(of Proposition 6.1)First we want to get boundedness of $G$, using the second equation. We can write the second equation as

(6.3)

$$\det(\phi_{\alpha\bar{\beta}})\phi^{i\bar{j}}\partial_{i\bar{j}}G=e^{G}G.$$

which is equivalent to:

(6.4)

$$Re\big(\partial_{i}(\det(\phi_{\alpha\bar{\beta}})\phi^{i\bar{j}}\partial_{\bar{j}}G)\big)=e^{G}G.$$

Denote $a_{i\bar{j}}=\det(\phi_{\alpha\bar{\beta}})\phi^{i\bar{j}}$, which is a hermitian matrix, then for some constant $c_{n}>0$, we have

(6.5)

$$\frac{1}{c_{n}\big(\sum_{i}\frac{1}{\phi_{i\bar{i}}}\big)^{n-1}}I\leq a_{i\bar{j}}\leq c_{n}(\Delta\phi)^{n-1}I.$$

The left hand side of (6.4) is a real elliptic operator in divergence form, which satisfies an ellipticity condition same as (6.5). We wish to apply Lemma 6.3 to the equation (6.4). Using (6.5), we can take $\lambda=\big(\Delta\phi+\sum_{i}\frac{1}{\phi_{i\bar{i}}}\big)^{n-1}$, and $f=e^{G}G$. In order to apply Lemma 6.3, we need to show $(\Delta\phi)^{n-1},\,(\sum_{i}\frac{1}{\phi_{i\bar{i}}})^{n-1}\in L^{p}(B_{1})$, and $e^{G}G\in L^{p/2}(B_{1})$ for some $p>3n$. The desired integrability for $\Delta\phi$ and $\sum_{i}\frac{1}{\phi_{i\bar{i}}}$ is clear from assumption, while for $e^{G}G$, since $e^{G}\leq(\Delta\phi)^{n}$, we just need to make sure $(\Delta\phi)^{n}\in L^{p^{\prime}}$ for some $p^{\prime}>\frac{3n}{2}$. This is again clear from our assumption on $\Delta\phi$. So we can apply Lemma 6.3 to conclude $G$ is bounded(with the said dependence) on any interior ball of $B_{1}$. In the following we assume $G$ is bounded on $B_{1}$ without loss of generality.

The estimate for $\Delta\phi$ is really similar to our calculation in section 4, so we will be suitably brief here.

Choose any point $p$ and we can do a unitary coordinate transform so that $\phi_{i\bar{j}}(p)=\phi_{i\bar{i}}(p)\delta_{ij}$. We can compute

(6.6)

$$\Delta_{\phi}(|\nabla_{\phi}G|^{2})=\frac{1}{\phi_{i\bar{i}}\phi_{\alpha\bar{\alpha}}}|\phi_{i\alpha}-\sum_{p}\frac{\phi_{i\alpha\bar{p}}G_{p}}{\phi_{p\bar{p}}}|^{2}+\frac{|G_{p\bar{i}}|^{2}}{\phi_{i\bar{i}}\phi_{p\bar{p}}}-\frac{G_{q\bar{p}}G_{p}G_{\bar{q}}}{\phi_{p\bar{p}}\phi_{q\bar{q}}}.$$

Here $|\nabla_{\phi}G|^{2}=\phi^{p\bar{q}}G_{p}G_{\bar{q}}$.

(6.7)

$$\Delta_{\phi}(e^{\frac{1}{2}G}|\nabla_{\phi}G|^{2})=\Delta_{\phi}(e^{\frac{1}{2}G})|\nabla_{\phi}G|^{2}+e^{\frac{1}{2}G}\Delta_{\phi}(|\nabla_{\phi}G|^{2})+\frac{1}{2}e^{\frac{1}{2}G}\frac{G_{i}(|\nabla_{\phi}G|^{2})_{\bar{i}}+G_{\bar{i}}(|\nabla_{\phi}G|^{2})_{i}}{\phi_{i\bar{i}}}.$$

