mathrm{{d}}d^c u)^n = f \mathrm{{d}}\mu$.

Proof

Let h be the maximal function defined in (2.3). Then by the comparison principle [23, Corollary 3.4], we have $h+ v \le u_j \le h$. Thus, the sequence $\{u_j\}$ is uniformly bounded and this allows to apply [24, Lemma 3.5 c)] and conclude that there is a subsequence $\{u_{j_s}\}$ such that

$$\begin{aligned} \lim _{j_s\rightarrow \infty } \int _\Omega |u_{j_s} -u| (\omega + \mathrm{{d}}d^c u_{j_s})^n =0, \end{aligned}$$

(2.5)

and moreover $\omega _{u_{j_s}}^n$ converges weakly to $\omega _u^n= f \mathrm{{d}}\mu$, see [24, Lemma 3.6]. This gives the result. $\square$

In this statement, we would like to have also that a subsequence of $u_j$ converges to u in capacity. Near the end of this section (Corollary 2.10), we are able to do this for $n=2.$ Recall that the Bedford-Taylor capacity for a Borel set $E\subset \Omega$ is defined by

$$\begin{aligned} cap (E) = cap (E,\Omega ) = \sup \left\{ \int _E (\mathrm{{d}}d^c v)^n: v\in PSH(\Omega ), -1 \le v\le 0 \right\} . \end{aligned}$$

(2.6)

Note that there is another (natural) capacity associated with the metric $\omega$ which is equivalent to the one above (see [24, Lemma 5.6]).

Proposition 2.8

Let $\{u_j\}$ be the sequence in Theorem 2.7 in the case of smooth $\varphi$. Then, $u_j$ converges to u in capacity if and only if

$$\begin{aligned} \lim _{j\rightarrow +\infty }\int _{\Omega } |u_j -u| \omega _{u_j}^k \wedge \omega ^{n-k} =0, \quad k=0,...,n. \end{aligned}$$

(2.7)

Proof

Suppose that $u_j \rightarrow u$ in capacity, i.e., for a fixed $\varepsilon >0$, we have

$$\begin{aligned} \lim _{j\rightarrow +\infty } cap (|u_j -u|>\varepsilon ) =0. \end{aligned}$$

Let g be a strictly plurisubharmonic function in a neighborhood of $\overline{\Omega }$ such that $\mathrm{{d}}d^c g \ge \omega$ and $g= -\varphi$ on $\partial \Omega$. Denoting $\widehat{u}_j = u_j+ g$, we have $\widehat{u}_j \rightarrow \widehat{u}= u+g$ in capacity. Let $\rho$ be a strictly plurisubharmonic defining function of $\Omega$ such that $\mathrm{{d}}d^c \rho \ge \omega$ on $\overline{\Omega }$. By [24, Lemma 3.3, Corollary 3.4],

$$\begin{aligned} \sup _{j\ge 1} \int _{\Omega } (\mathrm{{d}}d^c \widehat{u}_j)^k \wedge (\mathrm{{d}}d^c \rho )^{n-k} \le C \end{aligned}$$

(2.8)

for some C. We can repeat the argument of [24, Lemma 2.3], with the sequence $\{w_j\}_{j\ge 1}$ in the place of $\{\widehat{u}_j\}_{j\ge 1}$, to obtain

$$\begin{aligned} \lim _{j\rightarrow +\infty } \int |u_j -u| (\mathrm{{d}}d^c \widehat{u}_j)^k \wedge (\mathrm{{d}}d^c \rho )^{n-k} = 0, \end{aligned}$$

This gives the proof the necessary condition.

It remains to prove the other implication. Suppose that we have the limit (2.7) for all $k=0,...,n$. Note that

$$\begin{aligned} cap(|u_j-u|>\varepsilon ) = cap (|\widehat{u}_j - \widehat{u}|>\varepsilon ) \le cap (\widehat{u}_j - \widehat{u}>\varepsilon ) + cap (\widehat{u} - \widehat{u}_j >\varepsilon ). \end{aligned}$$

By Hartogs’ Lemma, it is easy to see that $\max \{\widehat{u}_j, \widehat{u}\} \rightarrow \widehat{u}$ in capacity. Hence, $\lim _{j\rightarrow +\infty } cap (\widehat{u}_j -\widehat{u} > \varepsilon ) =0.$ The remaining term is estimated as in [8, page 718] via the comparison principle for plurisubharmonic functions:

$$\begin{aligned} \begin{aligned} cap (\widehat{u}_j< \widehat{u} -\varepsilon ) \le \left( 2/\varepsilon \right) ^n \int _{\{u_j < u -\varepsilon /2\}} (\mathrm{{d}}d^c \widehat{u}_j)^n \\ \le (2/\varepsilon )^{n+1} \int _{\Omega } |u_j-u| (\mathrm{{d}}d^c \widehat{u}_j)^n. \end{aligned} \end{aligned}$$

