Asymptotics and Convergence for the Complex Monge-Ampère Equation

Abstract

We study the asymptotics of complete Kähler-Einstein metrics on strictly pseudoconvex domains in $\mathbb {C}^n$ and derive a convergence theorem for solutions to the corresponding Monge-Ampère equation. If only a portion of the boundary is analytic, the solutions satisfy Gevrey type estimates for tangential derivatives. A counterexample for the model linearized equation suggests that there is no local convergence theorem for the complex Monge-Ampère equation.

Appendix A. Analyticity Type Estimates

If $p<0$, we assume that $p!=0.$ The following result is due to Friedman [7]. Also see [12].

Lemma A.1

Let $\Omega$ be a domain in $\mathbb R^n$ and p be a positive integer. Assume that $\Phi$ is a $C^p$-function in $\Omega \times \mathbb R^N$ satisfying, for any $(x,y)\in \Omega \times \mathbb R^N$ and any nonnegative integers j and k with $j+k\le p$,

$$\begin{aligned} \Big |\frac{\partial ^{j+k}\Phi (x,y)}{\partial x^j\partial y^k}\Big | \le A_0A_1^jA_2^k(j-2)!(k-2)!, \end{aligned}$$

(A.1)

for some positive constants $A_0$, $A_1$, and $A_2$. Then, there exist positive constants $B_0$, $\widetilde{B}_0$, and $B_1$, depending only on n, N, $A_0$, $A_1$ and $A_2$, such that, for any $C^p$-function $y=(y_1, \cdots , y_N): \Omega \rightarrow \mathbb R^N$, if for any $x\in \Omega$ and any nonnegative integer $k\le p$,

$$\begin{aligned} \sum _{i=1}^N|\partial ^k_xy_i(x)|\le B_0B_1^{(k-2)^+}(k-2)!, \end{aligned}$$

(A.2)

then, for any $x\in \Omega$,

$$\begin{aligned} |\partial _x^p[\Phi (x,y(x))]|\le \widetilde{B}_0B_1^{(p-2)^+}(p-2)!. \end{aligned}$$

(A.3)

Remark A.2

Write $x=(x',x_n)$. In Lemma A.1, if we assume (A.1) and (A.2) hold only for $D_{x'}$ instead of $D_x$, then (A.3) holds for $D_{x'}$.

Lemma A.3

Let p be any integer and $p\ge 2$. Then, there is a universal constant C such that

$$\begin{aligned} \sum _{l=1}^{p-1} {\left( {\begin{array}{c}p-1\\ l\end{array}}\right) }(l-2)! \cdot (p-2-l)! \le C (p-2)!, \end{aligned}$$

(A.4)

and

$$\begin{aligned} \sum _{l=0}^{p-1} {\left( {\begin{array}{c}p-1\\ l\end{array}}\right) } (l-2) ! (p-3-l)!\le C(p-3)!. \end{aligned}$$

(A.5)

Proof

If $l=p-1$, we have

$$\begin{aligned} {\left( {\begin{array}{c}p-1\\ l\end{array}}\right) }(l-2)! \cdot (p-l-2)! = (p-3)!. \end{aligned}$$

Moreover, we get

$$\begin{aligned} \sum _{l=1}^{p-2} {\left( {\begin{array}{c}p-1\\ l\end{array}}\right) }(l-2)! \cdot (p-2-l)!&\le \sum _{l=2}^{p-2} \frac{p-1}{l(l-1)(p-l-1)}(p-2)! \\&\le C(p-2)!, \end{aligned}$$

as

$$\begin{aligned} \sum _{l=2}^{p-2} \frac{p-1}{l(l-1)(p-l-1)}\&=\sum _{l=2}^{p-2} \frac{1}{l-1}\left( \frac{1}{l}+\frac{1}{p-l-1}\right) \\&=\sum _{l=2}^{p-2} \frac{1}{(l-1)l}+\frac{1}{p-2}\sum _{l=2}^{p-2}\left( \frac{1}{l-1}+\frac{1}{p-l-1}\right) , \end{aligned}$$

which is bounded by a universal constant C. Notice that $\sum _{l=2}^{p-2} \frac{1}{l-2} < 1+\ln p$ and $\frac{1+\ln p}{p-2}$ is bounded by a universal constant when $p \ge 3$. Then we derive (A.4).

For (A.5), first we observe that, for $l=0, 1, p-1$ or $p-2$,

$$\begin{aligned} {\left( {\begin{array}{c}p-1\\ l\end{array}}\right) } (l-2) ! (p-3-l)!\le C(p-3)!. \end{aligned}$$

Moreover, we get

$$\begin{aligned}&\sum _{l=2}^{p-3} {\left( {\begin{array}{c}p-1\\ l\end{array}}\right) } (l-2) ! (p-3-l)!\\&= (p-3)! \sum _{l=2}^{p-3} \frac{(p-1)(p-2)}{l(l-1)(p-1-l)(p-2-l)} \\&\le 2(p-3)! \sum _{l=2}^{p-3} \left( \frac{1}{l} + \frac{1}{p-2-l}\right) \left( \frac{1}{l-1} + \frac{1}{p-1-l}\right) , \end{aligned}$$

where the summation over l is bounded by a universal constant. Then, we get (A.5). $\square$