(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. ∎

(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. ∎

The proof follows the same argument as the uniformly elliptic case. From the inequality we know that for any $\zeta\in C_{c}^{\infty}(B_{1})$, with $\zeta\geq 0$, the following holds:

(6.15)

$$\int_{B_{1}}a^{ij}\partial_{j}u\partial_{i}\zeta dx\leq\int_{B_{1}}fu\zeta+g\zeta dx.$$

Now let $\eta\in C^{\infty}_{c}(B_{1})$, define $\bar{u}=u+1$. Take $\zeta=\eta^{2}\bar{u}^{\beta}$, for some $\beta>0$. We plug in this $\zeta$ and obtain

(6.16)

$$\begin{split}\int_{B_{1}}&\beta a_{ij}\partial_{i}\bar{u}\partial_{j}\bar{u}\bar{u}^{\beta-1}\eta^{2}\leq\int-a_{ij}\partial_{j}u\partial_{i}\eta\bar{u}^{\beta}2\eta+|f|\bar{u}^{\beta+1}\eta^{2}+|g|\bar{u}^{\beta}\eta^{2}\\ &\leq\int_{B_{1}}\frac{\beta}{2}a_{ij}\partial_{i}\bar{u}\partial_{j}\bar{u}\bar{u}^{\beta-1}\eta^{2}+\frac{4}{\beta}a_{ij}\partial_{i}\eta\partial_{j}\eta\bar{u}^{\beta+1}+(|f|+|g|)\bar{u}^{\beta+1}\eta^{2}.\end{split}$$

Use the ellipticity condition to get:

(6.17)

$$\int_{B_{1}}\frac{\beta}{\lambda}|\nabla\bar{u}|^{2}\bar{u}^{\beta-1}\eta^{2}\leq\int_{B_{1}}\big(\frac{4\lambda}{\beta}|\nabla\eta|^{2}+|f|\eta^{2}+|g|\eta^{2}\big)\bar{u}^{\beta+1}.$$

This is equivalent to:

(6.18)

$$\int_{B_{1}}|\nabla(\bar{u}^{\frac{\beta+1}{2}})|^{2}\eta^{2}\frac{1}{\lambda}\leq\frac{(\beta+1)^{2}}{\beta^{2}}\int_{B_{1}}\lambda|\nabla\eta|^{2}\bar{u}^{\beta+1}+\frac{(\beta+1)^{2}}{4\beta}\int_{B_{1}}(|f|+|g|)\bar{u}^{\beta+1}\eta^{2}.$$

Next observe

$$|\nabla(\bar{u}^{\frac{\beta+1}{2}}\eta)|^{2}\leq 2|\nabla(\bar{u}^{\frac{\beta+1}{2}})|^{2}\eta^{2}+2\bar{u}^{\beta+1}|\nabla\eta|^{2}.$$

Hence it follows from (6.18) that if $\beta\geq 1$,

(6.19)

$$\begin{split}\int_{B_{1}}|\nabla(\bar{u}^{\frac{\beta+1}{2}}\eta)|^{2}\frac{1}{\lambda}&\leq\int_{B_{1}}\big(\frac{2\lambda(\beta+1)^{2}}{\beta^{2}}+\frac{2}{\lambda}\big)\bar{u}^{\beta+1}|\nabla\eta|^{2}+\frac{(\beta+1)^{2}}{2\beta}\int_{B_{1}}(|f|+|g|)\bar{u}^{\beta+1}\eta^{2}\\ &\leq\int_{B_{1}}10\lambda\bar{u}^{\beta+1}|\nabla\eta|^{2}+2\beta\int_{B_{1}}(|f|+|g|)\bar{u}^{\beta+1}\eta^{2}.\end{split}$$

We would like to get rid of the $\lambda$ in the above estimate. Let $\varepsilon>0$ to be determined, then we have

(6.20)

$$||\nabla(\bar{u}^{\frac{\beta+1}{2}}\eta)||_{L^{2-\varepsilon}}^{2}\leq||\lambda||_{L^{\frac{2}{\varepsilon}-1}}^{\frac{2-\varepsilon}{2}}\int_{B_{1}}\frac{1}{\lambda}|\nabla(\bar{u}^{\frac{\beta+1}{2}}\eta)|^{2}.$$