Denote $\tau = \mathrm{{d}}d^c g -\omega \le C \omega$. Then, $\mathrm{{d}}d^c \widehat{u}_j = \omega _{u_j} + \tau$. It follows that

$$\begin{aligned} \int _{\Omega }|u_j -u| (\mathrm{{d}}d^c \widehat{u}_j)^n = \sum _{k=0}^n \left( {\begin{array}{c}n\\ k\end{array}}\right) \int _\Omega |u_j -u| \omega _{u_j}^k \wedge \tau ^{n-k} \end{aligned}$$

which goes to zero by the assumption. Thus, the convergence in capacity is proved by the previous inequality. $\square$

Lemma 2.9

Suppose $n=2$. Let $u,v \in PSH(\Omega , \omega ) \cap L^\infty (\overline{\Omega })$ be such that $u\le v$ in $\Omega$ and $u = v$ near $\partial \Omega$. Let $-1 \le \rho \le 0$ be a plurisubharmonic function in $\Omega$. Then,

$$\begin{aligned} \int _\Omega (v-u)^3 (\mathrm{{d}}d^c \rho )^2 \le 6 \int _\Omega (v-u) \omega _u^2 + C \int _\Omega (v-u)^2 \omega ^2 +C E, \end{aligned}$$

where C is a uniform constant depending only on $\Omega$ and $\omega$ and

$$\begin{aligned} E = \left( \int _{\Omega } (v-u) \omega _u \wedge \omega \right) ^\frac{1}{2} \left( \int _{\Omega } (v-u)^2\,\omega ^2 \right) ^\frac{1}{2}. \end{aligned}$$

Proof

By quasi-continuity of plurisubharmonic and $\omega$-psh functions and the convergence theorems in [3] and [13], we may assume that all functions are smooth. Let us denote $h = v-u \ge 0$. Then, $\mathrm{{d}}d^c h = \omega _v - \omega _u$ and $\mathrm{{d}}h\wedge \mathrm{{d}}^ch$ is a positive (1,1)-current. By integration by parts

$$\begin{aligned} \begin{aligned} \int _\Omega h^3 (\mathrm{{d}}d^c \rho )^2&= \int _{\Omega } \rho \mathrm{{d}}d^c h^3 \wedge \mathrm{{d}}d^c \rho \\&= \int _\Omega \rho (3 h^2 \mathrm{{d}}d^c h + 6 h \mathrm{{d}}h \wedge d^c h) \wedge \mathrm{{d}}d^c \rho \\&\le 3\Vert \rho \Vert _\infty \int _{\Omega } h^2 \omega _u \wedge \mathrm{{d}}d^c \rho . \end{aligned} \end{aligned}$$

Using the integration by parts again

$$\begin{aligned} \int _\Omega h^2 \omega _u \wedge \mathrm{{d}}d^c \rho = \int _\Omega \rho \mathrm{{d}}d^c ( h^2 \omega _u). \end{aligned}$$

Here,

$$\begin{aligned} \begin{aligned} \rho \mathrm{{d}}d^c (h^2 \omega _u)&= \rho \left[ \mathrm{{d}}d^c h^2 \wedge \omega _u+ 2 \mathrm{{d}}h^2 \wedge d^c \omega + h^2 \mathrm{{d}}d^c \omega \right] \\&= \rho \left[ 2 h \mathrm{{d}}d^c h \wedge \omega _u + 2 \mathrm{{d}}h \wedge d^c h \wedge \omega _u + 2 \mathrm{{d}}h^2 \wedge \mathrm{{d}}^c \omega + h^2 \mathrm{{d}}d^c \omega \right] \\&\le 2 |\rho | h\, \omega _u^2 + 4\rho h \,\mathrm{{d}}h \wedge d^c \omega + \rho h^2 \mathrm{{d}}d^c \omega . \end{aligned} \end{aligned}$$

because $\rho$ is negative. Using $-C \omega ^2\le \mathrm{{d}}d^c \omega \le C\omega ^2$ and $-1\le \rho \le 0$, we have

$$\begin{aligned} \int _\Omega h^2 \omega _u \wedge \mathrm{{d}}d^c \rho \le 2 \int _\Omega h \omega _u^2 + 4 \left| \int _\Omega \rho h\, \mathrm{{d}}h \wedge d^c \omega \right| + C \int _\Omega h^2 \omega ^2. \end{aligned}$$

Thus, to complete the proof,we need to estimate the middle integral on the right-hand side. In fact, using [25, Proposition 1.4], we have

$$\begin{aligned} \begin{aligned} \left| \int _{\Omega } \rho h\, \mathrm{{d}} h \wedge d^c \omega \right|&\le C \left( \int _{\Omega } \mathrm{{d}} h \wedge d^c h \wedge \omega \right) ^\frac{1}{2} \left( \int _{\Omega } |\rho |^2 h^2 \;\omega ^2 \right) ^\frac{1}{2} \\&\le C \left( \int _{\Omega } \mathrm{{d}} h \wedge d^c h \wedge \omega \right) ^\frac{1}{2} \left( \int _{\Omega } h^2\,\omega ^2 \right) ^\frac{1}{2}. \end{aligned} \end{aligned}$$

