Subsolution theorem and the Dirichlet problem for the quaternionic Monge–Ampère equation

Abstract

In this paper, we obtain the existence of solution to the Dirichlet problem for quaternionic Monge–Ampère equation. After showing a stability theorem, we prove the subsolution theorem for quaternionic Monge–Ampère equation by combining our stability theorem and Kołodziej’s approach for the complex case.

Appendix A: Proof of Lemma 3.7

Proof

For $C^2$ smooth plurisubharmonic functions $u_i$, $i=1,\ldots ,n$, their quaternionic Hessian $\left[ \frac{\partial ^2u_i}{\partial {\overline{q}}_j\partial q_k}(q)\right]$ are positive definite hyperhermitian matrices. By Theorem 1.1.15 in [1],

$$\begin{aligned} det(u_1,\ldots ,u_n):=det\left( \left( \frac{\partial ^2u_1}{\partial {\bar{q}}_j\partial q_k}\right) ,\ldots ,\left( \frac{\partial ^2u_n}{\partial {\bar{q}}_j\partial q_k}\right) \right) >0. \end{aligned}$$

It follows from Aleksandrov inequality (cf. Corollary 1.1.16 in [1], or Corollary 2.16 in [51]) that

$$\begin{aligned} det\left( u_1,u_2,\ldots ,u_n\right) \ge det\left( u_1,u_1,u_3,\ldots ,u_n\right) ^{\frac{1}{2}}\cdot det\left( u_2,u_2,u_3,\ldots ,u_n\right) ^{\frac{1}{2}}. \end{aligned}$$

(A.1)

We claim that when all functions are in $QPSH\cap C^2(\Omega )$,

$$\begin{aligned} \begin{aligned}&det\left( \underbrace{u_1,\ldots ,u_1}_p,\underbrace{u_2,\ldots ,u_2}_q,v_1,\ldots ,v_{n-p-q}\right) \\&\quad \ge det\left( \underbrace{u_1,\ldots ,u_1}_{p+q},v_1,\ldots ,v_{n-p-q}\right) ^{\frac{p}{p+q}}\cdot det\left( \underbrace{u_2,\ldots ,u_2}_{p+q},v_1,\ldots ,v_{n-p-q}\right) ^{\frac{q}{p+q}}. \end{aligned} \end{aligned}$$

(A.2)

We prove this claim by induction. The case for $p=q=1$ holds by (A.1). Assume by induction that the case for $p+q\le m$ has already been proved. It suffices to prove it for $p+q\le m+1$. First we need the following inequality.

$$\begin{aligned} det\left( u_1,\underbrace{u_2,\ldots ,u_2}_{p+q},v,\ldots \right) \ge det\left( \underbrace{u_1,\ldots ,u_1}_{p+q+1},v,\ldots \right) ^{\frac{1}{p+q+1}}\cdot det\left( \underbrace{u_2,\ldots ,u_2}_{p+q+1},v,\ldots \right) ^{\frac{p+q}{p+q+1}}. \end{aligned}$$

(A.3)

By induction assumption we have

$$\begin{aligned}&det\left( u_1,\underbrace{u_2,\ldots ,u_2}_{p+q},v,\ldots \right) \\&\quad \ge det\left( \underbrace{u_1,\ldots ,u_1}_{p+q},u_2,v,\ldots \right) ^{\frac{1}{p+q}}\cdot det\left( \underbrace{u_2,\ldots ,u_2}_{p+q},u_2,v,\ldots \right) ^{\frac{p+q-1}{p+q}}\\&\quad =det\left( \underbrace{u_1,\ldots ,u_1}_{p+q-1},u_2,u_1,v,\ldots \right) ^{\frac{1}{p+q}}\cdot det\left( \underbrace{u_2,\ldots ,u_2}_{p+q+1},v,\ldots \right) ^{\frac{p+q-1}{p+q}}\\&\quad \ge \left[ det\left( \underbrace{u_1,\ldots ,u_1}_{p+q},u_1,v,\ldots \right) ^{\frac{p+q-1}{p+q}}\cdot det\left( \underbrace{u_2,\ldots ,u_2}_{p+q},u_1,v,\ldots \right) ^{\frac{1}{p+q}}\right] ^{\frac{1}{p+q}}\\&\qquad \cdot det\left( \underbrace{u_2,\ldots ,u_2}_{p+q+1},v,\ldots \right) ^{\frac{p+q-1}{p+q}}. \end{aligned}$$