On the other hand, we estimate the right hand side by Hölder’s inequality:

(6.21)			$\displaystyle\int_{B_{1}}\lambda\bar{u}^{\beta+1}|\nabla\eta|^{2}\leq||\lambda||_{L^{\frac{2}{\varepsilon}-1}}||\bar{u}^{\frac{\beta+1}{2}}|\nabla\eta|||_{L^{\frac{2-\varepsilon}{1-\varepsilon}}}^{2},$
(6.22)			$\displaystyle\int_{B_{1}}(|f|+|g|)\bar{u}^{\beta+1}\eta^{2}\leq\big(||f||_{L^{p/2}}+||g||_{L^{p/2}}\big)||\bar{u}^{\frac{\beta+1}{2}}\eta||_{L^{\frac{2p}{p-2}}}^{2}.$

Therefore,

(6.23)

$$||\nabla(\bar{u}^{\frac{\beta+1}{2}}\eta)||_{L^{2-\varepsilon}}^{2}\leq||\lambda||_{L^{\frac{2}{\varepsilon}-1}}^{\frac{2-\varepsilon}{2}}\big(10||\lambda||_{L^{\frac{2}{\varepsilon}-1}}||\bar{u}^{\frac{\beta+1}{2}}|\nabla\eta|||^{2}_{L^{\frac{2-\varepsilon}{1-\varepsilon}}}+2\beta(||f||_{L^{p/2}}+||g||_{L^{p/2}})||\bar{u}^{\frac{\beta+1}{2}}\eta||_{L^{\frac{2p}{p-2}}}^{2}\big).$$

Now we choose $\varepsilon=\frac{2}{p+1}$, then $\frac{2}{\varepsilon}-1=p$. With this choice, we have $\frac{2-\varepsilon}{1-\varepsilon}<\frac{2p}{p-2}$ in the above, then we find for some constant $C_{6.1}$, depending on $||\lambda||_{L^{p}}$, $||f||_{L^{p/2}}$, $||g||_{L^{p/2}}$, such that

(6.24)

$$||\nabla(\bar{u}^{\frac{\beta+1}{2}}\eta)||_{L^{\frac{2p}{p+1}}}^{2}\leq C_{6.1}\big(||\bar{u}^{\frac{\beta+1}{2}}|\nabla\eta|||^{2}_{L^{\frac{2p}{p-2}}}+\beta||\bar{u}^{\frac{\beta+1}{2}}\eta||_{L^{\frac{2p}{p-2}}}^{2}\big).$$

Fix $\frac{1}{2}\leq r<R\leq 1$, Denote $r_{i}=r+2^{-i}(R-r)$, for $i\geq 0$. Note that $r_{0}=R$, and $r_{i}\rightarrow r$ as $i\rightarrow\infty$. We choose the cut-off function $\eta$ so that $0\leq\eta\leq 1$, $\eta\equiv 1$ on $B_{r_{i+1}}$, $supp\,\eta\subset B_{r_{i}}$, and $|\nabla\eta|\leq\frac{2}{r_{i}-r_{i+1}}=\frac{2^{i+2}}{R-r}$. Denote $\theta$ to be such that $\frac{1}{\theta}=\frac{p+1}{2p}-\frac{1}{d}$. Since $p>\frac{3d}{2}$, it follows that $\theta>\frac{2p}{p-2}$. Then apply the Sobolev inequality to get

(6.25)

$$||\bar{u}^{\frac{\beta+1}{2}}||_{L^{\theta}(B_{r_{i+1}})}\leq\frac{C_{6.2}\sqrt{\beta}2^{i+2}}{R-r}||\bar{u}^{\frac{\beta+1}{2}}||_{L^{\frac{2p}{p-2}}(B_{r_{i}})}.$$

This is equivalent to:

(6.26)

$$||\bar{u}||_{L^{\frac{\theta(\beta+1)}{2}}(B_{r_{i+1}})}\leq\frac{C_{6.2}^{\frac{2}{\beta+1}}\beta^{\frac{1}{\beta+1}}2^{\frac{2(i+2)}{\beta+1}}}{(R-r)^{\frac{2}{\beta+1}}}||\bar{u}||_{L^{\frac{2p}{p-2}\cdot\frac{\beta+1}{2}}(B_{r_{i}})}.$$

Now denote $\theta=\chi\cdot\frac{2p}{p-2}$ for some $\chi>1$, and choose $\beta$ to be $\frac{\beta+1}{2}=\chi^{i}$, then we obtain from (6.26):

(6.27)

$$||\bar{u}||_{L^{\frac{2p\chi^{i+1}}{p-2}}(B_{r_{i+1}})}\leq\big(\frac{2C_{6.2}}{R-r}\big)^{\frac{1}{\chi^{i}}}\chi^{\frac{i}{\chi^{i}}}2^{\frac{i+2}{\chi^{i}}}||\bar{u}||_{L^{\frac{2p\chi^{i}}{p-2}}(B_{r_{i}})},\textrm{ for $i\geq 0$.}$$

Iterating this inequality we obtain for any $\frac{1}{2}\leq r<R\leq 1$, and for some constant $C_{6.3}$ independent of $r$, $R$,

(6.28)

$$\begin{split}&||\bar{u}||_{L^{\infty}(B_{r})}\leq\frac{C_{6.3}}{(R-r)^{\sum_{i\geq 0}\chi^{-i}}}||\bar{u}||_{L^{\frac{2p}{p-2}}(B_{R})}=\frac{C_{6.3}}{(R-r)^{\frac{\chi}{\chi-1}}}||\bar{u}||_{L^{\frac{2p}{p-2}}(B_{R})}\\ &\leq\frac{C_{6.3}}{(R-r)^{\frac{\chi}{\chi-1}}}||\bar{u}||_{L^{1}(B_{R})}^{\frac{p-2}{2p}}||\bar{u}||_{L^{\infty}(B_{R})}^{\frac{p+2}{2p}}\leq\frac{1}{2}||\bar{u}||_{L^{\infty}(B_{R})}+\frac{2^{\frac{p+2}{p-2}}C_{6.3}^{\frac{2p}{p-2}}}{(R-r)^{\frac{2\chi p}{(\chi-1)(p-2)}}}||\bar{u}||_{L^{1}(B_{R})}.\end{split}$$

The desired conclusion now follows from the following lemma applied to $f(r)=||\bar{u}||_{L^{\infty}(B_{r})}$, which is a special case of Lemma 4.3 in [22]. ∎

Let $0<\delta<1$ and $K>1$ to be determined. We will compute $\Delta_{\phi}(e^{\delta G}(|\nabla\phi|^{2}+K))$. As before, for any point $p\in B_{1}$ we are considering, we can always do a unitary coordinate transform which makes $\phi_{i\bar{j}}(p)=\phi_{i\bar{i}}(p)\delta_{ij}$. Under this coordinate, we can compute:

(6.30)

$$\begin{split}&\Delta_{\phi}(e^{\delta G}(|\nabla\phi|^{2}+K))=e^{\delta G}(\delta^{2}|\nabla_{\phi}G|^{2}-\delta\underline{R})(|\nabla\phi|^{2}+K)+e^{\delta G}\Delta_{\phi}(|\nabla\phi|^{2})\\ &+e^{\delta G}\delta\frac{G_{i}(|\nabla\phi|^{2})_{\bar{i}}+G_{\bar{i}}(|\nabla\phi|^{2})_{i}}{\phi_{i\bar{i}}}.\end{split}$$

Similar to the calculation in Theorem 2.1, we can find:

(6.31)			$\displaystyle\Delta_{\phi}(|\nabla\phi|^{2})=\frac{|\phi_{ij}|^{2}}{\phi_{i\bar{i}}}+\Delta\phi+G_{i}\phi_{\bar{i}}+G_{\bar{i}}\phi_{i}$
(6.32)			$\displaystyle(|\nabla\phi|^{2})_{i}=\sum_{j}\phi_{ij}\phi_{\bar{j}}+\phi_{i\bar{i}}\phi_{\bar{i}}.$