Note that $2 \mathrm{{d}}h \wedge d^c h = \mathrm{{d}}d^c h^2 - 2h \mathrm{{d}}d^c h \le \mathrm{{d}}d^c h^2 + 2h \omega _u$. Therefore,

$$\begin{aligned} 2\int _\Omega \mathrm{{d}} h \wedge d^c h \wedge \omega \le \int _\Omega (\mathrm{{d}}d^c h^2 +2h \omega _u) \wedge \omega . \end{aligned}$$

By integration by parts

$$\begin{aligned} \int _\Omega \mathrm{{d}}d^c h^2 \wedge \omega = \int _\Omega h^2 \mathrm{{d}}d^c \omega ^2 \le C \int _\Omega h^2 \omega ^2. \end{aligned}$$

Combining these inequalities, we get

$$\begin{aligned} \begin{aligned} \left| \int _{\Omega } \rho h\, \mathrm{{d}} h \wedge d^c \omega \right|&\le C \left( \int _{\Omega } h^2\omega ^2 + h \omega _u \wedge \omega \right) ^\frac{1}{2} \left( \int _{\Omega } h^2\,\omega ^2 \right) ^\frac{1}{2} \\&\le C\int _{\Omega } h^2\,\omega ^2 + C\left( \int _{\Omega } h\omega _u \wedge \omega \right) ^\frac{1}{2} \left( \int _{\Omega } h^2\,\omega ^2 \right) ^\frac{1}{2}, \end{aligned} \end{aligned}$$

where we used an elementary inequality $(x+y)^\frac{1}{2} \le x^\frac{1}{2} + y^\frac{1}{2}$ with $x, y\ge 0$ in the second inequality. This completes the proof. $\square$

Corollary 2.10

If $n=2$, then there is a subsequence $\{u_{j_s}\}$ of $\{ u_j\}$ in Theorem 2.7 that converges to u in capacity.

Proof

Fix $a>0$. Let us denote $w_s = \max \{u_{j_s}, u-1/s\}$. By Hartogs’ Lemma $w_s\rightarrow u$ in capacity. Moreover, $\{|u-u_j|>2a\} \subset \{|u-w_s|>a\} \cup \{ |w_s - u_{j_s}|>a \}$. Therefore, it is enough to show that $cap(|w_s - u_{j_s}|>a) \rightarrow 0$ as $s \rightarrow +\infty .$ By definition, we have $w_s = \max \{u_{j_s}, u- 1/s\} \ge u_{j_s}$. Then, we need to show

$$\begin{aligned} cap(\{u_{j_s} < w_s - a\}) \rightarrow 0 \quad \text {as} \quad s\rightarrow +\infty . \end{aligned}$$

To this end, by Lemma 2.9,

$$\begin{aligned} \left( \frac{a}{2} \right) ^3 cap (\{ u_{j_s} < w_s -a \}) \le \int _\Omega ( w_s - u_{j_s} ) \omega _{u_{j_s}}^2 + C \int _\Omega (w_s - u_{j_s})^2 \omega ^2 + CE_s, \end{aligned}$$

where

$$\begin{aligned} E_s = \left( \int _{\Omega } (w_s-u_{j_s}) \omega _{u_{j_s}} \wedge \omega \right) ^\frac{1}{2} \left( \int _{\Omega } (w_s-u_{j_s})^2\,\omega ^2 \right) ^\frac{1}{2}. \end{aligned}$$

By the assumption $u_j \rightarrow u$ in $L^1(\Omega )$ (they are uniformly bounded) and (2.5), the first and second integrals on the right-hand side go to zero as s goes to infinity. Note that the first factor of $E_s$ is uniformly bounded by (2.8). Hence, the last term $E_s$ also goes to zero as $s\rightarrow +\infty$. Therefore, the conclusion follows. $\square$

Proposition 2.8 and Corollary 2.10 have their analogues on a compact Hermitian manifold without boundary. They allow, for example, to provide another proof of the following result of Guedj and Lu [17, Proposition 3.4].

Lemma 2.11

Let $(X, \omega )$ be a compact n-dimensional Hermitian manifold (without boundary). Then for any $A>0$,

$$\begin{aligned} \inf \left\{ \int _X (\omega + \mathrm{{d}}d^c v)^n: v \in PSH(X,\omega ), -A \le v \le 0\right\} >0. \end{aligned}$$

Proof

By replacing $\omega$ with $\omega /A$ and v with v/A, we may assume that $A =1$. We argue by contradiction. Suppose that there was a sequence $\{u_j\}_{j\ge 1} \subset PSH(X, \omega )$ such that $-1\le u_j \le 0$ and

$$\begin{aligned} \lim _{j\rightarrow \infty } \int _{X} \omega _{u_j}^n =0, \quad \sup _X u_j =0. \end{aligned}$$

By passing to a subsequence, we assume that $u_j \rightarrow u$ in $L^1(X)$ and $u_j \rightarrow u$ a.e, where

$$\begin{aligned} u = (\limsup _{j\rightarrow \infty } u_j)^* = \lim _{j\rightarrow \infty } (\sup _{\ell \ge j} u_\ell )^*. \end{aligned}$$