Then we have

$$\begin{aligned}&det\left( u_1,\underbrace{u_2,\ldots ,u_2}_{p+q},v,\ldots \right) ^{\frac{(p+q)^2-1}{(p+q)^2}}\\&\quad \ge det\left( \underbrace{u_1,\ldots ,u_1}_{p+q+1},v,\ldots \right) ^{\frac{p+q-1}{(p+q)^2}}\cdot det\left( \underbrace{u_2,\ldots ,u_2}_{p+q+1},v,\ldots \right) ^{\frac{p+q-1}{p+q}}. \end{aligned}$$

It follows (A.3). Now we complete the induction by using (A.3).

$$\begin{aligned} \begin{aligned}&det\left( \underbrace{u_1,\ldots ,u_1}_{p+1},\underbrace{u_2,\ldots ,u_2}_{q},v,\ldots \right) \\&\quad \ge det\left( \underbrace{u_1,\ldots ,u_1}_{p+q},u_1,v,\ldots \right) ^{\frac{p}{p+q}}\cdot det\left( \underbrace{u_2,\ldots ,u_2}_{p+q},u_1,v,\ldots \right) ^{\frac{q}{p+q}}\\&\quad \ge det\left( \underbrace{u_1,\ldots ,u_1}_{p+q+1},v,\ldots \right) ^{\frac{p}{p+q}}\cdot \\&\qquad \left[ det\left( \underbrace{u_2,\ldots ,u_2}_{p+q+1},v,\ldots \right) ^{\frac{p+q}{p+q+1}}\cdot det\left( \underbrace{u_1,\ldots ,u_1}_{p+q+1},v,\ldots \right) ^{\frac{1}{p+q+1}}\right] ^{\frac{q}{p+q}}\\&\quad =det\left( \underbrace{u_1,\ldots ,u_1}_{p+q+1},v,\ldots \right) ^{\frac{p+1}{p+q+1}}\cdot det\left( \underbrace{u_2,\ldots ,u_2}_{p+q+1},v,\ldots \right) ^{\frac{q}{p+q+1}}. \end{aligned} \end{aligned}$$

Then (A.2) is proved. It follows that

$$\begin{aligned} det\left( u_1,\underbrace{u_2,\ldots ,u_2}_{n-1}\right) \ge (det u_1)^{\frac{1}{n}}(det u_2)^{\frac{n-1}{n}}. \end{aligned}$$

This is the case of (3.4) for $u_2=\cdots =u_n=u$. Assume that (3.4) is proved for $u_{p+1}=\cdots =u_n=u$. We now prove it for $u_{p+2}=\cdots =u_n=u$. By (A.2) we have

$$\begin{aligned} \begin{aligned}&det\left( u_1,\ldots ,u_p,u_{p+1},\underbrace{u,\ldots ,u}_{n-p-1}\right) \\&\quad \ge det\left( u_1,\ldots ,u_p,\underbrace{u_{p+1},\ldots ,u_{p+1}}_{n-p}\right) ^{\frac{1}{n-p}} \cdot det\left( u_1,\ldots ,u_p,\underbrace{u,\ldots ,u}_{n-p}\right) ^{\frac{n-p-1}{n-p}}\\&\quad \ge \left[ (det u_1)^{\frac{1}{n}}\cdots (det u_p)^{\frac{1}{n}}(det u_{p+1})^{\frac{n-p}{n}}\right] ^{\frac{1}{n-p}}\cdot \left[ (det u_1)^{\frac{1}{n}}\cdots (det u_p)^{\frac{1}{n}}(det u)^{\frac{n-p}{n}}\right] ^{\frac{n-p-1}{n-p}}\\&\quad =(det u_1)^{\frac{1}{n}}\cdots (det u_p)^{\frac{1}{n}}(det u_{p+1})^{\frac{1}{n}}(det u)^{\frac{n-p-1}{n}}. \end{aligned} \end{aligned}$$

The induction is complete. $\square$