One can also compute

(6.8)

$$\frac{G_{i}(|\nabla_{\phi}G|^{2})_{\bar{i}}}{\phi_{i\bar{i}}}=\frac{G_{i}G_{p}}{\phi_{i\bar{i}}\phi_{p\bar{p}}}(G_{\bar{p}\bar{i}}-\sum_{t}\frac{\phi_{t\bar{q}\bar{i}}G_{\bar{t}}}{\phi_{t\bar{t}}})+\frac{G_{p\bar{i}}G_{\bar{p}}G_{i}}{\phi_{p\bar{p}}\phi_{i\bar{i}}}.$$

Combining (6.6), (6.7), (6.8), we obtain

(6.9)

$$\Delta_{\phi}(e^{\frac{1}{2}G}|\nabla_{\phi}G|^{2})e^{-\frac{1}{2}G}=-\frac{1}{2}\underline{R}|\nabla_{\phi}G|^{2}+\frac{1}{\phi_{i\bar{i}}\phi_{\alpha\bar{\alpha}}}|G_{i\alpha}-\sum_{p}\frac{\phi_{i\alpha\bar{p}}G_{p}}{\phi_{p\bar{p}}}-\frac{1}{2}\phi_{i}\phi_{\alpha}|^{2}+\frac{|G_{p\bar{i}}|^{2}}{\phi_{i\bar{i}}\phi_{p\bar{p}}}.$$

Also we can compute

(6.10)

$$\Delta_{\phi}(\Delta\phi)=\frac{|\phi_{i\bar{j}p}|^{2}}{\phi_{i\bar{i}}\phi_{j\bar{j}}}+\Delta G.$$

Hence

(6.11)

$$\begin{split}\Delta_{\phi}&(e^{\frac{1}{2}G}|\nabla_{\phi}G|^{2}+\Delta\phi)\geq-\frac{1}{2}\underline{R}e^{\frac{1}{2}G}|\nabla_{\phi}G|^{2}+\frac{|G_{p\bar{i}}|^{2}e^{\frac{1}{2}G}}{\phi_{i\bar{i}}\phi_{p\bar{p}}}+\Delta G\\ &\geq-\frac{1}{2}\underline{R}e^{\frac{1}{2}G}|\nabla_{\phi}G|^{2}+\frac{e^{\frac{1}{2}G}G_{i\bar{i}}^{2}}{\phi_{i\bar{i}}^{2}}-\frac{1}{2}\frac{G_{i\bar{i}}^{2}e^{\frac{1}{2}G}}{\phi_{i\bar{i}}^{2}}-\frac{1}{2}(\Delta\phi)^{2}e^{-\frac{1}{2}G}.\end{split}$$

In the last inequality above, we noticed

$$\Delta G=\sum_{i}G_{i\bar{i}}\leq\frac{1}{2}\frac{G_{i\bar{i}}^{2}e^{\frac{1}{2}G}}{\phi_{i\bar{i}}^{2}}+\frac{1}{2}\sum_{i}\phi_{i\bar{i}}^{2}e^{-\frac{1}{2}G}\leq\frac{1}{2}\frac{G_{i\bar{i}}^{2}e^{\frac{1}{2}G}}{\phi_{i\bar{i}}^{2}}+\frac{1}{2}(\Delta\phi)^{2}e^{-\frac{1}{2}G}.$$

Denote $u=e^{\frac{1}{2}G}|\nabla_{\phi}G|^{2}+\Delta\phi$, then we know

(6.12)

$$\Delta_{\phi}(u)\geq-(\frac{1}{2}\underline{R}+\frac{1}{2}\Delta\phi e^{-\frac{1}{2}G})u.$$

Denote $f=\frac{1}{2}\underline{R}+\frac{1}{2}\Delta\phi e^{-\frac{1}{2}G}$. Recall that we now already know $G$ is bounded. Our assumption implies $f\in L^{p}$ for some $p>3n$. Hence we may invoke Lemma 6.3 to get the desired result. ∎