Then, $-1\le u \le 0$ and $u \in PSH(X, \omega )$. We will show that there exists a subsequence $\{u_{j_s}\}$ of $\{u_j\}$ such that $\omega _{u_{j_s}}^n$ converges weakly to $\omega _u^n$. Indeed, set

$$\begin{aligned} w_{j} = \{u_j, u-1/j\}. \end{aligned}$$

By the Hartogs lemma,$w_j$ converges to u in capacity. Therefore, by the convergence theorem in [BT82] and [DK12] $\lim _{j\rightarrow \infty } \omega _{w_j}^n = \omega _u^n.$ Next, we will show that

$$\begin{aligned} \int _X |u_j -u| (\omega + \mathrm{{d}}d^c u_j)^n \rightarrow 0 \quad \text {and}\quad \int _{X} |u_j -u| (\omega + \mathrm{{d}}d^c w_j)^n \rightarrow 0 \end{aligned}$$

as $j \rightarrow +\infty$. The first statement holds because $\int _X |u_j -u| \omega _{u_j}^n \le 2 \int _X \omega _{u_j}^n \rightarrow 0$ by the assumption. The second convergence also holds because $w_j \rightarrow u$ in capacity and so the proofs of Lemma 2.1, Corollary 2.2, Lemma 2.3 in [24] can be applied because all considered functions are uniformly bounded and X is compact.

Thus, again by [24, Lemma 3.6], there exists a subsequence $u_{j_s}$ such that $\omega _{u_{j_s}}^n$ converges weakly to $\omega _u^n$. In particular, $\int _X \omega _u^n = \lim _{j_s\rightarrow +\infty } \int _X \omega _{u_{j_s}}^n = 0$. Hence, $\omega _u^n \equiv 0$ on X. This is a contradiction with the fact that the Monge–Ampère mass of a bounded $\omega$-psh function is always positive [23, Remark 5.7]. $\square$

3 The Dirichlet Problem

In this section, we solve the Dirichlet problem for Monge–Ampère type equation under the existence of a bounded subsolution. Let $\mu$ be a positive Radon measure on $M = \overline{M}\setminus \partial M$. Let $\varphi \in C^0(\partial M)$. Let $F(t,z): \mathbb {R}\times M \rightarrow \mathbb {R}^+$ be a non-negative function which is non-decreasing in t and $d\mu$-measurable in z. We consider the Dirichlet problem

$$\begin{aligned} {\left\{ \begin{array}{ll} u \in PSH(M, \omega ) \cap L^\infty (\overline{M}), \\ (\omega + \mathrm{{d}}d^c u)^n = F(u, z)\mu , \\ \lim _{z \rightarrow x} u(z) = \varphi (x) \quad \text {for } x\in \partial M. \end{array}\right. } \end{aligned}$$

(3.1)

A necessary condition to solve the Dirichlet problem is the existence of a subsolution, i.e., a function $\underline{u} \in PSH(M, \omega )$ satisfying $\lim _{z\rightarrow x} u(z) = \varphi (x)$ for $x\in \partial M$ and

$$\begin{aligned} (\omega + \mathrm{{d}}d^c \underline{u})^n \ge F(\underline{u}, z) \mu . \end{aligned}$$

(3.2)

We say that $\mu$ is locally dominated by Monge–Ampère measures of bounded plurisubharmonic functions if for each $p\in M$, there exists a coordinate ball $B \subset \subset M$ centered at p and $v \in PSH(B) \cap L^\infty (B)$ such that

$$\begin{aligned} \mu _{|_B} \le (\mathrm{{d}}d^c v)^n. \end{aligned}$$

(3.3)

By the subsolution theorem from [21] on a smaller ball $B' \subset \subset B$, we can choose v in the Cegrell class $\mathcal {E}_0(B')$, i.e., $\lim _{z\rightarrow \partial B'} v(z) =0$ and $\int _{B'} (\mathrm{{d}}d^cv)^n <+\infty$.

Remark 3.1

The latter property often follows from the existence of a subsolution $\underline{u}$. For example, in geometrically interesting cases: if either $F(u,z)= e^{\lambda u}$ with $\lambda \in \mathbb {R}$, or if $\mu$ is the volume form.

Our main result is as follows.

Theorem 3.2

Suppose that F(t, z) is a bounded non-negative function which is continuous and non-decreasing in the first variable and $\mu$-measurable in the second one. Let $\mu$ be a positive Radon measure satisfying (3.3). Suppose that there exists a bounded subsolution $\underline{u}$ as in (3.2). Then there exists a bounded solution to the Dirichlet problem (3.1).

