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 $p$, $\eta\leq v$.

Suppose at $p$, we have $\frac{1}{4}e^{\delta G}\leq C_{6.7}e^{-(1-3\delta)G}$, this immediately gives a bound for $e^{G}$, hence $v$ at $p$. Then we are done.

Suppose otherwise, then we have at $p$

(6.36)

$$0\geq e^{\delta G}\frac{1}{4}\Delta\phi-e^{\delta G}C_{6.5}\geq e^{\delta G}(\frac{1}{4}e^{\frac{G}{2}}-C_{6.5}).$$

Then we also get an estimate for $e^{G}$ at $p$. So we are done as well. ∎