Hence we obtain

(6.33)

$$\begin{split}&\Delta_{\phi}(e^{\delta G}(|\nabla\phi|^{2}+K))=\frac{e^{\delta G}}{\phi_{i\bar{i}}}|\delta\phi_{j}G_{i}+\phi_{ij}|^{2}+Ke^{\delta G}(\delta^{2}|\nabla_{\phi}G|^{2}-\delta\underline{R})-e^{\delta G}\delta\underline{R}|\nabla\phi|^{2}\\ &+e^{\delta G}\Delta\phi+e^{\delta G}(1+\delta)(G_{i}\phi_{\bar{i}}+G_{\bar{i}}\phi)\geq e^{\delta G}K\delta^{2}|\nabla_{\phi}G|^{2}+e^{\delta G}\Delta\phi-\delta\underline{R}e^{\delta G}|\nabla\phi|^{2}\\ &-\delta\underline{R}Ke^{\delta G}-\frac{1}{2}K\delta^{2}e^{\delta G}|\nabla_{\phi}G|^{2}-\frac{1}{2}\frac{(1+\delta)^{2}e^{\delta G}|\nabla\phi|^{2}\Delta\phi}{K\delta^{2}}.\end{split}$$

Now we choose $\delta=\frac{1}{8}$, and we choose $K$ sufficiently large so that $\frac{(1+\delta)^{2}|\nabla\phi|^{2}}{K\delta^{2}}<1$. Hence we obtain from (6.33):

(6.34)

$$\Delta_{\phi}(e^{\delta G}(|\nabla\phi|^{2}+K))\geq\frac{1}{2}e^{\delta G}\Delta\phi-e^{\delta G}C_{6.5}.$$

Here $C_{9}$ depends only on $\underline{R}$ and $||\nabla\phi||_{0}$. Define $\eta(z)=(1-|z|^{2})^{-1}$ for $z\in B_{1}$. We show that $|\Delta_{\phi}\eta|\leq C_{6.6}\eta^{3}\sum_{i}\frac{1}{\phi_{i\bar{i}}}$.Indeed,

$$\begin{split}\Delta_{\phi}\eta&=\phi^{i\bar{j}}\partial_{i\bar{j}}(\eta)=\phi^{i\bar{j}}\bigg((1-|z|^{2})^{-2}\delta_{ij}+2(1-|z|^{2})^{-3}\bar{z}_{i}z_{j}\bigg)\\ &=\phi^{i\bar{j}}\eta^{3}\big((1-|z|^{2})\delta_{ij}+2\bar{z}_{i}z_{j}\big).\end{split}$$

From this the claim follows easily. Denote $v=e^{\delta G}(|\nabla\phi|^{2}+K)$. Suppose the function $v-\eta$ achieves maximum at $p\in B_{1}$. There are two possibilities:

Suppose $v(p)-\eta(p)\leq 0$, then we immediately conclude that

$$v(z)\leq\eta(z)\leq\frac{4}{3},\textrm{ for any $z\in B_{\frac{1}{2}}$.}$$

Then we are done.

Suppose otherwise $v(p)-\eta(p)\geq 0$, then we know at $p$:

(6.35)

$$\begin{split}0&\geq\Delta_{\phi}(v-\eta)(p)\geq\frac{1}{2}e^{\delta G}\Delta\phi-e^{\delta G}C_{6.5}-C_{6.6}\eta^{3}\sum_{i}\frac{1}{\phi_{i\bar{i}}}\\ &\geq\frac{1}{2}e^{\delta G}\Delta\phi-C_{6.6}v^{3}e^{-G}\Delta\phi-e^{\delta G}C_{6.5}\geq\frac{1}{2}e^{\delta G}\Delta\phi-C_{6.7}e^{-(1-3\delta)G}\Delta\phi-e^{\delta G}C_{6.5}.\end{split}$$

In the third inequality above, we used that $\sum_{i}\frac{1}{\phi_{i\bar{i}}}=e^{-G}\Delta\phi$, which is true only in dimension 2. Also we used that at $p$, $\eta\leq v$.