Proof

We use the Perron envelope method as in [24, Theorem 1.2]. Let $\mathcal {B}(\varphi , \mu )$ be the set

$$\begin{aligned} \left\{ w \in PSH(M, \omega ) \cap L^\infty (M): (\omega + \mathrm{{d}}d^c w)^n \ge F(w,z)\mu , w^*_{|_{\partial M}} \le \varphi \right\} , \end{aligned}$$

(3.4)

where $w^*(x) = \limsup _{M \ni z\rightarrow x} w(z)$ for every $x\in \partial M$. Then, $\mathcal {B}(\varphi , \mu )$ is non-empty as it contains $\underline{u}$. Let $u_0 \in C^0(\overline{M})$ be a $\omega$-subharmonic solution to

$$\begin{aligned} (\omega + \mathrm{{d}}d^c u_0) \wedge \omega ^{n-1} = 0, \quad u_0 = \varphi \quad \text {on } \partial M \end{aligned}$$

(see, e.g., Corollary 4.1). By the comparison principle for $\omega$-subharmonic functions, we have

$$\begin{aligned} v \le u_0 \quad \text {for every }v\in \mathcal {B}(\varphi , \mu ). \end{aligned}$$

(3.5)

Thus, the function

$$\begin{aligned} u (z) = \sup \{v(z) : v \in \mathcal {B}(\varphi , \mu )\} \end{aligned}$$

is well defined. We know that $u^* \in PSH(M, \omega ) \cap L^\infty (\overline{M})$ and $u= u^*$ almost everywhere, outside a pluripolar set. Moreover, if $v_1, v_2 \in \mathcal {B}(\varphi , \mu )$, then so is $\max \{v_1,v_2\}$. Indeed, by an inequality of Demailly [12], we have

$$\begin{aligned} \begin{aligned} (\omega + \mathrm{{d}}d^c \max \{v_1, v_2\})^n&\ge \mathbf{1}_{\{v_1>v_2\}} (\omega + \mathrm{{d}}d^c v_1)^n + \mathbf{1}_{\{v_1\le v_2\}} (\omega + \mathrm{{d}}d^c v_2)^n \\&\ge \mathbf{1}_{\{v_1>v_2\}} F(v_1, z) \mu + \mathbf{1}_{\{v_1\le v_2 \}} F(v_2, z) \mu \\&= F (\max \{v_1,v_2\},z) \mu . \end{aligned} \end{aligned}$$

By this property and Choquet’s lemma, we can write $u= \lim _{j\rightarrow +\infty } u_j$, where $\{u_j\}_{j\ge 1} \subset \mathcal {B}(\varphi , \mu )$ is an increasing sequence. Therefore,

$$\begin{aligned} \begin{aligned} (\omega + \mathrm{{d}}d^c u^*)^n&= \lim _{j \rightarrow +\infty } (\omega + \mathrm{{d}}d^c u_j)^n, \\&\ge \lim _{j \rightarrow +\infty } F (u_j, z) \mu \\&= F (u, z) \mu = F(u^*,z) \mu , \end{aligned} \end{aligned}$$

where the last equality follows the fact that $\mu$ does not charge pluripolar sets. Thus, $u=u^* \in \mathcal {B}(\varphi ,\mu )$ is $\omega$-psh in M. It also follows from the definition and (3.5) that $\underline{u}\le u \le u_0$. Hence, $u= \varphi$ is continuous on $\partial M$.

It remains to show that $\omega _u^n = F(u,z)\mu$ in M. To see this, let $B \subset \subset M$ be a coordinate ball in M. Following the same argument as in [24, Lemma 3.7], given the local solvability of the Dirichlet problem (Theorem 2.3), there exists $\widetilde{u} \in \mathcal {B}(\varphi , \mu )$ such that $u \le \widetilde{u}$ and $(\omega +\mathrm{{d}}d^c \widetilde{u})^n = F(\widetilde{u}, z) \mu$ in this small coordinate ball. By definition of u we must have $\widetilde{u} \le u$. So, $\widetilde{u} = u$ in B. In other words, $\omega _u^n = F(u, z) \mu$. Since B is arbitrary, this proves our claim. $\square$

For $F(t, z) = e^{\lambda t}$ with $\lambda >0$, we obtain a stronger statement.

Corollary 3.3

Let $\lambda >0$. There exists a unique solution to the Dirichlet problem

$$\begin{aligned} {\left\{ \begin{array}{ll} u \in PSH(M, \omega ) \cap L^\infty (\overline{M}), \\ \omega _u^n = e^{\lambda u} \mu , \\ \lim _{z\rightarrow x} u(z) = \varphi (x) \quad \text {for } x\in \partial M. \end{array}\right. } \end{aligned}$$

if and only if there exists a bounded subsolution.

Moreover, if the subsolution is Hölder continuous, so is the solution.

Proof

The uniqueness of the solution under the hypothesis of the existence of a subsolution follows by the same arguments as in [25, Lemma 2.3]. Next, to prove the equivalence, we only need to verify the local domination by Monge–Ampère measures of bounded plurisubharmonic functions for $\mu$ (see condition (3.3)). This is straightforward.

For the second conclusion, we first observe as in [24, Lemma 6.5] that u is Hölder continuous on the boundary $\partial M$. Since u is bounded,

$$\begin{aligned} \omega _u^n = e^{\lambda u}\mu \le C (\omega + \mathrm{{d}}d^c \underline{u})^n. \end{aligned}$$

Thus, the Hölder continuity of u follows from the proof of [24, Theorem 1.4]. $\square$

Thanks to this, we can also obtain the solution of the Monge–Ampère equation as the limit of the solutions of the Monge–Ampère type equations.

