Proof.

The argument uses maximum principle again. This time we will calculate $\Delta_{\varphi}(e^{F-\lambda\varphi}(K+|\nabla\varphi|^{2}))$ where $\lambda,K>0$ are constants to be determined later. We choose a normal coordinate (equation (2.1)) and do the following calculations. We have

(2.4)		$\displaystyle\qquad\Delta_{\varphi}(e^{F-\lambda\varphi}(K+\|\nabla\varphi\|^{2}))$	$\displaystyle=$	$\displaystyle\Delta_{\varphi}(e^{F-\lambda\varphi})(K+\|\nabla\varphi\|^{2})+e^{F-\lambda\varphi}\Delta_{\varphi}(K+\|\nabla\varphi\|^{2})$
(2.4)				$\displaystyle+e^{F-\lambda\varphi}\cdot\frac{(F_{i}-\lambda\varphi_{i})(\|\nabla\varphi\|^{2})_{\bar{i}}+(F_{\bar{i}}-\lambda\varphi_{\bar{i}})(\|\nabla\varphi\|^{2})_{i}}{1+\varphi_{i\bar{i}}}.$

We can first calculate:

(2.5)

\begin{split}\Delta_{\varphi}(e^{F-\lambda\varphi})&=e^{F-C\varphi}\frac{|F_{i}-\lambda\varphi_{i}|^{2}}{1+\varphi_{i\bar{i}}}+e^{F-\lambda\varphi}(\Delta_{\varphi}F-\frac{\lambda\varphi_{i\bar{i}}}{1+\varphi_{i\bar{i}}})\\ &=e^{F-\lambda\varphi}\frac{|F_{i}-\lambda\varphi_{i}|^{2}}{1+\varphi_{i\bar{i}}}+e^{F-\lambda\varphi}(-\underline{R}-\lambda n+\frac{\lambda+R_{i\bar{i}}}{1+\varphi_{i\bar{i}}}).\end{split}

By differentiating equation (1.1) in $z_{\alpha}$ direction, we obtain

(2.6)

\displaystyle\sum_{i}\frac{\varphi_{i\bar{i}\alpha}}{1+\varphi_{i\bar{i}}}=F_{\alpha}\;\;{\rm and}\;\;\displaystyle\sum_{i}\frac{\varphi_{i\bar{i}\bar{\alpha}}}{1+\varphi_{i\bar{i}}}=F_{\bar{\alpha}}.

Then we calculate

(2.7)

\begin{split}\Delta_{\varphi}(|\nabla\varphi|^{2})&=\frac{R_{\alpha\bar{\beta}i\bar{i}}\varphi_{\alpha}\varphi_{\bar{\beta}}}{1+\varphi_{i\bar{i}}}+\frac{|\varphi_{i\alpha}|^{2}}{1+\varphi_{i\bar{i}}}+\frac{\varphi^{2}_{i\bar{i}}}{1+\varphi_{i\bar{i}}}+\frac{\varphi_{\alpha}\varphi_{\bar{\alpha}i\bar{i}}+\varphi_{\bar{\alpha}}\varphi_{\alpha i\bar{i}}}{1+\varphi_{i\bar{i}}}\\ &\geq-\frac{C_{2.21}|\nabla\varphi|^{2}}{1+\varphi_{i\bar{i}}}+\frac{|\varphi_{i\alpha}|^{2}}{1+\varphi_{i\bar{i}}}+\frac{\varphi^{2}_{i\bar{i}}}{1+\varphi_{i\bar{i}}}+\varphi_{\alpha}F_{\bar{\alpha}}+\varphi_{\bar{\alpha}}F_{\alpha}\\ &=-\frac{C_{2.21}|\nabla\varphi|^{2}}{1+\varphi_{i\bar{i}}}+\frac{|\varphi_{i\alpha}|^{2}}{1+\varphi_{i\bar{i}}}+\frac{\varphi^{2}_{i\bar{i}}}{1+\varphi_{i\bar{i}}}+\varphi_{\alpha}(F_{\bar{\alpha}}-\lambda\varphi_{\bar{\alpha}})+\varphi_{\bar{\alpha}}(F_{\alpha}-\lambda\varphi_{\alpha})+2\lambda|\varphi_{\alpha}|^{2}\\ &\geq-\frac{C_{2.21}|\nabla\varphi|^{2}}{1+\varphi_{i\bar{i}}}+\frac{|\varphi_{i\alpha}|^{2}}{1+\varphi_{i\bar{i}}}+\frac{\varphi^{2}_{i\bar{i}}}{1+\varphi_{i\bar{i}}}-\varepsilon(n+\Delta\varphi)-\frac{|F_{\alpha}-\lambda\varphi_{\alpha}|^{2}|\varphi_{\alpha}|^{2}}{\varepsilon(1+\varphi_{\alpha\bar{\alpha}})}+2\lambda|\varphi_{\alpha}|^{2}.\end{split}

