(of Corollary 1.5) Define $F=\log\det u_{\alpha\bar{\beta}}$, first we show that $F$ is a constant. By the assumption, we can take a sequence of $r_{s}\rightarrow\infty$, and a constant $M$, such that

(6.13)

$$\sup_{s\geq 1}\frac{1}{r_{s}^{2n}}\int_{B_{r_{s}}(0)}(\Delta u)^{p}+\big(\sum_{k}\frac{1}{u_{k\bar{k}}}\big)^{p}\leq M.$$

Define $u_{s}(z)=\frac{1}{r_{s}^{2}}u(r_{s}z)$. Let $\Delta_{s}:=u_{s}^{i\bar{j}}\partial_{i\bar{j}}$, the Laplace operator in $\mathbb{C}^{n}$ defined by the metric $\sqrt{-1}\partial\bar{\partial}u_{s}$. Also we denote $F_{s}:=\log\det\partial_{\alpha\bar{\beta}}u_{s}$, then $F_{s}(z)=F(r_{s}z)$. Hence $\Delta_{s}F_{s}=0$ in $B_{1}(0)$, and (6.13) implies

(6.14)

$$\sup_{s\geq 1}\int_{B_{1}}(\Delta u_{s})^{p}+\big(\sum_{k}\frac{1}{(u_{s})_{k\bar{k}}}\big)^{p}\leq M.$$

Proposition 6.1 shows that there exists a positive constant $C_{7}$, independent of $s$, such that $\frac{1}{C_{7}}\leq(u_{s})_{i\bar{j}}\leq C_{7}$ and $|\nabla F_{s}|\leq C_{7}$ in $B_{\frac{1}{2}}(0)$. Rescaling back, we find that $|\nabla F|\leq\frac{C_{7}}{r_{s}}$ in $B_{\frac{1}{2}r_{s}}(0)$. Sending $s\rightarrow\infty$, we get $\nabla F\equiv 0$ on $\mathbb{C}^{n}$. Namely we have $\det\partial_{\alpha\bar{\beta}}u_{s}=c$ for some $c>0$ on $B_{1}$. Then we may use Evans-Krylov theorem to conclude for some $\alpha>0$

$$\sup_{k}[\sqrt{-1}\partial\bar{\partial}u_{s}]_{\alpha,B_{\frac{1}{4}}}\leq M_{1}.$$

In terms of $u$, this implies

$$r_{s}^{\alpha}\frac{|u_{i\bar{j}}(z_{1})-u_{i\bar{j}}(z_{2})|}{|z_{1}-z_{2}|^{\alpha}}\leq M,\textrm{ for any $z_{1},\,z_{2}\in B_{\frac{r_{s}}{4}}(0)$.}$$

Letting $s\rightarrow\infty$, we obtain $u_{i\bar{j}}(z_{1})=u_{i\bar{j}}(z_{2})$, for any $z_{1},\,z_{2}\in\mathbb{C}^{n}$. This implies the Levi Hessian of $u$ is constant. ∎