Corollary 3.4

Suppose that there exists a function $\underline{u} \in PSH(M, \omega ) \cap L^\infty (M)$ which satisfies: $\lim _{z\rightarrow x}\underline{u}(z) = \varphi (x)$ for $x\in \partial M$ and $(\omega + \mathrm{{d}}d^c \underline{u})^n \ge \mu$ in M. Then, the sequence of solutions

$$\begin{aligned} (\omega + \mathrm{{d}}d^c u_\lambda )^n = e^{\lambda u} \mu \quad \text {with } \quad \lambda >0 \end{aligned}$$

converges to a solution u of $\omega _u^n = \mu$ and $u = \varphi - \sup _{M} \underline{u}$ on $\partial M$ as $\lambda \rightarrow 0$.

Proof

Let $b = \sup _{M} \underline{u}$. Then, $v:=\underline{u} - b \le 0$ on M. For every $\lambda >0$, the function v satisfies $v = \varphi - b$ on $\partial M$ and $(\omega + \mathrm{{d}}d^c v)^n \ge e^{\lambda v} \mu$ in M. Applying Theorem 3.2, we obtain the family of solutions $\{u_\lambda \}_{0< \lambda \le 1}$ of

$$\begin{aligned} (\omega +\mathrm{{d}}d^c u_\lambda )^n = e^{\lambda u} \mu , \quad u_\lambda = \varphi - b \quad \text {on } \partial M. \end{aligned}$$

By the domination principle, the family is increasing in $\lambda >0$ and $u_\lambda \ge v$ for every $0<\lambda \le 1$. Set $u = \lim _{\lambda \rightarrow 0} u_\lambda .$ Then, $u + b$ is a solution to $\omega _u^n = \mu$ in M and $u = \varphi$ on $\partial M$. $\square$

Appendix

In the proofs in this paper as well as in [24], we use the existence and regularity of solutions of the linear elliptic equation on a manifold with boundary. Those statements are known, for instance, as consequences of general results for harmonic maps [20, Theorem 6] (see also [27, Theorem 5.3]). However, since the case of Hölder continuous boundary data seems not to be available in literature, we include the proof here for the sake of completeness.

Let $(\overline{M},\omega )$ be a compact n-dimensional Hermitian manifold with non-empty boundary $\partial M$. Then $\overline{M} = M \cup \partial M$, where M is a (open) Hermitian manifold. Suppose that in local coordinate, we have

$$\begin{aligned} \omega = \sqrt{-1}g_{i{\bar{j}}} (z) \mathrm{{d}}z^i \wedge \mathrm{{d}}{\bar{z}}^j. \end{aligned}$$

Define $\Delta _g = g^{{\bar{j}} i} \partial _i \partial _{{\bar{j}}}$ be the Laplace operator associated to $\omega$ and denote by $\mathrm{dist}(z,w)$ the distance function induced by $\omega$.

Proposition 4.1

Let $\varphi \in C^0(\partial M)$. Then, there exists a unique continuous solution to

$$\begin{aligned} (\omega + \mathrm{{d}}d^c u) \wedge \omega ^{n-1} =0\quad \text {in }M, \quad u= \varphi \quad \text {on } \partial M. \end{aligned}$$

Moreover, if $\varphi$ is Hölder continuous on $\partial M$, then so is the solution u.

Since

$$\begin{aligned} \Delta _g u = \mathrm{tr}_\omega u = \frac{n\mathrm{{d}}d^c u \wedge \omega ^{n-1}}{\omega ^n}, \end{aligned}$$

we can separate the equation into two problems $\Delta _g u_1 = 0$ in M with $u_1 = \varphi$ on $\partial M$ and $\Delta _g u_2 = -n$ in M with $u_2 =0$ on $\partial M$. The latter solution $u_2$ is smooth by the classical PDEs.

We are thus reduced to proving the following.

Proposition 4.2

Let $\varphi \in C^0(\partial M)$. Then, there exists a unique continuous solution to

$$\begin{aligned} \Delta _g u = 0 \quad \text {in M}, \quad u = \varphi \quad \text {on } \partial M. \end{aligned}$$

(4.1)

Moreover, if $\varphi$ is Hölder continuous on $\partial M$, then so is u.

The existence of continuous solutions follows exactly as in [15, pp. 24–25] by using the Perron envelope

$$\begin{aligned} \mathcal {S}_\varphi = \{ v \in SH_\omega (M) \cap C^0(\overline{M}): v_{|_{\partial M}} \le \varphi \}, \end{aligned}$$

where $SH_\omega (M)$ is the set of all $\Delta _g$-subharmonic functions in M. The function

$$\begin{aligned} u(x) = \sup _{v \in S_\varphi } v(x) \quad \text {for } x\in \overline{M} \end{aligned}$$

is the solution to (4.1) by the Perron method using harmonic liftings:

Lemma 4.3

Let $v \in S_\varphi$. Then there exists a function $\widetilde{v}$ called a lift of v in B such that $\widetilde{v} \ge v$ on $\overline{M}$ and satisfying $\Delta _g \widetilde{v} = 0$ in B and $\widetilde{v}= v$ on $\partial B$.