Here $C_{2.21}$ depends only on lower bound of bisectional curvature of $g$ .

For the last term in (2.4), we estimate in the following way:

(2.8)		$\displaystyle\frac{\|(F_{\bar{i}}-\lambda\varphi_{\bar{i}})(\|\nabla\varphi\|^{2})_{i}\|}{1+\varphi_{i\bar{i}}}$
	$\displaystyle=$	$\displaystyle\frac{\|(F_{\bar{i}}-\lambda\varphi_{\bar{i}})(\varphi_{\alpha}\varphi_{\bar{\alpha}i}+\varphi_{\bar{\alpha}}\varphi_{\alpha i})\|}{1+\varphi_{i\bar{i}}}$
	$\displaystyle\leq$	$\displaystyle\frac{\|(F_{\bar{i}}-\lambda\varphi_{\bar{i}})\varphi_{i}\varphi_{i\bar{i}}\|}{1+\varphi_{i\bar{i}}}+\frac{\|(F_{\bar{i}}-\lambda\varphi_{\bar{i}})\varphi_{\bar{\alpha}}\varphi_{\alpha i}\|}{1+\varphi_{i\bar{i}}}$
	$\displaystyle\leq$	$\displaystyle\frac{\|F_{i}-\lambda\varphi_{i}\|^{2}\|\varphi_{i}\|^{2}}{2\varepsilon(1+\varphi_{i\bar{i}})}+\frac{\|F_{i}-\lambda\varphi_{i}\|^{2}\|\varphi_{\alpha}\|^{2}}{2\varepsilon(1+\varphi_{i\bar{i}})}+\frac{\varepsilon\varphi_{i\bar{i}}^{2}}{2(1+\varphi_{i\bar{i}})}+\frac{\varepsilon\|\varphi_{i\alpha}\|^{2}}{2(1+\varphi_{i\bar{i}})}.$

The other conjugate term satisfies the same estimate as above. Combining above calculations, we obtain:

(2.9)		$\displaystyle\frac{\Delta_{\varphi}(e^{F-\lambda\varphi}(K+\|\nabla\varphi\|^{2}))}{e^{F-\lambda\varphi}}$
	$\displaystyle\geq$	$\displaystyle\big(K+\|\nabla\varphi\|^{2}-3\varepsilon^{-1}\|\nabla\varphi\|^{2}\big)\frac{\|F_{i}-\lambda\varphi_{i}\|^{2}}{1+\varphi_{i\bar{i}}}$
		$\displaystyle\qquad\qquad+\frac{(\lambda+R_{i\bar{i}})(K+\|\nabla\varphi\|^{2})-C_{2.21}\|\nabla\varphi\|^{2}}{1+\varphi_{i\bar{i}}}+\frac{(1-\varepsilon)\varphi_{i\bar{i}}^{2}}{1+\varphi_{i\bar{i}}}-\varepsilon(n+\Delta\varphi)$
		$\displaystyle\qquad\qquad\qquad\qquad+\frac{\|\varphi_{i\alpha}\|^{2}(1-\varepsilon)}{1+\varphi_{i\bar{i}}}+(-\underline{R}-\lambda n)(K+\|\nabla\varphi\|^{2}).$

Now it’s time to choose the constants $\varepsilon$ , $\lambda$ , and $K$ appearing above.