The boundary condition is satisfied since using the regularity of the boundary of the domain, one can easily construct, as in [15], first the local barriers, and then the global one.

We now assume further that $\varphi$ belongs to $C^{0,\alpha }(\overline{M})$ with $0<\alpha <1$. Then, we wish to show that the solution also belongs to a Hölder space.

First, we will construct a Hölder continuous local barrier similarly as in [1, Theorem 6.2] on the coordinate half-ball at each boundary point.

Lemma 4.4

Suppose the origin $0 \in \partial M \cap B(0,R)$, $\rho$ is the defining function of $\partial M \cap B(0,R)$ in the coordinate ball B(0, R) and $U_R = \{ z\in B(0,R): \rho (z)\le 0 \}$ is the coordinate half-ball centered at 0. Denote $\Vert \varphi \Vert _{\alpha }=c_1$ for the Hölder norm of $\varphi$ on $\partial M$. Let $0<\tau \le \alpha < 1$. Then, there exists a constant $k = k(\varphi , U_R)$ and a neighborhood W of 0 such that the function

$$\begin{aligned} v(z) = k |\rho |^\tau (z) + c_1 |z|^{\alpha } + \varphi (0) \end{aligned}$$

is $\Delta _g$-superharmonic in $W \cap U_R$. Moreover,

$$\begin{aligned} v(0) = \varphi (0), \quad v(x) \ge \varphi (x) \quad \text {for every } x\in \partial M \cap B(0,R). \end{aligned}$$

Proof

We compute in B(0, R),

$$\begin{aligned} \mathrm{{d}}d^c |\rho |^\tau = - \tau |\rho |^{\tau -1} \mathrm{{d}}d^c \rho - \tau (1-\tau ) |\rho |^{\tau -2} \mathrm{{d}}\rho \wedge d^c \rho , \end{aligned}$$

and for $\alpha ' = \alpha /2$,

$$\begin{aligned} \mathrm{{d}}d^c |z|^{2\alpha '} = \alpha ' |z|^{2(\alpha '-1)} \mathrm{{d}}d^c |z|^2 - \alpha '(\alpha '-1) |z|^{2(\alpha '-2)} \mathrm{{d}} |z|^2 \wedge d^c|z|^2. \end{aligned}$$

Hence,

$$\begin{aligned} \mathrm{{d}}d^c v (z) \wedge \omega ^{n-1}/\omega ^n \le - k\tau (1-\tau ) |\rho |^{\tau -2} |\nabla \rho |_g^2 + \frac{c_1 \alpha }{2} |z|^{\alpha -2}. \end{aligned}$$

Furthermore, $|\rho (z)|= |\rho (z) - \rho (0)| \le c_2 |z|$ for every $z \in \overline{B(0,R) }$. Since $\tau -2<0$, it implies that

$$\begin{aligned} \begin{aligned} \frac{1}{n} \Delta _g v (z)&\le - c_3 k |z|^{\tau -2} |\nabla \rho |_g^2 + \frac{c_1 \alpha }{2} |z|^{\alpha -2} \\&= |z|^{\alpha -2} \left( c_1 \alpha /2 - c_3 k |z|^{\tau -\alpha } |\nabla \rho |_g^2\right) \\&\le |z|^{\alpha -2} \left( c_1 \alpha - c_3 k R^{\tau -\alpha } |\nabla \rho |_g^2\right) , \end{aligned} \end{aligned}$$

where $c_3 = \tau (1-\tau ) c_2^{\tau -2}$. We used the fact $\tau \le \alpha$ for the last inequality. Since $\rho$ is the defining function for $\partial M \cap B(0,R)$, it follows that $|\nabla \rho |_g> \varepsilon _0 >0$ in a neighborhood W of 0. Then, we can choose large $k>0$, independent of the boundary point 0, so that $\Delta _g v \le 0$. Finally, the remaining properties of v hold because $\rho (x) =0$ for every $x\in \partial M \cap B(0,R)$. $\square$

Proposition 4.5

There exists a function $\overline{u}(x) \in C^{0,\alpha }(\overline{M})$ that is $\Delta _g$-superharmonic in M and $\overline{u}(\xi ) = \varphi (\xi )$ for $\xi \in \partial M$.

Proof

We first show that at each point $\xi \in \partial M$, there exist a $\Delta _g$-superharmonic function $v_\xi$ in M such that $v_\xi \in C^{0,\alpha }(\overline{M})$ and

$$\begin{aligned}&v_{\xi } (x) \ge \varphi (x) \quad \text {on } \partial M, \end{aligned}$$

(4.2)

$$\begin{aligned}&v_\xi (\xi ) = \varphi (\xi ), \end{aligned}$$

(4.3)

$$\begin{aligned}&\Vert v_\xi \Vert _{\alpha } \le C \Vert \varphi \Vert _{2\alpha }, \end{aligned}$$

(4.4)

where $\Vert \cdot \Vert _\alpha$ denotes the $\alpha$-Hölder norm of the function and the constant C depends only on $\overline{M}$ and the metric $\omega$.

Without loss of generality, we may assume $\xi$ is the origin and $\varphi (0) =0$. By Lemma 4.4 for the boundary point $0 \in \partial M$ there exists a function $v = k |\rho |^\alpha + c_1 |z|^{\alpha }$ and a neighborhood W of 0 such that v satisfies (4.2), (4.3) and (4.4) on $W \cap U_R$. Note that the constants $k,c_1$ are independent of the boundary points. We can extend this function to a global one as follows. Set $k_1 = \sup _{\partial M} \varphi +1$. Then, for large $k_2\ge 1$ (depending on $k_1$ and R, but independent of the boundary point), the function

$$\begin{aligned} \widehat{v} = \min \{k_1, k_2 v\} \end{aligned}$$

is $\Delta _g$-superharmonic on M and satisfies the list of required properties.

Now let us define

$$\begin{aligned} \overline{u} = \inf \{ v_\xi : \xi \in \partial M\}. \end{aligned}$$

From $|v_\xi (z) - v_\xi (w)| \le C[\mathrm{dist}(z,w)]^\alpha$, we deduce that $|\overline{u}(z) -\overline{u}(w)| \le C[\mathrm{dist}(z,w)]^\alpha$. Thus, $\overline{u}$ is Hölder continuous $\Delta _g$-superharmonic, and clearly $\overline{u}(\xi ) = \varphi (\xi )$ for every $\xi \in \partial M$. This completes the proof. $\square$

By the similar argument, we can find a global Hölder continuous $\Delta _g$-subharmonic barrier $\underline{u}\in C^{0,\alpha }(\overline{M})$ such that

$$\begin{aligned} \underline{u}(\xi ) = \varphi (\xi ) \quad \text {for }\xi \in \partial M. \end{aligned}$$

(4.5)

Hence, by the maximum principle, $\underline{u} \le u \le \overline{u}$ on $\overline{M}.$ Consequently, we get

$$\begin{aligned} |u(x) - u(\xi )| \le C [\mathrm{dist}( x, \xi )]^\alpha \quad \text {for every } x\in \overline{M} \text{ and } \xi \in \partial M. \end{aligned}$$

(4.6)

Now we are going to show the global Hölder continuity of the solution.

Lemma 4.6

There exists a constant $C = C(\varphi , M, \omega )$ such that

$$\begin{aligned} |u(x) - u(y)| \le C [\mathrm{dist}(x,y)]^\alpha \quad \text {for every } x, y \in \overline{M}. \end{aligned}$$

(4.7)

Proof

By maximum principle, we get $\inf _{\partial M} \varphi \le u \le \sup _{\partial M}\varphi$. Denote $d_x = \mathrm{dist}(x, \partial M)$ and $d_y = \mathrm{dist}(y, \partial M)$. Suppose that $d_y \le d_x$, and take $x_0, y_0 \in \partial M$ such that $\mathrm{dist}(x, x_0) = d_x$ and $\mathrm{dist}(y, y_0) = d_y$.

Case 1 $\mathrm{dist}(x,y) \le d_x/2$. Since $\Delta _g$ is uniformly elliptic, we have the interior Hölder estimates (see Corollary 9.24 and Lemma 8.23 in [15])

$$\begin{aligned} \Vert u\Vert _{C^{0,\alpha }(\overline{B}_{R/2})} \le C R^{-\alpha } \Vert u\Vert _{L^\infty (B_R)}, \end{aligned}$$

where $B_R \subset M$ is a coordinate ball of small radius $0< R<1$. Since $y \in \overline{B}_{d_x/2} (x) \subset B_{d_x}(x) \subset M$. Applying the interior inequality to $u-u(x_0)$ in $B_{d_x}(x)$, we get

$$\begin{aligned} d_x^\alpha \frac{|u(x) - u(y)|}{[\mathrm{dist}(x,y)]^\alpha } \le C \Vert u - u(x_0)\Vert _{L^\infty (B_{d_x}(x))}. \end{aligned}$$

By (4.6), the right-hand side is less than $C d_x^{\alpha }$. It follows that

$$\begin{aligned} |u(x) - u(y)| \le C [\mathrm{dist}(x,y)]^\alpha . \end{aligned}$$

Case 2 $d_y \le d_x \le 2 \,\mathrm{dist}(x,y)$. Then,

$$\begin{aligned} \begin{aligned} |u(x) - u(y)|&\le |u(x) - u(x_0)| + |u(x_0)-u(y_0)| + |u(y)- u(y_0)| \\&\le C(d_x^\alpha + [\mathrm{dist} (x_0,y_0)]^\alpha + d_y^\alpha ) \\&\le C [\mathrm{dist}(x,y)]^\alpha \end{aligned} \end{aligned}$$

since $\mathrm{dist} (x_0 ,y_0) \le d_x + \mathrm{dist}(x,y) + d_y \le 5 \, \mathrm{dist}(x,y)$. $\square$