First we choose $\varepsilon=\frac{1}{4}$ . With this choice, we have

(2.10)

\begin{split}\sum_{i}\frac{(1-\varepsilon)\varphi_{i\bar{i}}^{2}}{1+\varphi_{i\bar{i}}}-\varepsilon(n+\Delta\varphi)&=\frac{3}{4}(n+\Delta\varphi)-\frac{3n}{2}+\frac{3}{4}\sum_{i}\frac{1}{1+\varphi_{i\bar{i}}}-\frac{1}{4}(1+\varphi_{i\bar{i}})\\ &\geq\frac{1}{2}(n+\Delta\varphi)-\frac{3n}{2}.\end{split}

Then we choose $\lambda$ so large that $\lambda+R_{i\bar{i}}>1$ . Finally, we choose $K$ so large that

(2.11)			$\displaystyle K>C_{2.21}\max_{M}\|\nabla\varphi\|^{2}$
(2.12)			$\displaystyle K>3\varepsilon^{-1}\max_{M}\|\nabla\varphi\|^{2}=12\max_{M}\|\nabla\varphi\|^{2}.$

With above choices for $\lambda$ and $K$ , we have

(2.13)

(\lambda+R_{i\bar{i}})(K+|\nabla\varphi|^{2})-C_{2.21}|\nabla\varphi|^{2}\geq K-C_{2.21}|\nabla\varphi|^{2}>0.

and also

(2.14)

K-3\varepsilon^{-1}|\nabla\varphi|^{2}>0.

Hence we conclude from (2.9) that

(2.15)

\Delta_{\varphi}(e^{F-\lambda\varphi}(K+|\nabla\varphi|^{2}))\geq e^{F-\lambda\varphi}\big(-(|\underline{R}|+\lambda n)(K+\max_{M}|\nabla\varphi|^{2})-C_{2.22}+\frac{1}{4}(n+\Delta\varphi)\big).

Denote $v=e^{F-C\varphi}(K+|\nabla\varphi|^{2})$ , it is enough to show $v$ has an upper bound. we see from (2.15) that there exists constants $C_{2.23}>0$ , $C_{2.24}>0$ , possibly depending on $\displaystyle\max_{M}|\nabla\varphi|^{2}$ , such that

(2.16)

\Delta_{\varphi}(v)\geq v(-C_{2.23}+\frac{1}{C_{2.24}}(n+\Delta\varphi)).

Here we notice that $n+\Delta\varphi\geq ne^{\frac{F}{n}}$ . Hence we obtain from (2.16) that

(2.17)

\Delta_{\varphi}(v)\geq v(-C_{2.23}+\frac{1}{C_{2.24}}e^{\frac{F}{n}}).

Let the maximum of $v$ be achieved at point $p$ , then we know $-C_{2.23}+\frac{e^{\frac{F}{n}(p)}}{C_{2.24}}\leq 0$ . This gives an upper bound of $F$ at $p_{0}$ , hence an upper bound for $v$ , where this bound depends on $\displaystyle\max_{M}|\nabla\varphi|$ . ∎

(2.9)		$\displaystyle\frac{\Delta_{\varphi}(e^{F-\lambda\varphi}(K+\|\nabla\varphi\|^{2}))}{e^{F-\lambda\varphi}}$
	$\displaystyle\geq$	$\displaystyle\big(K+\|\nabla\varphi\|^{2}-3\varepsilon^{-1}\|\nabla\varphi\|^{2}\big)\frac{\|F_{i}-\lambda\varphi_{i}\|^{2}}{1+\varphi_{i\bar{i}}}$
		$\displaystyle\qquad\qquad+\frac{(\lambda+R_{i\bar{i}})(K+\|\nabla\varphi\|^{2})-C_{2.21}\|\nabla\varphi\|^{2}}{1+\varphi_{i\bar{i}}}+\frac{(1-\varepsilon)\varphi_{i\bar{i}}^{2}}{1+\varphi_{i\bar{i}}}-\varepsilon(n+\Delta\varphi)$
		$\displaystyle\qquad\qquad\qquad\qquad+\frac{\|\varphi_{i\alpha}\|^{2}(1-\varepsilon)}{1+\varphi_{i\bar{i}}}+(-\underline{R}-\lambda n)(K+\|\nabla\varphi\|^{2}).$

Proof. [020V]